There are 26 choices for each letter and 10 choices for each digit.

(a)

With repetition allowed, \[ 26^2\,10^5=67{,}600{,}000. \]

(b)

Without repetition, the two letter positions can be filled in \(26\cdot 25\) ways and the five digit positions in \(10\cdot 9\cdot 8\cdot 7\cdot 6\) ways. Thus \[ 26\cdot 25\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6 =196{,}560{,}000. \]

For \(n\) letters with repeated-letter multiplicities \(n_1,\ldots ,n_r\), the number of distinct arrangements is \[ \frac {n!}{n_1!\cdots n_r!}. \]

(a)

FLUKE has five distinct letters: \[ 5!=120. \]

(b)

PROPOSE has seven letters, with \(P\) repeated twice and \(O\) repeated twice: \[ \frac {7!}{2!\,2!}=1260. \]

(c)

MISSISSIPPI has 11 letters, with \(I\) repeated four times, \(S\) four times, \(P\) twice, and \(M\) once: \[ \frac {11!}{4!\,4!\,2!}=34{,}650. \]

(d)

ARRANGE has seven letters, with \(A\) repeated twice and \(R\) repeated twice: \[ \frac {7!}{2!\,2!}=1260. \]

The binomial theorem states \[ (a+b)^n=\sum _{k=0}^{n}\binom {n}{k}a^{n-k}b^k. \]

(a)

Set \(a=1\) and \(b=2\): \[ \sum _{k=0}^{n}\binom {n}{k}2^k = (1+2)^n = 3^n. \]

(b)

Set \(a=1\) and \(b=x\): \[ \sum _{k=0}^{n}\binom {n}{k}x^k = (1+x)^n. \]

Choose 3 of the 12 people for the first committee, then 4 of the remaining 9 for the second; the final 5 form the third: \[ \binom {12}{3}\binom {9}{4}\binom {5}{5} = \frac {12!}{3!\,4!\,5!} = 27{,}720. \]

(1) By the binomial theorem, \[ (3x^2+y)^5 = \sum _{k=0}^{5}\binom {5}{k}(3x^2)^{5-k}y^k. \] Therefore, \[ { (3x^2+y)^5 = 243x^{10} +405x^8y +270x^6y^2 +90x^4y^3 +15x^2y^4 +y^5. } \]

(2) The \((k+1)^{th}\) term is given by \(\binom {n}{k} a^{k}b^{n-k}\). Thus, the third term is given by \(\binom {5}{2} (y)^2(3x^2)^{5-2}.\)

Count the number of ways to form a group of \(r\) people from \(n\) men and \(m\) women.

Directly, there are \[ \binom {n+m}{r} \] such groups.

Alternatively, classify the group according to the number \(k\) of men selected. If exactly \(k\) men are selected, then \(r-k\) women are selected, giving \[ \binom {n}{k}\binom {m}{r-k} \] possibilities. Summing over all feasible values of \(k\) gives \[ \binom {n+m}{r} = \sum _{k=0}^{r}\binom {n}{k}\binom {m}{r-k}. \] Terms with impossible choices are interpreted as zero.

(a)

Count pairs consisting of a nonempty committee and its chairperson.

If the committee has size \(k\), choose its members in \(\binom nk\) ways and its chairperson in \(k\) ways. This gives \[ \sum _{k=1}^{n}k\binom nk. \]

Alternatively, first choose the chairperson in \(n\) ways. Each of the other \(n-1\) people may independently be included or excluded, giving \(2^{n-1}\) choices. Hence \[ \sum _{k=1}^{n}k\binom nk=n2^{n-1}. \] Count selections of a committee, a chairperson, and a secretary, where the two officers may coincide. A committee of size \(k\) contributes \(k^2\binom nk\), so the total is the left-hand side.

Split into two cases.

If the chairperson and secretary are the same, choose that person in \(n\) ways and choose any subset of the remaining \(n-1\) people: \[ n2^{n-1}. \]

If they are different, choose the chairperson and secretary as an ordered pair in \(n(n-1)\) ways, and choose any subset of the remaining \(n-2\) people: \[ n(n-1)2^{n-2}. \]

Therefore, \[ \sum _{k=1}^{n}\binom nk k^2 = n2^{n-1}+n(n-1)2^{n-2} = 2^{n-2}n(n+1). \]

(b)

Count selections of a committee together with three ordered offices, where a person may hold more than one office. A committee of size \(k\) contributes \(k^3\binom nk\).

Classify according to the number of distinct officeholders.

One distinct officeholder: \[ n2^{n-1}. \]

Exactly two distinct officeholders: choose which one of the three offices is held by the person who holds only one office (\(3\) choices), then choose the two distinct people in order (\(n(n-1)\) choices), and choose any subset of the remaining \(n-2\) people: \[ 3n(n-1)2^{n-2}. \]

Three distinct officeholders: \[ n(n-1)(n-2)2^{n-3}. \]

Adding, \begin {align*} \sum _{k=1}^{n}\binom nk k^3 &= n2^{n-1} +3n(n-1)2^{n-2} +n(n-1)(n-2)2^{n-3}\\ &= 2^{n-3}n^2(n+3). \end {align*}

Apply the binomial theorem to \((1-1)^n\): \[ (1-1)^n = \sum _{i=0}^{n}\binom ni1^{n-i}(-1)^i = \sum _{i=0}^{n}(-1)^i\binom ni. \] For \(n>0\), the left-hand side is \(0^n=0\). Therefore, \[ \sum _{i=0}^{n}(-1)^i\binom ni=0. \]

Let \(C\) be the event of smoking cigarettes and \(G\) the event of smoking cigars. Then \[ P(C)=0.28,\qquad P(G)=0.07,\qquad P(C\cap G)=0.05. \]

(a)

\[ P(C\cup G)=0.28+0.07-0.05=0.30. \] Thus \[ P((C\cup G)^c)=1-0.30=0.70. \] The answer is \( {70\%}\).

(b)

\[ P(G\cap C^c)=P(G)-P(G\cap C)=0.07-0.05=0.02. \] The answer is \( {2\%}\).

There are \(\binom {52}{2}\) unordered pairs of cards.

(a)

There are \(\binom 42\) pairs of aces: \[ P(\text {both aces}) = \frac {\binom 42}{\binom {52}{2}} = \frac {6}{1326} = {\frac 1{221}}. \]

(b)

For each of the 13 ranks, choose 2 of the 4 suits: \[ P(\text {same rank}) = \frac {13\binom 42}{\binom {52}{2}} = \frac {78}{1326} = {\frac 1{17}}. \]

(a)

Only \(E\): \(E\cap F^c\cap G^c.\)

(b)

\(E\) and \(G\), but not \(F\): \( E\cap F^c\cap G.\)

(c)

At least one: \( E\cup F\cup G. \)

(d)

At least two: \( (E\cap F)\cup (E\cap G)\cup (F\cap G). \)

(e)

All three: \( E\cap F\cap G.\)

(f)

None: \( E^c\cap F^c\cap G^c=(E\cup F\cup G)^c.\)

(g)

At most one: \[ (E^c\cap F^c\cap G^c) \cup (E\cap F^c\cap G^c) \cup (E^c\cap F\cap G^c) \cup (E^c\cap F^c\cap G). \] Equivalently, \[ \bigl [(E\cap F)\cup (E\cap G)\cup (F\cap G)\bigr ]^c. \]

(h)

At most two: \[ (E\cap F\cap G)^c. \]

(i)

Exactly two: \[ (E\cap F\cap G^c) \cup (E\cap F^c\cap G) \cup (E^c\cap F\cap G). \]

(j)

At most three: \[ \Omega , \] because there are only three events.

The addition rule gives \[ P(E\cup F)=P(E)+P(F)-P(E\cap F). \] Because \(P(E\cup F)\leq 1\), \[ P(E)+P(F)-P(E\cap F)\leq 1, \] and therefore \[ P(E\cap F)\geq P(E)+P(F)-1. \] For \(P(E)=0.9\) and \(P(F)=0.8\), \[ P(E\cap F)\geq 0.9+0.8-1=0.7. \]

The event \(E\) is the disjoint union \[ E=(E\cap F)\cup (E\cap F^c). \] Hence, by additivity, \[ P(E)=P(E\cap F)+P(E\cap F^c). \] Rearranging gives \[ P(E\cap F^c)=P(E)-P(E\cap F). \]

Let \(D\) be the event that the person has the disease and \(+\) the event that the test is positive. Then \[ P(D)=0.005,\quad P(D^c)=0.995,\quad P(+\mid D)=0.95,\quad P(+\mid D^c)=0.01. \] By Bayes’ rule, \begin {align*} P(D\mid +) &= \frac {P(+\mid D)P(D)} {P(+\mid D)P(D)+P(+\mid D^c)P(D^c)}\\ &= \frac {(0.95)(0.005)} {(0.95)(0.005)+(0.01)(0.995)}\\ &= \frac {0.00475}{0.01470} = \frac {95}{294} \approx {0.3231}. \end {align*}

Thus, given a positive result, the probability of having the disease is about \(32.3\%\).

Because \(A\subseteq B\), we have \(A\cap B=A\), \(A\cap B^c=\varnothing \), and \(B\cap A=A\).

Provided \(P(B)>0\), \[ P(A\mid B) = \frac {P(A\cap B)}{P(B)} = {\frac {P(A)}{P(B)}}. \]

Provided \(P(B^c)>0\), \[ P(A\mid B^c) = \frac {P(A\cap B^c)}{P(B^c)} = {0}. \]

Provided \(P(A)>0\), \[ P(B\mid A) = \frac {P(B\cap A)}{P(A)} = {1}. \]

Provided \(P(A^c)>0\), \[ P(B\mid A^c) = \frac {P(B\cap A^c)}{P(A^c)} = {\frac {P(B)-P(A)}{1-P(A)}}. \]

The 36 ordered outcomes are equally likely. The nonzero probabilities are \[ \begin {array}{c|cccccccccccccccccc} i &1&2&3&4&5&6&8&9&10&12&15&16&18&20&24&25&30&36\\ \hline 36P(X=i) &1&2&2&3&2&4&2&1&2&4&2&1&2&2&2&1&2&1 \end {array} \] Thus \[ P(X=i)= \begin {cases} \dfrac {1}{36}, &i\in \{1,9,16,25,36\},\\[1mm] \dfrac {1}{18}, &i\in \{2,3,5,8,10,15,18,20,24,30\},\\[1mm] \dfrac {1}{12}, &i=4,\\[1mm] \dfrac {1}{9}, &i\in \{6,12\},\\[1mm] 0, &\text {otherwise}. \end {cases} \]

For any point \(a\), \[ P(X=a)=F(a)-F(a^-), \] the size of the jump of the CDF at \(a\).

(a)

At \(1\), \[ P(X=1) = F(1)-F(1^-) = \frac 12-\frac 14 = {\frac 14}. \]

At \(2\), \[ P(X=2) = F(2)-F(2^-) = \frac {11}{12}-\frac 34 = {\frac 16}. \]

At \(3\), \[ P(X=3) = F(3)-F(3^-) = 1-\frac {11}{12} = {\frac 1{12}}. \]

(b)

Since there is no point mass at \(1/2\) or \(3/2\), \begin {align*} P\left (\frac 12<X<\frac 32\right ) &= F\left (\frac 32\right )-F\left (\frac 12\right )\\ &= \left (\frac 12+\frac {(3/2)-1}{4}\right ) -\frac {1/2}{4}\\ &= \frac 58-\frac 18 = {\frac 12}. \end {align*}

The probability at each support point is the jump of the CDF: \[ p_X(x)=P(X=x)=F(x)-F(x^-). \] Therefore, \[ P(X=0)=\frac 12, \] \[ P(X=1)=\frac 35-\frac 12=\frac 1{10}, \] \[ P(X=2)=\frac 45-\frac 35=\frac 15, \] \[ P(X=3)=\frac 9{10}-\frac 45=\frac 1{10}, \] and \[ P(X=3.5)=1-\frac 9{10}=\frac 1{10}. \] Hence \[ { p_X(x)= \begin {cases} \dfrac 12, & x=0,\\[1mm] \dfrac 1{10}, & x=1,\\[1mm] \dfrac 15, & x=2,\\[1mm] \dfrac 1{10}, & x=3,\\[1mm] \dfrac 1{10}, & x=3.5,\\[1mm] 0, & \text {otherwise}. \end {cases}} \] The probabilities sum to \[ \frac 12+\frac 1{10}+\frac 15+\frac 1{10}+\frac 1{10}=1. \]

Let \(X\) denote the number of typographical errors on the next page. Under the Poisson assumption,

\[ X\sim \operatorname {Pois}(0.2). \]

Thus,

\[ P(X=k)=e^{-0.2}\frac {(0.2)^k}{k!}. \]

(a)

\[ P(X=0) = e^{-0.2} \approx {0.8187}. \]

(b)

Using the complement,

\[ \begin {aligned} P(X\geq 2) &= 1-P(X=0)-P(X=1)\\ &= 1-e^{-0.2}-(0.2)e^{-0.2}\\ &= 1-1.2e^{-0.2}\\ &\approx {0.0175}. \end {aligned} \]

The expected value alone does not uniquely determine these probabilities. The calculation requires the additional assumption that the number of errors follows a Poisson distribution.

Let \(X\) denote the number of accidents during the next month. Under the Poisson assumption,

\[ X\sim \operatorname {Pois}(3.5). \]

Therefore,

\[ P(X=k)=e^{-3.5}\frac {3.5^k}{k!}. \]

(a)

\[ \begin {aligned} P(X\geq 2) &= 1-P(X=0)-P(X=1)\\ &= 1-e^{-3.5}-3.5e^{-3.5}\\ &= 1-4.5e^{-3.5}\\ &\approx {0.8641}. \end {aligned} \]

(b)

\[ \begin {aligned} P(X\leq 1) &= P(X=0)+P(X=1)\\ &= e^{-3.5}+3.5e^{-3.5}\\ &= 4.5e^{-3.5}\\ &\approx {0.1359}. \end {aligned} \]

Notice that the answers in parts (a) and (b) are complements.

Assume that:

• birthdays are uniformly distributed among the \(365\) days of the year;
• leap days are ignored;
• the birthdays of the two partners are independent;
• different couples may be treated independently.

Let \(n=80{,}000\).

(a)

For a particular couple, the probability that both partners were born on April 30 is

\[ p=\left (\frac {1}{365}\right )^2. \]

Thus, the exact probability that at least one couple has this property is

\[ \begin {aligned} P(\text {at least one}) &= 1-(1-p)^{80{,}000}\\ &= 1-\left (1-\frac {1}{365^2}\right )^{80{,}000}. \end {aligned} \]

Because \(n\) is large and \(p\) is small, a Poisson approximation with

\[ \lambda =np = \frac {80{,}000}{365^2} \approx 0.60049 \]

gives

\[ P(\text {at least one}) \approx 1-e^{-0.60049} \approx {0.4515}. \]

(b)

For a particular couple, condition on the birthday of the first partner. The probability that the second partner has the same birthday is

\[ p=\frac {1}{365}. \]

Therefore,

\[ P(\text {at least one matching couple}) = 1-\left (1-\frac {1}{365}\right )^{80{,}000}. \]

Using a Poisson approximation,

\[ \lambda = \frac {80{,}000}{365} \approx 219.18, \]

so

\[ P(\text {at least one matching couple}) \approx 1-e^{-219.18}\approx 1. \]

If \(X\sim \operatorname {Bin}(n,p), \) then the exact probability is

\[ P(X=k)=\binom nkp^k(1-p)^{n-k}. \]

The usual Poisson approximation uses

\[ Y\sim \operatorname {Pois}(\lambda ), \qquad \lambda =np, \]

and

\[ P(X=k)\approx P(Y=k) = e^{-\lambda }\frac {\lambda ^k}{k!}. \]

(a)

Here \(n=8\), \(p=0.1\), \(k=2\), and \(\lambda =0.8\). The exact probability is

\[ \begin {aligned} P(X=2) &= \binom 82(0.1)^2(0.9)^6\\ &\approx {0.1488}. \end {aligned} \] The Poisson approximation is reasonably accurate: \[ \begin {aligned} P(Y=2) &= e^{-0.8}\frac {0.8^2}{2!}\\ &\approx {0.1438}. \end {aligned} \]

(b)

Here \(n=10\), \(p=0.95\), \(k=9\), and \(\lambda =9.5\). The exact probability is

\[ \begin {aligned} P(X=9) &= \binom {10}{9}(0.95)^9(0.05)\\ &\approx {0.3151}. \end {aligned} \]

The direct Poisson approximation is poor because \(p=0.95\) is not small. \[ \begin {aligned} P(Y=9) &= e^{-9.5}\frac {9.5^9}{9!}\\ &\approx {0.1300}. \end {aligned} \]

A better approach is to let \(Z=10-X,\) the number of failures. Then \(Z\sim \operatorname {Bin}(10,0.05), \) and \(X=9\) is equivalent to \(Z=1\). Approximating \(Z\) by \(\operatorname {Pois}(0.5)\) gives the following, which is much closer to the exact value: \[ P(X=9) = P(Z=1) \approx e^{-0.5}(0.5) \approx 0.3033, \]

(c)

Here \(n=10\), \(p=0.1\), \(k=0\), and \(\lambda =1\). The exact probability is \[ P(X=0) = (0.9)^{10} \approx {0.3487}. \]

The Poisson approximation is reasonably accurate: \(P(Y=0) = e^{-1} \approx {0.3679}.\)

(d)

Here \(n=9\), \(p=0.2\), \(k=4\), and \(\lambda =1.8\). The exact probability is

\[ \begin {aligned} P(X=4) &= \binom 94(0.2)^4(0.8)^5\\ &\approx {0.0661}. \end {aligned} \]

The Poisson approximation is

\[ \begin {aligned} P(Y=4) &= e^{-1.8}\frac {1.8^4}{4!}\\ &\approx {0.0723}. \end {aligned} \]

Assume that monthly event counts are independent and identically distributed Poisson random variables and that the rate remains constant over time.

For a city of \(400{,}000\) inhabitants, the expected monthly number is

\[ \lambda = 400{,}000\left (\frac {1}{100{,}000}\right ) = 4. \]

Thus, if \(X\) is the number of events in a given month,

\[ X\sim \operatorname {Pois}(4). \]

(a)

Let \(q=P(X\geq 8).\) Then

\[ \begin {aligned} q &= 1-P(X\leq 7)\\ &= 1-\sum _{k=0}^{7}e^{-4}\frac {4^k}{k!}\\ &\approx {0.05113}. \end {aligned} \]

(b)

Let \(Y\) be the number of months, among the next 12 months, that have at least eight events. Under the independence assumption,

\[ Y\sim \operatorname {Bin}(12,q), \]

where \(q\approx 0.05113\).

Therefore,

\[ \begin {aligned} P(Y\geq 2) &= 1-P(Y=0)-P(Y=1)\\ &= 1-(1-q)^{12} -12q(1-q)^{11}\\ &\approx {0.1229}. \end {aligned} \]

(c)

Let \(T\) denote the first month with at least eight events. Each month is a success with probability \(q\), independently of other months. Therefore,

\[ T\sim \operatorname {Geom}(q), \]

using the convention that \(T\) is the trial number of the first success.

Hence,

\[ { P(T=i)=(1-q)^{i-1}q, \qquad i=1,2,\ldots , } \]

where

\[ q = 1-\sum _{k=0}^{7}e^{-4}\frac {4^k}{k!} \approx 0.05113. \]

Numerically, \(P(T=i) \approx (0.94887)^{i-1}(0.05113).\)

Let \(X\) denote the number of typographical errors on the next page. Under the Poisson assumption,

\[ X\sim \operatorname {Pois}(0.2). \]

Thus,

\[ P(X=k)=e^{-0.2}\frac {(0.2)^k}{k!}. \]

(a)

\[ P(X=0) = e^{-0.2} \approx {0.8187}. \]

(b)

Using the complement,

\[ \begin {aligned} P(X\geq 2) &= 1-P(X=0)-P(X=1)\\ &= 1-e^{-0.2}-(0.2)e^{-0.2}\\ &= 1-1.2e^{-0.2}\\ &\approx {0.0175}. \end {aligned} \]

The expected value alone does not uniquely determine these probabilities. The calculation requires the additional assumption that the number of errors follows a Poisson distribution.

Using linearity of expectation,

\begin {align*} E[Y] &= E\left [\frac {X-\mu }{\sigma }\right ]\\ &= \frac {1}{\sigma } \left (E[X]-\mu \right )\\ &= \frac {1}{\sigma }(\mu -\mu )\\ &= {0}. \end {align*}

For the variance, use

\[ \operatorname {Var}(aX+b) = a^2\operatorname {Var}(X). \]

Thus,

\begin {align*} \operatorname {Var}(Y) &= \operatorname {Var} \left ( \frac {X-\mu }{\sigma } \right )\\ &= \frac {1}{\sigma ^2}\operatorname {Var}(X)\\ &= \frac {\sigma ^2}{\sigma ^2}\\ &= {1}. \end {align*}

Therefore, the standardized random variable \(Y\) has mean \(0\) and variance \(1\).

First suppose that \(x+y>0\), and define

\[ p=\frac {x}{x+y}. \]

Then

\[ 1-p=\frac {y}{x+y}. \]

If \(X\sim \operatorname {Bin}(n,p)\), the probabilities of all possible values of \(X\) sum to one:

\[ \sum _{i=0}^{n} \binom ni p^i(1-p)^{n-i} = 1. \]

Substituting the expressions for \(p\) and \(1-p\),

\[ \sum _{i=0}^{n} \binom ni \left (\frac {x}{x+y}\right )^i \left (\frac {y}{x+y}\right )^{n-i} = 1. \]

Because every term has denominator \((x+y)^n\),

\[ \frac {1}{(x+y)^n} \sum _{i=0}^{n} \binom ni x^iy^{n-i} = 1. \]

Multiplying by \((x+y)^n\) gives

\[ { (x+y)^n = \sum _{i=0}^{n} \binom ni x^iy^{n-i}. } \]

If \(x=y=0\), the identity also holds directly for \(n>0\), because both sides equal zero.

The Poisson probability mass function is

\[ p_i=P(X=i)=e^{-\lambda }\frac {\lambda ^i}{i!}, \qquad i=0,1,2,\ldots \]

For \(i\geq 1\), consider the ratio of consecutive probabilities:

\begin {align*} \frac {p_i}{p_{i-1}} &= \frac {e^{-\lambda }\lambda ^i/i!} {e^{-\lambda }\lambda ^{i-1}/(i-1)!}\\ &= \frac {\lambda }{i}. \end {align*}

Therefore,

\[ p_i>p_{i-1} \quad \Longleftrightarrow \quad i<\lambda , \]

and

\[ p_i<p_{i-1} \quad \Longleftrightarrow \quad i>\lambda . \]

Thus the probabilities increase while \(i<\lambda \) and decrease once \(i>\lambda \).

If \(\lambda \) is not an integer, the unique mode is

\[ {\lfloor \lambda \rfloor }. \]

If \(\lambda =m\) is a positive integer, then

\[ \frac {p_m}{p_{m-1}} = \frac {m}{m} = 1, \]

so

\[ p_{m-1}=p_m. \]

Hence there are two modes:

\[ {m-1\ \text {and}\ m}. \]

In particular, \(\lfloor \lambda \rfloor \) is always a mode, but when \(\lambda \) is an integer, it is not the only mode.

Because \(X\) is geometric,

\[ P(X=j)=(1-p)^{j-1}p, \qquad j=1,2,\ldots \]

Also,

\[ P(X>n)=(1-p)^n. \]

Since the event \(\{X=n+k\}\) is contained in the event \(\{X>n\}\),

\begin {align*} P(X=n+k\mid X>n) &= \frac {P(X=n+k)}{P(X>n)}\\ &= \frac {(1-p)^{n+k-1}p}{(1-p)^n}\\ &= (1-p)^{k-1}p\\ &= P(X=k). \end {align*}

Therefore,

\[ { P(X=n+k\mid X>n)=P(X=k). } \]

For the verbal argument, the condition \(X>n\) means that the first \(n\) trials were failures. Because the trials are independent, those failures do not affect the outcomes of future trials. Starting after trial \(n\), the waiting time until the first success has the same geometric distribution as it had at the beginning. This property is called the memoryless property.

(a)

By definition,

\[ E[X] = \int _{0}^{\infty }x f(x)\,dx. \]

Therefore,

\begin {align*} E[X] &= \int _{0}^{\infty } x\left (\frac 14xe^{-x/2}\right )\,dx\\ &= \frac 14 \int _{0}^{\infty }x^2e^{-x/2}\,dx. \end {align*}

Using integration by parts, we have

\[ \int _{0}^{\infty }x^2e^{-x/2}\,dx = \frac {2!}{(1/2)^3} = 16. \]

Equivalently, \(X\) has a gamma distribution with shape \(2\) and scale \(2\), whose mean is \(2\cdot 2=4\).

(b)

First determine \(c\) by requiring the density to integrate to one:

\begin {align*} 1 &= \int _{-1}^{1}c(1-x^2)\,dx\\ &= c\left [x-\frac {x^3}{3}\right ]_{-1}^{1}\\ &= c\left (\frac 43\right )\Rightarrow c=\frac {3}{4} \end {align*}

Now,

\[ E[X] = \int _{-1}^{1}x\frac 34(1-x^2)\,dx = 0 \ \ \text {(symmetric interval around 0 of an odd function is zero)}. \]

(c)

First verify that \(f\) is a density:

\[ \int _{5}^{\infty }\frac {5}{x^2}\,dx = 5\left [-\frac 1x\right ]_{5}^{\infty } = 1. \]

The expected value is

\begin {align*} E[X] &= \int _{5}^{\infty }x\frac {5}{x^2}\,dx\\ &= 5\int _{5}^{\infty }\frac 1x\,dx. \end {align*}

The integral diverges. Hence, \( {E[X]=\infty }.\)

Because \(f\) is a probability density function,

\[ \int _{0}^{1}(a+bx^2)\,dx=1. \]

Thus,

\[ a+\frac {b}{3}=1. \]

Also,

\begin {align*} E[X] &= \int _{0}^{1}x(a+bx^2)\,dx\\ &= \frac {a}{2}+\frac {b}{4}. \end {align*}

Since \(E[X]=\frac 35\),

\[ \frac {a}{2}+\frac {b}{4} = \frac 35. \]

We therefore solve

\[ \begin {cases} a+\dfrac {b}{3}=1,\\[2mm] \dfrac {a}{2}+\dfrac {b}{4}=\dfrac 35. \end {cases} \]

From the first equation,

\[ a=1-\frac {b}{3}. \]

Substituting into the second equation gives

\begin {align*} \frac 12\left (1-\frac {b}{3}\right )+\frac {b}{4} &= \frac 35,\\ \frac 12+\frac {b}{12} &= \frac 35,\\ \frac {b}{12} &= \frac 1{10}. \end {align*}

Hence,

\[ b=\frac 65. \]

Then

\[ a = 1-\frac {1}{3}\left (\frac 65\right ) = 1-\frac 25 = \frac 35. \]

Therefore,

\[ { a=\frac 35, \qquad b=\frac 65. } \]

The resulting density is nonnegative on \([0,1]\), as required.

The expected lifetime is

\begin {align*} E[X] &= \int _{0}^{\infty }x f(x)\,dx\\ &= \int _{0}^{\infty }x^2e^{-x}\,dx. \end {align*}

Using

\[ \int _{0}^{\infty }x^ne^{-x}\,dx=n!, \]

we obtain

\[ E[X] = 2! = {2}. \]

Thus the expected lifetime is

\[ {2\text { hours}}. \]

Let \(X\) be the distance from the left endpoint of the segment to the randomly selected point.

Choosing the point uniformly at random means that

\[ X\sim \operatorname {Unif}(0,L), \]

with density

\[ f_X(x)=\frac 1L, \qquad 0<x<L. \]

The two resulting segment lengths are

\[ X \qquad \text {and}\qquad L-X. \]

By symmetry, first consider \(0<X\leq L/2\). In this case, \(X\) is the shorter segment and \(L-X\) is the longer segment. The desired condition is

\[ \frac {X}{L-X}<\frac 14. \]

Solving,

\begin {align*} 4X&<L-X,\\ 5X&<L,\\ X&<\frac {L}{5}. \end {align*}

By symmetry, the same condition occurs near the right endpoint when

\[ X>\frac {4L}{5}. \]

Thus the favorable set is

\[ \left (0,\frac {L}{5}\right ) \cup \left (\frac {4L}{5},L\right ). \]

Its total length is

\[ \frac {L}{5}+\frac {L}{5} = \frac {2L}{5}. \]

Therefore,

\[ P\left ( \frac {\text {shorter segment}}{\text {longer segment}}<\frac 14 \right ) = \frac {2L/5}{L} = {\frac 25}. \]

Since

\[ X\sim N(10,36), \]

the mean is

\[ \mu =10 \]

and the standard deviation is

\[ \sigma =6. \]

Define the standardized random variable

\[ Z=\frac {X-10}{6}, \]

so that \(Z\sim N(0,1)\). Let \(\Phi \) denote the standard normal CDF.

(a)

\begin {align*} P(X>5) &= P\left ( Z>\frac {5-10}{6} \right )\\ &= P\left (Z>-\frac 56\right )\\ &= \Phi \left (\frac 56\right )\\ &\approx {0.7977}. \end {align*}

(b)

\begin {align*} P(4<X<16) &= P\left ( \frac {4-10}{6}<Z< \frac {16-10}{6} \right )\\ &= P(-1<Z<1)\\ &= \Phi (1)-\Phi (-1)\\ &\approx {0.6827}. \end {align*}

(c)

\begin {align*} P(X<8) &= P\left ( Z<\frac {8-10}{6} \right )\\ &= P\left (Z<-\frac 13\right )\\ &= \Phi \left (-\frac 13\right )\\ &\approx {0.3694}. \end {align*}

(d)

\begin {align*} P(X<20) &= P\left ( Z<\frac {20-10}{6} \right )\\ &= P\left (Z<\frac 53\right )\\ &= \Phi \left (\frac 53\right ) \approx {0.9522}. \end {align*}

(e)

\begin {align*} P(X>16) &= P\left ( Z>\frac {16-10}{6} \right )\\ &= P(Z>1)\\ &= 1-\Phi (1)\\ &\approx {0.1587}. \end {align*}

The lower and upper quartiles are

\[ Q_1=180{,}000, \qquad Q_3=320{,}000. \]

Because the normal distribution is symmetric, its mean is the midpoint:

\[ \mu = \frac {180{,}000+320{,}000}{2} = 250{,}000. \]

For a standard normal random variable,

\[ \Phi ^{-1}(0.75)\approx 0.67449. \]

Thus,

\[ \frac {320{,}000-250{,}000}{\sigma } = 0.67449. \]

Therefore,

\[ \sigma = \frac {70{,}000}{0.67449} \approx 103{,}782. \]

(a)

Standardize \(200{,}000\):

\[ z = \frac {200{,}000-250{,}000}{103{,}782} \approx -0.4818. \]

Hence,

\[ \begin {aligned} P(X<200{,}000) &= \Phi (-0.4818)\\ &\approx {0.3150}. \end {aligned} \]

Thus approximately \(31.5\%\) of physicians earn less than \(\$200{,}000\).

(b)

Standardize \(280{,}000\):

\[ z = \frac {280{,}000-250{,}000}{103{,}782} \approx 0.2891. \]

Also, because \(320{,}000\) is the upper quartile,

\[ P(X<320{,}000)=0.75. \]

Therefore,

\begin {align*} P(280{,}000<X<320{,}000) &= P(X<320{,}000)-P(X<280{,}000)\\ &= 0.75-\Phi (0.2891)\\ &\approx 0.75-0.6137\\ &= {0.1363}. \end {align*}

Thus approximately \(13.6\%\) of physicians earn between \(\$280{,}000\) and \(\$320{,}000\).

Write \(Y=Y^+-Y^-,\) where \(Y^+=\max (Y,0), \text { and } Y^-=\max (-Y,0).\) Then \(E[Y]=E[Y^+]-E[Y^-]. \) We first consider the positive part. Since

\[ P(Y>y) = \int _y^\infty f_Y(x)\,dx, \]

Tonelli’s theorem gives

\begin {align*} \int _0^\infty P(Y>y)\,dy &= \int _0^\infty \int _y^\infty f_Y(x)\,dx\,dy\\ &= \int _0^\infty \int _0^x dy\, f_Y(x)\,dx\\ &= \int _0^\infty x f_Y(x)\,dx\\ &= E[Y^+]. \end {align*}

Similarly,

\[ P(Y<-y) = \int _{-\infty }^{-y}f_Y(x)\,dx. \]

Hence,

\begin {align*} \int _0^\infty P(Y<-y)\,dy &= \int _0^\infty \int _{-\infty }^{-y}f_Y(x)\,dx\,dy. \end {align*}

For a fixed \(x<0\), the condition \(x<-y\) is equivalent to

\[ 0<y<-x. \]

Therefore, changing the order of integration gives

\begin {align*} \int _0^\infty P(Y<-y)\,dy &= \int _{-\infty }^{0} \int _0^{-x}dy\, f_Y(x)\,dx\\ &= \int _{-\infty }^{0}(-x)f_Y(x)\,dx\\ &= -\int _{-\infty }^{0}x f_Y(x)\,dx\\ &= E[Y^-]. \end {align*}

Combining the two identities,

\begin {align*} E[Y] &= E[Y^+]-E[Y^-]\\ &= \int _{0}^{\infty }P(Y>y)\,dy - \int _{0}^{\infty }P(Y<-y)\,dy. \end {align*}

Thus,

\[ { E[Y] = \int _{0}^{\infty }P(Y>y)\,dy - \int _{0}^{\infty }P(Y<-y)\,dy. } \]

The standard normal density is

\[ \phi (z) = \frac {1}{\sqrt {2\pi }}e^{-z^2/2}. \]

Because

\[ \phi (-z)=\phi (z), \]

the density is symmetric about zero.

(a)

\begin {align*} P(Z<-x) &= \int _{-\infty }^{-x}\phi (z)\,dz. \end {align*}

Using the substitution \(u=-z\),

\begin {align*} P(Z<-x) &= \int _x^\infty \phi (-u)\,du\\ &= \int _x^\infty \phi (u)\,du\\ &= P(Z>x). \end {align*}

Therefore,

\[ {P(Z>x)=P(Z<-x)}. \]

(b)

The events \(\{Z>x\}\) and \(\{Z<-x\}\) are disjoint, so

\begin {align*} P(|Z|>x) &= P(Z>x)+P(Z<-x)\\ &= 2P(Z>x). \end {align*}

Thus,

\[ {P(|Z|>x)=2P(Z>x)}. \]

(c)

By symmetry,

\[ P(Z\leq -x)=P(Z\geq x)=1-\Phi (x). \]

Therefore,

\begin {align*} P(|Z|<x) &= P(-x<Z<x)\\ &= \Phi (x)-\Phi (-x)\\ &= \Phi (x)-\bigl (1-\Phi (x)\bigr )\\ &= 2\Phi (x)-1. \end {align*}

Since \(\Phi (x)=P(Z<x)\), \(P(|Z|<x)=2P(Z<x)-1.\)

Differentiate \(f\). By the chain rule,

\begin {align*} f'(x) &= f(x) \left ( -\frac {x-\mu }{\sigma ^2} \right )\\ &= -\frac {x-\mu }{\sigma ^2}f(x). \end {align*}

Differentiating again,

\begin {align*} f''(x) &= -\frac {1}{\sigma ^2}f(x) - \frac {x-\mu }{\sigma ^2}f'(x)\\ &= -\frac {1}{\sigma ^2}f(x) + \frac {(x-\mu )^2}{\sigma ^4}f(x)\\ &= \frac {(x-\mu )^2-\sigma ^2}{\sigma ^4}f(x). \end {align*}

Since \(f(x)>0\) for every \(x\),

\[ f''(x)=0 \]

if and only if

\[ (x-\mu )^2-\sigma ^2=0. \]

Thus,

\[ (x-\mu )^2=\sigma ^2, \]

which gives

\[ x-\mu =\pm \sigma . \]

Therefore,

\[ {x=\mu -\sigma \quad \text {or}\quad x=\mu +\sigma }. \]

Moreover,

\[ f''(x)>0 \]

when \(|x-\mu |>\sigma \), while

\[ f''(x)<0 \]

when \(|x-\mu |<\sigma \). Hence the concavity changes at both points, so they are indeed inflection points.

(a)

If \(X\sim \operatorname {Unif}(a,b)\), then for \(a\leq x\leq b\),

\[ F(x)=\frac {x-a}{b-a}. \]

Set \(F(m)=1/2\):

\begin {align*} \frac {m-a}{b-a} &= \frac 12,\\ m-a &= \frac {b-a}{2},\\ m &= \frac {a+b}{2}. \end {align*}

Thus,

\[ {m=\frac {a+b}{2}}. \]

(b)

A normal distribution is symmetric about its mean \(\mu \). Hence

\[ P(X<\mu )=\frac 12. \]

Therefore,

\[ {m=\mu }. \]

(c)

If \(X\sim \operatorname {Exp}(\lambda )\), then for \(x\geq 0\),

\[ F(x)=1-e^{-\lambda x}. \]

Set \(F(m)=1/2\):

\begin {align*} 1-e^{-\lambda m} &= \frac 12,\\ e^{-\lambda m} &= \frac 12,\\ -\lambda m &= -\log 2. \end {align*}

Therefore,

\[ {m=\frac {\log 2}{\lambda }}. \]

Let \(0<y<1\), and let

\[ F^{-1}(y) = \inf \{x:F(x)\geq y\} \]

denote the generalized inverse of \(F\).

Because \(F\) is continuous,

\[ F(F^{-1}(y))=y. \]

Now,

\begin {align*} P(Y\leq y) &= P(F(X)\leq y)\\ &= P(X\leq F^{-1}(y))\\ &= F(F^{-1}(y))\\ &= y. \end {align*}

Therefore, the CDF of \(Y\) is

\[ F_Y(y) = \begin {cases} 0, & y\leq 0,\\ y, & 0<y<1,\\ 1, & y\geq 1. \end {cases} \]

This is the CDF of a uniform random variable on \((0,1)\). Hence,

\[ {F(X)\sim \operatorname {Unif}(0,1)}. \]

By symmetry of the standard normal distribution,

\[ P(Z\leq -x)=P(Z\geq x). \]

Since the normal distribution is continuous,

\[ P(Z\geq x)=1-P(Z\leq x)=1-\Phi (x). \]

Therefore,

\[ {\Phi (-x)=1-\Phi (x)}. \]

It follows that:

(a)

The statement

\[ \Phi (-x)=\Phi (x) \]

is generally false. It holds only when \(x=0\).

(b)

\[ \Phi (x)+\Phi (-x) = \Phi (x)+1-\Phi (x) = 1. \]

Thus this statement is

\[ {\text {true}}. \]

(c)

The statement

\[ \Phi (-x)=\frac {1}{\Phi (x)} \]

is false. In fact,

\[ \Phi (-x)=1-\Phi (x). \]

Hence, only statement \( {\text {(b)}}\) is true for every real \(x\).

(a)

Clearly,

\[ f(x)\geq 0 \]

for every \(x\). Also,

\begin {align*} \int _{-\infty }^{\infty }f(x)\,dx &= \frac 13\int _{-\infty }^{0}e^x\,dx + \frac 13\int _{0}^{1}1\,dx + \frac 13\int _{1}^{\infty }e^{-(x-1)}\,dx\\ &= \frac 13(1) + \frac 13(1) + \frac 13(1)\\ &= 1. \end {align*}

Therefore, \(f\) is a probability density function.

(b)

\begin {align*} E[X] &= \frac 13\int _{-\infty }^{0}xe^x\,dx + \frac 13\int _{0}^{1}x\,dx + \frac 13\int _{1}^{\infty }xe^{-(x-1)}\,dx. \end {align*}

For the first integral,

\[ \int xe^x\,dx=(x-1)e^x, \]

so

\[ \int _{-\infty }^{0}xe^x\,dx=-1. \]

For the second integral,

\[ \int _0^1x\,dx=\frac 12. \]

For the third integral, let \(u=x-1\). Then \(x=u+1\), so

\begin {align*} \int _{1}^{\infty }xe^{-(x-1)}\,dx &= \int _{0}^{\infty }(u+1)e^{-u}\,du\\ &= \int _0^\infty ue^{-u}\,du + \int _0^\infty e^{-u}\,du\\ &= 1+1\\ &= 2. \end {align*}

Therefore,

\begin {align*} E[X] &= \frac 13(-1) + \frac 13\left (\frac 12\right ) + \frac 13(2) = {\frac 12}. \end {align*}

Since \(\theta >0\), \(f(x)\geq 0.\) We will use the gamma-integral identity

\[ \int _0^\infty x^k e^{-\theta x}\,dx = \frac {k!}{\theta ^{k+1}}, \qquad k=0,1,2,\ldots \]

(a)

\begin {align*} \int _0^\infty f(x)\,dx &= \frac {\theta ^2}{1+\theta } \int _0^\infty (1+x)e^{-\theta x}\,dx\\ &= \frac {\theta ^2}{1+\theta } \left ( \frac 1\theta +\frac 1{\theta ^2} \right )\\ &= \frac {\theta ^2}{1+\theta } \left ( \frac {\theta +1}{\theta ^2} \right )\\ &= 1. \end {align*}

(b)

\begin {align*} E[X] &= \int _0^\infty x f(x)\,dx\\ &= \frac {\theta ^2}{1+\theta } \int _0^\infty x(1+x)e^{-\theta x}\,dx\\ &= \frac {\theta ^2}{1+\theta } \left ( \int _0^\infty xe^{-\theta x}\,dx + \int _0^\infty x^2e^{-\theta x}\,dx \right )\\ &= \frac {\theta ^2}{1+\theta } \left ( \frac 1{\theta ^2} + \frac 2{\theta ^3} \right ) = { \frac {\theta +2}{\theta (\theta +1)} }. \end {align*}

(c)

First compute the second moment:

\begin {align*} E[X^2] &= \int _0^\infty x^2f(x)\,dx\\ &= \frac {\theta ^2}{1+\theta } \int _0^\infty x^2(1+x)e^{-\theta x}\,dx\\ &= \frac {\theta ^2}{1+\theta } \left ( \frac 2{\theta ^3} + \frac 6{\theta ^4} \right )\\ &= \frac {2(\theta +3)} {\theta ^2(\theta +1)}. \end {align*}

Hence,

\begin {align*} \operatorname {Var}(X) &= E[X^2]-E[X]^2\\ &= \frac {2(\theta +3)} {\theta ^2(\theta +1)} - \left ( \frac {\theta +2} {\theta (\theta +1)} \right )^2\\ &= \frac { 2(\theta +3)(\theta +1)-(\theta +2)^2 } {\theta ^2(\theta +1)^2}\\ &= \frac { 2\theta ^2+8\theta +6 - (\theta ^2+4\theta +4) } {\theta ^2(\theta +1)^2} = { \frac {\theta ^2+4\theta +2} {\theta ^2(\theta +1)^2} }. \end {align*}

Define disjoint events

\[ B_1=A_1 \]

and, for \(n\geq 2\),

\[ B_n = A_n\setminus \bigcup _{j=1}^{n-1}A_j. \]

Then the events \(B_1,B_2,\ldots \) are pairwise disjoint and

\[ \bigcup _{n=1}^{\infty }B_n = \bigcup _{n=1}^{\infty }A_n. \]

Also,

\[ B_n\subseteq A_n, \]

so

\[ P(B_n)\leq P(A_n). \]

By countable additivity,

\begin {align*} P\left (\bigcup _{n=1}^{\infty }A_n\right ) &= P\left (\bigcup _{n=1}^{\infty }B_n\right )\\ &= \sum _{n=1}^{\infty }P(B_n)\\ &\leq \sum _{n=1}^{\infty }P(A_n). \end {align*}

Thus,

\[ { P\left (\bigcup _{n=1}^{\infty }A_n\right ) \leq \sum _{n=1}^{\infty }P(A_n). } \]

By De Morgan’s law,

\[ \left (\bigcap _{n=1}^{\infty }A_n\right )^c = \bigcup _{n=1}^{\infty }A_n^c. \]

Therefore,

\begin {align*} P\left (\bigcap _{n=1}^{\infty }A_n\right ) &= 1- P\left ( \bigcup _{n=1}^{\infty }A_n^c \right ). \end {align*}

By Boole’s inequality,

\[ P\left ( \bigcup _{n=1}^{\infty }A_n^c \right ) \leq \sum _{n=1}^{\infty }P(A_n^c). \]

Hence,

\[ { P\left (\bigcap _{n=1}^{\infty }A_n\right ) \geq 1-\sum _{n=1}^{\infty }P(A_n^c). } \]

Both statements are false in general.

(a)

Consider

\[ \Omega =[0,1] \]

with the uniform probability distribution, and let

\[ A=(0,1]. \]

Then

\[ A\neq \Omega , \]

because \(0\notin A\), but

\[ P(A)=1. \]

Thus, an event may have probability \(1\) without being the entire sample space.

(b)

In the same probability space, let

\[ B=\left \{\frac 12\right \}. \]

Then

\[ B\neq \varnothing , \]

but a singleton has probability zero under the uniform distribution:

\[ P(B)=0. \]

Thus, a nonempty event may have probability zero.

The statements would be true in certain finite probability spaces in which every individual outcome has positive probability, but they are not true for arbitrary probability spaces.

By Bonferroni’s inequality,

\[ P(A\cap B) \geq P(A)+P(B)-1. \]

Therefore,

\[ P(A\cap B) \geq 1+1-1 = 1. \]

Since no probability can exceed \(1\),

\[ {P(A\cap B)=1}. \]

Alternatively,

\[ (A\cap B)^c=A^c\cup B^c. \]

Since

\[ P(A^c)=P(B^c)=0, \]

Boole’s inequality gives

\[ P((A\cap B)^c) \leq P(A^c)+P(B^c) = 0. \]

Hence \(P(A\cap B)=1\).

Let

\[ \Omega =\{\omega _1,\omega _2,\ldots \} \]

be countably infinite, and suppose every outcome has the same probability \(p\).

If \(p>0\), then by countable additivity,

\[ P(\Omega ) = \sum _{n=1}^{\infty }P(\{\omega _n\}) = \sum _{n=1}^{\infty }p = \infty , \]

which contradicts \(P(\Omega )=1\).

If \(p=0\), then

\[ P(\Omega ) = \sum _{n=1}^{\infty }0 = 0, \]

which again contradicts \(P(\Omega )=1\).

Therefore,

No countably infinite sample space can have equally probable outcomes under a countably additive probability measure

By De Morgan’s law,

\[ (A_1\cap \cdots \cap A_n)^c = A_1^c\cup \cdots \cup A_n^c. \]

Since \(P(A_i)=1\),

\[ P(A_i^c)=0 \]

for every \(i\).

By Boole’s inequality,

\[ P(A_1^c\cup \cdots \cup A_n^c) \leq \sum _{i=1}^{n}P(A_i^c) = 0. \]

Thus,

\[ P((A_1\cap \cdots \cap A_n)^c)=0, \]

and consequently,

\[ { P(A_1\cap \cdots \cap A_n)=1. } \]

An \(n\times m\) matrix contains

\[ nm \]

entries. Each entry can independently be chosen to be either \(0\) or \(1\), giving two choices per entry.

By the multiplication rule, the number of matrices is

\[ {2^{nm}}. \]

There are five choices for each of the four digit positions. Since none of the available digits is zero, every resulting string is a four-digit number.

Thus, the total number is

\[ 5^4=625. \]

To count numbers with at least one repeated digit, first count those with all four digits distinct.

There are

\[ 5\cdot 4\cdot 3\cdot 2 = 120 \]

such numbers.

Therefore, the number with at least one repeated digit is

\[ 625-120 = {505}. \]

Hence,

\[ {625\text { total numbers},\qquad 505\text { with repetition}.} \]

Let \(A\) be the event that at least two dice show the same face.

Its complement \(A^c\) is the event that all six dice show different faces.

There are

\[ 6^6 \]

equally likely ordered outcomes.

For all six faces to be different, every face \(1,\ldots ,6\) must appear exactly once. There are

\[ 6! \]

such outcomes.

Therefore,

\[ \begin {aligned} P(A) &= 1-P(A^c)\\ &= 1-\frac {6!}{6^6}\\ &= 1-\frac {720}{46656}\\ &= {\frac {899}{900}}. \end {aligned} \]

Since \(P(A)=1\),

\[ P(A^c)=0. \]

Now,

\[ B=(B\cap A)\cup (B\cap A^c), \]

where the union is disjoint. Therefore,

\[ P(B) = P(B\cap A)+P(B\cap A^c). \]

Since

\[ B\cap A^c\subseteq A^c, \]

we have

\[ P(B\cap A^c)=0. \]

Thus,

\[ P(B\cap A)=P(B). \]

Because \(P(A)=1>0\),

\[ P(B\mid A) = \frac {P(B\cap A)}{P(A)} = \frac {P(B)}{1} = {P(B)}. \]

There are exactly

\[ 5000 \]

odd integers between \(1\) and \(10{,}000\).

(a)

An odd integer divisible by \(3\) is an odd multiple of \(3\).

There are

\[ \left \lfloor \frac {10{,}000}{3}\right \rfloor = 3333 \]

multiples of \(3\). Among these, the odd multiples correspond to odd multipliers \(1,3,\ldots ,3333\), of which there are

\[ 1667. \]

Therefore,

\[ { P(3\mid \text {odd}) = \frac {1667}{5000}. } \]

(b)

Among the \(5000\) odd integers, count those divisible by \(3\) or \(5\).

The number of odd multiples of \(3\) is \(1667\).

There are

\[ \left \lfloor \frac {10{,}000}{5}\right \rfloor =2000 \]

multiples of \(5\), of which \(1000\) are odd.

The numbers divisible by both \(3\) and \(5\) are multiples of \(15\). There are

\[ \left \lfloor \frac {10{,}000}{15}\right \rfloor =666 \]

multiples of \(15\), of which \(333\) are odd.

By inclusion–exclusion, the number of odd integers divisible by \(3\) or \(5\) is

\[ 1667+1000-333=2334. \]

Thus, the number divisible by neither is

\[ 5000-2334=2666. \]

Therefore,

\[ { P(\text {neither }3\text { nor }5\mid \text {odd}) = \frac {2666}{5000} = \frac {1333}{2500}. } \]

Let \(G\) be the event that Susan is guilty and \(L\) the event that Robert lies.

We are given

\[ P(G)=0.65, \]

\[ P(L\mid G)=0.25, \]

and

\[ P(L\mid G^c)=0. \]

By the law of total probability,

\begin {align*} P(L) &= P(L\mid G)P(G) + P(L\mid G^c)P(G^c)\\ &= (0.25)(0.65) + (0)(0.35)\\ &= {0.1625}. \end {align*}

Thus, the probability that Robert commits perjury is

\[ {16.25\%}. \]

The random variable \(Y\) can be written as

\[ Y=\min (X,5). \]

Let

\[ F_Y(y)=P(Y\leq y). \]

If \(y<5\), then

\[ \{Y\leq y\}=\{X\leq y\}, \]

because whenever \(X>5\), we have \(Y=5>y\). Therefore,

\[ F_Y(y)=F(y), \qquad y<5. \]

If \(y\geq 5\), then \(Y\leq 5\leq y\) for every outcome, so

\[ F_Y(y)=1. \]

Hence,

\[ { F_Y(y)= \begin {cases} F(y), & y<5,\\[1mm] 1, & y\geq 5. \end {cases} } \]

Notice that \(Y\) has a point mass at \(5\) of size

\[ P(Y=5) = 1-F(5^-). \]

This mass includes both the event \(X=5\) and the event \(X>5\).

For a discrete random variable, the probability at a point is the size of the jump of the CDF:

\[ P(X=x)=F(x)-F(x^-). \]

At \(x=-2\),

\[ P(X=-2) = \frac 12-0 = \frac 12. \]

At \(x=2\),

\[ P(X=2) = \frac 35-\frac 12 = \frac 1{10}. \]

At \(x=4\),

\[ P(X=4) = \frac 89-\frac 35 = \frac {40-27}{45} = \frac {13}{45}. \]

At \(x=6\),

\[ P(X=6) = 1-\frac 89 = \frac 19. \]

Therefore,

\[ { p_X(x)= \begin {cases} \dfrac 12, & x=-2,\\[1mm] \dfrac 1{10}, & x=2,\\[1mm] \dfrac {13}{45}, & x=4,\\[1mm] \dfrac 19, & x=6,\\[1mm] 0, & \text {otherwise}. \end {cases} } \]

The probabilities sum to

\[ \frac 12+\frac 1{10}+\frac {13}{45}+\frac 19 = 1. \]

(a)

\[ 1 = k\sum _{x=1}^{5}x = k(1+2+3+4+5) = 15k\Rightarrow {k=\frac 1{15}}. \]

(b)

The values of \((1+x)^2\) are

\[ 1,\quad 1,\quad 4,\quad 9 \]

for \(x=-2,0,1,2\), respectively. Hence,

\[ 1 = k(1+1+4+9) = 15k\Rightarrow {k=\frac 1{15}}. \]

(c)

Using the geometric-series formula,

\begin {align*} 1 &= k\sum _{x=1}^{\infty }\left (\frac 19\right )^x\\ &= k\frac {1/9}{1-1/9}\\ &= \frac {k}{8} \Rightarrow k=8 \end {align*}

(d)

Since

\[ \sum _{x=1}^{n}x = \frac {n(n+1)}{2}, \]

we require

\[ 1 = k\frac {n(n+1)}{2} \Rightarrow k=\frac {2}{n(n+1)}. \]

(e)

Since

\[ \sum _{x=1}^{n}x^2 = \frac {n(n+1)(2n+1)}{6}, \]

we require

\[ 1 = k\frac {n(n+1)(2n+1)}{6}. \]

Hence,

\[ { k=\frac {6}{n(n+1)(2n+1)}. } \]

The probabilities are

\[ \begin {array}{c|rrrrrrr} x&-3&-2&-1&0&1&2&3\\ \hline 28p(x)&7&6&5&4&3&2&1. \end {array} \]

First compute the mean:

\begin {align*} E[X] &= \frac 1{28} \left [ (-3)(7)+(-2)(6)+(-1)(5) +0(4)+1(3)+2(2)+3(1) \right ]\\ &= \frac {-28}{28}\\ &= -1. \end {align*}

Next,

\begin {align*} E[X^2] &= \frac 1{28} \left [ 9(7)+4(6)+1(5)+0(4)+1(3)+4(2)+9(1) \right ]\\ &= \frac {112}{28}\\ &= 4. \end {align*}

Therefore,

\begin {align*} \operatorname {Var}(X) &= E[X^2]-E[X]^2\\ &= 4-(-1)^2\\ &= {3}. \end {align*}

The probability masses are the jumps of \(F\):

\[ P(X=-3)=\frac 38, \]

\[ P(X=0)=\frac 34-\frac 38=\frac 38, \]

and

\[ P(X=6)=1-\frac 34=\frac 14. \]

Thus,

\begin {align*} E[X] &= (-3)\frac 38 + 0\frac 38 + 6\frac 14\\ &= -\frac 98+\frac {12}{8}\\ &= \frac 38. \end {align*}

Also,

\begin {align*} E[X^2] &= 9\frac 38 + 0 + 36\frac 14\\ &= \frac {27}{8}+9\\ &= \frac {99}{8}. \end {align*}

Therefore,

\begin {align*} \operatorname {Var}(X) &= E[X^2]-E[X]^2\\ &= \frac {99}{8} - \left (\frac 38\right )^2\\ &= \frac {792}{64}-\frac 9{64}\\ &= {\frac {783}{64}}. \end {align*}

The standard deviation is

\[ { \sigma _X = \sqrt {\frac {783}{64}} = \frac {\sqrt {783}}{8} \approx 3.4978. } \]

Since

\[ X(X-2)=X^2-2X, \]

we have

\[ E[X^2]-2E[X]=3. \]

Using \(E[X]=1\),

\[ E[X^2]-2=3, \]

so

\[ E[X^2]=5. \]

Therefore,

\[ \operatorname {Var}(X) = E[X^2]-E[X]^2 = 5-1 = 4. \]

Now use

\[ \operatorname {Var}(aX+b) = a^2\operatorname {Var}(X). \]

Thus,

\[ \operatorname {Var}(-3X+5) = (-3)^2(4) = {36}. \]

For \(X\),

\begin {align*} E[X] &= 15.85(0.15)+15.9(0.21)+16(0.35)\\ &\qquad +16.1(0.15)+16.2(0.14)\\ &= {15.9995}. \end {align*}

Similarly,

\begin {align*} E[Y] &= 15.85(0.14)+15.9(0.05)+16(0.64)\\ &\qquad +16.1(0.08)+16.2(0.09)\\ &= {16.0000}. \end {align*}

The second moments are

\[ E[X^2] = \sum _xx^2P(X=x) = 256. - \text {computed through the distribution}, \]

and

\[ E[Y^2] = \sum _xx^2P(Y=x). \]

Using

\[ \operatorname {Var}(X)=E[X^2]-E[X]^2, \]

the resulting variances are

\[ { \operatorname {Var}(X)=0.01257475 } \]

and

\[ { \operatorname {Var}(Y)=0.00805. } \]

Thus, the standard deviations are approximately

\[ \sigma _X\approx 0.1121 \]

and

\[ \sigma _Y\approx 0.0897. \]

Both companies fill their bottles with an average of approximately \(16\) fluid ounces. Company \(B\), however, has the smaller variance and standard deviation, so its bottle amounts are more tightly concentrated around \(16\) ounces. Company \(B\) is therefore more consistent.

For \(n\) children, the complement of having at least one boy and at least one girl is the event that all children have the same sex.

Thus,

\begin {align*} P(\text {at least one boy and one girl}) &= 1-P(\text {all boys or all girls})\\ &= 1-\left (\frac 12\right )^n-\left (\frac 12\right )^n\\ &= 1-2^{1-n}. \end {align*}

We require

\[ 1-2^{1-n}\geq 0.95. \]

Therefore,

\[ 2^{1-n}\leq 0.05. \]

Testing integers,

\[ n=5: \qquad 1-2^{1-5} = 1-\frac 1{16} = 0.9375<0.95, \]

whereas

\[ n=6: \qquad 1-2^{1-6} = 1-\frac 1{32} = 0.96875>0.95. \]

Hence, the minimum number is

\[ {6}. \]

Let \(X\) be the number of people with the rare blood type. Then

\[ X\sim \operatorname {Bin}(3000,0.0005). \]

The desired probability is

\begin {align*} P(X\geq 2) &= 1-P(X=0)-P(X=1)\\ &= 1-(1-0.0005)^{3000}\\ &\qquad -3000(0.0005)(1-0.0005)^{2999}. \end {align*}

Thus,

\[ { P(X\geq 2) \approx 0.4422. } \]

Because \(n\) is large and \(p\) is small, a Poisson approximation is also appropriate. Here,

\[ \lambda =np=3000(0.0005)=1.5. \]

Thus,

\begin {align*} P(X\geq 2) &\approx 1-P(Y=0)-P(Y=1)\\ &= 1-e^{-1.5}(1+1.5)\\ &\approx 0.4422, \end {align*}

where \(Y\sim \operatorname {Pois}(1.5)\).

Let

\[ X\sim \operatorname {Pois}(\lambda ). \]

Then

\[ P(X=k) = e^{-\lambda }\frac {\lambda ^k}{k!}. \]

The condition gives

\[ e^{-\lambda }\lambda = e^{-\lambda }\frac {\lambda ^3}{3!}. \]

For \(\lambda >0\), canceling \(e^{-\lambda }\lambda \) yields

\[ 1=\frac {\lambda ^2}{6}. \]

Therefore,

\[ \lambda ^2=6 \]

and

\[ \lambda =\sqrt 6. \]

Consequently,

\begin {align*} P(X=5) &= e^{-\sqrt 6} \frac {(\sqrt 6)^5}{5!}\\ &= e^{-\sqrt 6} \frac {36\sqrt 6}{120}\\ &= { \frac {3\sqrt 6}{10}e^{-\sqrt 6}. } \end {align*}

(a)

For \(x>4\),

\[ f(x)=F'(x) = \frac {32}{x^3}. \]

Thus,

\[ { f(x)= \begin {cases} \dfrac {32}{x^3}, & x\geq 4,\\[2mm] 0, & x<4. \end {cases} } \]

There is no point mass at \(4\), because

\[ F(4)=1-\frac {16}{16}=0. \]

(b)

The CDF is zero for \(x<4\), begins at zero when \(x=4\), and increases continuously toward \(1\).

The density is zero for \(x<4\), has value

\[ f(4)=\frac {32}{64}=\frac 12, \]

and decreases toward zero as \(x\to \infty \).

(c)

Since \(X\) is measured in tens of hours, \(50\) hours corresponds to \(x=5\), \(60\) hours to \(x=6\), and so forth.

(i)

\[ P(X\leq 5) = F(5) = 1-\frac {16}{25} = {\frac 9{25}}. \]

(ii)

Since the distribution is continuous,

\begin {align*} P(X\geq 6) &= 1-F(6)\\ &= \frac {16}{36}\\ &= {\frac 49}. \end {align*}

(iii)

\begin {align*} P(5<X<7) &= F(7)-F(5)\\ &= \left (1-\frac {16}{49}\right ) - \left (1-\frac {16}{25}\right )\\ &= \frac {16}{25}-\frac {16}{49}\\ &= {\frac {384}{1225}}. \end {align*}

(iv)

The support of \(X\) is \([4,\infty )\), corresponding to durations of at least \(40\) hours. Hence,

\[ {P(1<X<3.5)=0}. \]

(a)

A lifetime below \(15{,}000\) miles corresponds to

\[ X<1.5. \]

Therefore,

\begin {align*} P(X<1.5) &= \int _1^{3/2}\frac {2}{x^2}\,dx\\ &= \left [-\frac 2x\right ]_1^{3/2}\\ &= 2-\frac 43\\ &= \frac 23. \end {align*}

Thus, \( {66\frac 23\%}\) of the tires last fewer than \(15{,}000\) miles.

(b)

The interval \(10{,}000\) to \(12{,}500\) miles corresponds to

\[ 1<X<1.25=\frac 54. \]

The required conditional probability is

\[ P\left ( 1<X<\frac 54 \,\middle |\, X<\frac 32 \right ). \]

Since \(\{1<X<5/4\}\subseteq \{X<3/2\}\),

\begin {align*} P\left ( 1<X<\frac 54 \,\middle |\, X<\frac 32 \right ) &= \frac { P(1<X<5/4) }{ P(X<3/2) }. \end {align*}

Now,

\begin {align*} P\left (1<X<\frac 54\right ) &= \int _1^{5/4}\frac 2{x^2}\,dx\\ &= 2-\frac {8}{5}\\ &= \frac 25. \end {align*}

Therefore, \(\frac {2/5}{2/3} = \frac 35.\) Thus, \( {60\%}\) of the tires lasting fewer than \(15{,}000\) miles last between \(10{,}000\) and \(12{,}500\) miles.

The density is

\[ f(x)= \frac {32}{x^3}, \qquad x\geq 4. \]

(a)

\begin {align*} E[X] &= \int _4^\infty x\frac {32}{x^3}\,dx\\ &= 32\int _4^\infty \frac 1{x^2}\,dx\\ &= 32\left [-\frac 1x\right ]_4^\infty \\ &= 32\left (\frac 14\right )\\ &= {8}. \end {align*}

Because \(X\) is measured in tens of hours, the expected duration is

\[ {80\text { hours}}. \]

(b)

\begin {align*} E[X^2] &= \int _4^\infty x^2\frac {32}{x^3}\,dx\\ &= 32\int _4^\infty \frac 1x\,dx. \end {align*}

The integral diverges, so

\[ E[X^2]=\infty . \]

Therefore,

\[ {\operatorname {Var}(X)\text { is not finite}.} \]

Although \(E[X]\) exists, the second moment and variance do not.

The density is symmetric about \(x=3/2\), so we expect the mean to be \(3/2\). Directly,

\begin {align*} E[X] &= \int _1^2x\,6(x-1)(2-x)\,dx\\ &= {\frac 32}. \end {align*}

Next,

\begin {align*} E[X^2] &= \int _1^2x^2\,6(x-1)(2-x)\,dx\\ &= \frac {23}{10}. \end {align*}

Therefore,

\begin {align*} \operatorname {Var}(X) &= E[X^2]-E[X]^2\\ &= \frac {23}{10} - \left (\frac 32\right )^2\\ &= \frac {23}{10}-\frac 94\\ &= \frac {46-45}{20}\\ &= \frac 1{20}. \end {align*}

Thus, the standard deviation is

\[ { \sigma _X = \sqrt {\frac 1{20}} = \frac 1{2\sqrt 5} \approx 0.2236\text { hours}. } \]

The mean is

\[ {E[X]=1.5\text { hours}}. \]

By definition,

\[ E[\ln X] = \int _{1}^{2}\ln (x)\frac {2}{x^2}\,dx. \]

Integrate by parts. Let

\[ u=\ln x, \qquad dv=\frac {2}{x^2}\,dx. \]

Then

\[ du=\frac 1x\,dx, \qquad v=-\frac 2x. \]

Therefore,

\begin {align*} E[\ln X] &= \left [ -\frac {2\ln x}{x} \right ]_{1}^{2} + 2\int _{1}^{2}\frac {1}{x^2}\,dx\\ &= -\ln 2 + 2\left [-\frac 1x\right ]_{1}^{2}\\ &= -\ln 2 + 2\left (1-\frac 12\right )\\ &= {1-\ln 2}. \end {align*}

Numerically,

\[ E[\ln X]\approx 0.3069. \]

The density is symmetric about zero, so

\[ E[X]=0. \]

Therefore,

\[ \operatorname {Var}(X)=E[X^2]. \]

Since \(x^2e^{-|x|}\) is an even function,

\begin {align*} E[X^2] &= \frac 12\int _{-\infty }^{\infty }x^2e^{-|x|}\,dx\\ &= \int _{0}^{\infty }x^2e^{-x}\,dx. \end {align*}

Using

\[ \int _{0}^{\infty }x^ne^{-x}\,dx=n!, \]

we obtain

\[ E[X^2]=2!=2. \]

Hence,

\[ {\operatorname {Var}(X)=2}. \]

Because the multiplicative constant

\[ \frac {\lambda ^r}{\Gamma (r)} \]

is positive, it is enough to maximize

\[ g(x)=x^{r-1}e^{-\lambda x}. \]

Take logarithms:

\[ \log g(x) = (r-1)\log x-\lambda x. \]

Differentiate:

\[ \frac {d}{dx}\log g(x) = \frac {r-1}{x}-\lambda . \]

The critical point satisfies

\[ \frac {r-1}{x}-\lambda =0, \]

so

\[ x=\frac {r-1}{\lambda }. \]

Moreover,

\[ \frac {d^2}{dx^2}\log g(x) = -\frac {r-1}{x^2}<0 \]

for \(x>0\). Thus \(\log g\), and hence \(g\), is strictly concave at the critical point.

Also,

\[ \frac {r-1}{x}-\lambda >0 \]

when \(x<(r-1)/\lambda \), and it is negative when \(x>(r-1)/\lambda \). Therefore, the density increases before this point and decreases afterward.

Hence the unique mode is

\[ {\frac {r-1}{\lambda }}. \]

For \(y\leq 0\),

\[ F_Y(y)=0. \]

For \(y>0\),

\begin {align*} F_Y(y) &= P(Y\leq y)\\ &= P(cX\leq y)\\ &= P\left (X\leq \frac {y}{c}\right )\\ &= F_X\left (\frac {y}{c}\right ). \end {align*}

Thus,

\[ { F_Y(y)= \begin {cases} 0, & y\leq 0,\\[1mm] F_X\left (\dfrac {y}{c}\right ), & y>0. \end {cases} } \]

Using the change-of-variables formula,

\begin {align*} f_Y(y) &= \frac 1c f_X\left (\frac {y}{c}\right )\\ &= \frac 1c \frac {\lambda ^r}{\Gamma (r)} \left (\frac {y}{c}\right )^{r-1} e^{-\lambda y/c}\\ &= \frac {(\lambda /c)^r}{\Gamma (r)} y^{r-1}e^{-(\lambda /c)y}, \qquad y>0. \end {align*}

Therefore,

\[ { Y\sim \operatorname {Gamma} \left (r,\frac {\lambda }{c}\right ) } \]

under the shape–rate parameterization.

Because \(f(x)=0\) for \(x\leq 0\),

\[ \int _{-\infty }^{\infty }f(x)\,dx = \frac {\lambda ^r}{\Gamma (r)} \int _{0}^{\infty } x^{r-1}e^{-\lambda x}\,dx. \]

Use the substitution

\[ u=\lambda x. \]

Then

\[ x=\frac {u}{\lambda }, \qquad dx=\frac {du}{\lambda }. \]

Therefore,

\begin {align*} \int _{-\infty }^{\infty }f(x)\,dx &= \frac {\lambda ^r}{\Gamma (r)} \int _{0}^{\infty } \left (\frac {u}{\lambda }\right )^{r-1} e^{-u}\frac {du}{\lambda }\\ &= \frac {\lambda ^r}{\Gamma (r)} \frac {1}{\lambda ^r} \int _{0}^{\infty } u^{r-1}e^{-u}\,du\\ &= \frac {1}{\Gamma (r)} \Gamma (r)\\ &= 1. \end {align*}

Thus,

\[ { \int _{-\infty }^{\infty }f(x)\,dx=1. } \]

A beta density with parameters \(\alpha ,\beta >0\) has the form

\[ f(x) = \frac {1}{B(\alpha ,\beta )} x^{\alpha -1}(1-x)^{\beta -1}, \qquad 0<x<1. \]

Here,

\[ x(1-x)^2 = x^{2-1}(1-x)^{3-1}, \]

so the possible parameters are

\[ \alpha =2, \qquad \beta =3. \]

The beta normalizing constant is

\[ \frac {1}{B(2,3)} = \frac {\Gamma (5)}{\Gamma (2)\Gamma (3)} = \frac {4!}{1!\,2!} = 12. \]

Thus \(f\) is exactly the density of a

\[ \operatorname {Beta}(2,3) \]

random variable.

For a beta random variable,

\[ E[X]=\frac {\alpha }{\alpha +\beta }, \]

so

\[ { E[X]=\frac 25. } \]

Also,

\[ \operatorname {Var}(X) = \frac {\alpha \beta } {(\alpha +\beta )^2(\alpha +\beta +1)}. \]

Therefore,

\begin {align*} \operatorname {Var}(X) &= \frac {(2)(3)} {5^2(6)}\\ &= {\frac 1{25}}. \end {align*}

The function is nonnegative. It remains to check whether it integrates to one.

The kernel

\[ x^2(1-x)^4 = x^{3-1}(1-x)^{5-1} \]

corresponds to a beta distribution with parameters

\[ \alpha =3, \qquad \beta =5. \]

Its required normalizing constant is

\begin {align*} \frac {1}{B(3,5)} &= \frac {\Gamma (8)} {\Gamma (3)\Gamma (5)}\\ &= \frac {7!}{2!\,4!}\\ &= 105. \end {align*}

The proposed coefficient is \(120\), not \(105\).

Indeed,

\begin {align*} \int _{0}^{1}120x^2(1-x)^4\,dx &= 120B(3,5)\\ &= \frac {120}{105}\\ &= \frac 87. \end {align*}

Since

\[ \frac 87\neq 1, \]

the proposed function is not a probability density function.

Thus,

\[ { f\text { is not a probability density function.} } \]

Replacing \(120\) by \(105\) would produce the density of a \(\operatorname {Beta}(3,5)\) random variable.

We have \[ X\sim \operatorname {Bin}(100,0.4). \]

To check whether the Normal approximation is appropriate, we compute \[ np=100(0.4)=40 \] and \[ n(1-p)=100(0.6)=60. \]

Since both quantities are greater than \(10\), the Normal approximation is appropriate.

The mean and variance are \[ \mu =np=40 \] and \[ \sigma ^2=np(1-p)=100(0.4)(0.6)=24. \]

Therefore, we approximate \(X\) by \[ Y\sim N(40,24). \]

Using the continuity correction, \[ \mathbb {P}(35\leq X\leq 45) \approx \mathbb {P}(34.5<Y<45.5). \]

Standardizing, \[ \begin {aligned} \mathbb {P}(34.5<Y<45.5) &= \mathbb {P}\left ( \frac {34.5-40}{\sqrt {24}} < Z < \frac {45.5-40}{\sqrt {24}} \right )\\ &= \mathbb {P}(-1.123<Z<1.123). \end {aligned} \]

Therefore, \[ \begin {aligned} \mathbb {P}(35\leq X\leq 45) &\approx \Phi (1.123)-\Phi (-1.123)\\ &= 2\Phi (1.123)-1\\ &\approx {0.739}. \end {aligned} \]

Let \[ X=\text {the number of customers who buy the book during the month}. \]

The total number of customers during the month is \[ n=30(800)=24000. \]

Because the odds in favor of buying the book are \(1\) to \(5000\), the probability that a customer buys the book is \[ p=\frac {1}{1+5000}=\frac {1}{5001}. \]

Therefore, \[ X\sim \operatorname {Bin}\left (24000,\frac {1}{5001}\right ). \]

We first determine the appropriate approximation.

The expected number of successes is \[ np = 24000\left (\frac {1}{5001}\right ) \approx 4.799. \]

Although \(n\) is large, \[ np\approx 4.799<10, \] so the usual Normal approximation is not appropriate.

However, \(n=24000\) is large and \[ p=\frac {1}{5001} \] is very small. Therefore, the Poisson approximation is appropriate, with \[ \lambda =np=\frac {24000}{5001}\approx 4.799. \]

Thus, \[ X\approx W, \qquad W\sim \operatorname {Pois}(4.799). \]

If the bookstore stocks \(m\) copies, then it does not run out when \[ X\leq m. \]

We seek the smallest integer \(m\) such that \[ \mathbb {P}(X\leq m)>0.98. \]

Using the Poisson approximation, \[ \mathbb {P}(X\leq 9) \approx \mathbb {P}(W\leq 9) = \sum _{k=0}^{9} e^{-4.799}\frac {4.799^k}{k!} \approx 0.9749. \]

Since \[ 0.9749<0.98, \] stocking \(9\) copies is not sufficient.

Next, \[ \mathbb {P}(X\leq 10) \approx \mathbb {P}(W\leq 10) = \sum _{k=0}^{10} e^{-4.799}\frac {4.799^k}{k!} \approx 0.9896. \]

Since \[ 0.9896>0.98, \] the smallest satisfactory value is \(m=10\).

Therefore, the bookstore should stock \[ {10\text { copies}.} \]

Let \[ X=\text {the number of Type I bulbs in the sample}. \]

Because the sample is selected without replacement, the exact distribution is \[ X\sim \operatorname {Hypergeo}(5000,2500,40). \]

The sampling fraction is \[ \frac {40}{5000}=0.008. \]

Since the sample size is very small relative to the population size, sampling without replacement is approximately equivalent to sampling with replacement. Therefore, \[ X \approx \operatorname {Bin}\left (40,\frac {2500}{5000}\right ) = \operatorname {Bin}\left (40,\frac 12\right ). \]

For this Binomial distribution, \[ np=40\left (\frac 12\right )=20 \] and \[ n(1-p)=40\left (\frac 12\right )=20. \]

Since both quantities are greater than \(10\), the Normal approximation is appropriate.

The mean and variance are \[ \mu =np=20 \] and \[ \sigma ^2=np(1-p) = 40\left (\frac 12\right )\left (\frac 12\right ) = 10. \]

Thus, we further approximate \(X\) by \[ Y\sim N(20,10). \]

If there are \(X\) Type I bulbs, then there are \[ 40-X \] Type II bulbs.

The sample contains at least \(18\) bulbs of each type when \[ X\geq 18 \] and \[ 40-X\geq 18. \]

The second inequality is equivalent to \[ X\leq 22. \]

Therefore, the desired event is \[ 18\leq X\leq 22. \]

Using the continuity correction, \[ \mathbb {P}(18\leq X\leq 22) \approx \mathbb {P}(17.5<Y<22.5). \]

Standardizing, \[ \begin {aligned} \mathbb {P}(17.5<Y<22.5) &= \mathbb {P}\left ( \frac {17.5-20}{\sqrt {10}} < Z < \frac {22.5-20}{\sqrt {10}} \right )\\ &= \mathbb {P}(-0.791<Z<0.791). \end {aligned} \]

Therefore, \[ \begin {aligned} \mathbb {P}(18\leq X\leq 22) &\approx \Phi (0.791)-\Phi (-0.791)\\ &= 2\Phi (0.791)-1\\ &\approx 2(0.785)-1\\ &= {0.570}. \end {aligned} \]

Thus, the approximate probability that the sample contains at least \(18\) bulbs of each type is \(0.570\).

Let \[ X=\text {the lifetime of a randomly selected light bulb}. \]

Then \[ X\sim N(1000,100^2). \]

The proportion of bulbs that last at least \(900\) hours is \[ \mathbb {P}(X\geq 900). \]

Standardizing, \[ \begin {aligned} \mathbb {P}(X\geq 900) &= \mathbb {P}\left ( Z\geq \frac {900-1000}{100} \right )\\ &= \mathbb {P}(Z\geq -1). \end {aligned} \]

By symmetry of the standard Normal distribution, \[ \mathbb {P}(Z\geq -1) = \mathbb {P}(Z\leq 1) = \Phi (1). \]

Using the standard Normal table, \[ \Phi (1)\approx 0.8413. \]

Therefore, \[ \mathbb {P}(X\geq 900)\approx 0.8413. \]

Thus, approximately \(84.13\%\), rather than \(95\%\), of the bulbs last at least \(900\) hours. Since \[ 0.8413<0.95, \] the company’s claim is incorrect.

\[ {\text {No, the company's claim is not correct.}} \]

Let \[ X=\text {the lifetime of the bulb from the first company} \] and \[ Y=\text {the lifetime of the bulb from the second company}. \]

Then \[ X\sim N(1000,100^2) \] and \[ Y\sim N(900,150^2). \]

We want \[ \mathbb {P}(X\geq 980\text { or }Y\geq 980). \]

It is easier to use the complementary event: \[ \mathbb {P}(X\geq 980\text { or }Y\geq 980) = 1-\mathbb {P}(X<980,\ Y<980). \]

Because \(X\) and \(Y\) are independent, \[ \mathbb {P}(X<980,\ Y<980) = \mathbb {P}(X<980)\mathbb {P}(Y<980). \]

For the first bulb, \[ \begin {aligned} \mathbb {P}(X<980) &= \mathbb {P}\left ( Z< \frac {980-1000}{100} \right )\\ &= \mathbb {P}(Z<-0.20)\\ &= \Phi (-0.20)\\ &\approx 0.4207. \end {aligned} \]

For the second bulb, \[ \begin {aligned} \mathbb {P}(Y<980) &= \mathbb {P}\left ( Z< \frac {980-900}{150} \right )\\ &= \mathbb {P}(Z<0.533)\\ &= \Phi (0.533)\\ &\approx 0.7031. \end {aligned} \]

Therefore, \[ \begin {aligned} \mathbb {P}(X<980,\ Y<980) &\approx (0.4207)(0.7031)\\ &\approx 0.2958. \end {aligned} \]

Consequently, \[ \begin {aligned} \mathbb {P}(X\geq 980\text { or }Y\geq 980) &\approx 1-0.2958\\ &= {0.7042}. \end {aligned} \]

Thus, the probability that at least one of the two bulbs lasts \(980\) hours or more is approximately \(70.42\%\).