There are \(25\) choices for the first letter because it cannot be X. Each of the next two letters has \(26\) choices.

For the four digits, count all digit strings and subtract those containing only odd digits. There are \(10^4\) total digit strings and \(5^4\) strings in which all four digits are odd. Therefore the number of digit strings containing at least one even digit is \[ 10^4-5^4. \] By the multiplication principle, the required number is \[ 25\cdot 26^2\bigl (10^4-5^4\bigr ). \] Since \(10^4-5^4=9375\), this equals \[ 25\cdot 676\cdot 9375=158{,}437{,}500. \]

Let \(D\) be the event that the selected item is defective. The given probabilities are \[ P(A)=0.40,\qquad P(B)=0.35,\qquad P(C)=0.25, \] and \[ P(D\mid A)=0.02,\qquad P(D\mid B)=0.05,\qquad P(D\mid C)=0.08. \] By the law of total probability, \begin {align*} P(D) &=P(D\mid A)P(A)+P(D\mid B)P(B)+P(D\mid C)P(C)\\ &=(0.02)(0.40)+(0.05)(0.35)+(0.08)(0.25)\\ &=0.008+0.0175+0.020=0.0455. \end {align*}

(a) By Bayes’ rule, \[ P(C\mid D)=\frac {P(D\mid C)P(C)}{P(D)} =\frac {(0.08)(0.25)}{0.0455} =\frac {40}{91}. \]

(b) Since the factories are mutually exclusive, \begin {align*} P(A^c\mid D) &=1-P(A\mid D)\\ &=1-\frac {P(D\mid A)P(A)}{P(D)}\\ &=1-\frac {0.008}{0.0455} =1-\frac {16}{91} =\frac {75}{91}. \end {align*}

Equivalently, one may add \(P(B\mid D)\) and \(P(C\mid D)\).

(a) Apply the two-event addition rule to \((A\cup B)\) and \(C\): \[ P(A\cup B\cup C)=P(A\cup B)+P(C)-P\bigl ((A\cup B)\cap C\bigr ). \] Now \[ P(A\cup B)=P(A)+P(B)-P(A\cap B), \] and, by distributivity, \[ (A\cup B)\cap C=(A\cap C)\cup (B\cap C). \] Applying the addition rule again, \begin {align*} P\bigl ((A\cup B)\cap C\bigr ) &=P\bigl ((A\cap C)\cup (B\cap C)\bigr )\\ &=P(A\cap C)+P(B\cap C)-P(A\cap B\cap C). \end {align*}

Substitution gives \begin {align*} P(A\cup B\cup C) &=P(A)+P(B)+P(C)\\ &\quad -P(A\cap B)-P(A\cap C)-P(B\cap C)\\ &\quad +P(A\cap B\cap C). \end {align*}

(b) Let \[ N=\mathbf {1}_A+\mathbf {1}_B+\mathbf {1}_C. \] Pointwise, \(\mathbf {1}_{A\cup B\cup C}\leq N\). Taking expectations gives \[ P(A\cup B\cup C)\leq P(A)+P(B)+P(C). \]

(c) Equality holds exactly when no outcome belongs to two or more of the events, except possibly on a probability-zero set. Equivalently, \[ P(A\cap B)=P(A\cap C)=P(B\cap C)=0. \] Indeed, the union bound overcounts an outcome precisely when it lies in at least two of the events.

(a) Since \(X\) is continuous, differentiate \(F_X\) on each interval: \[ f_X(x)=F_X'(x)= \begin {cases} \dfrac {x}{2}, & 0<x<1,\\[2mm] \dfrac 12, & 1<x<2,\\[2mm] \dfrac {3-x}{2}, & 2<x<3,\\[2mm] 0, & \text {otherwise}. \end {cases} \] The values at the endpoints do not affect the density.

(b) To verify that \(f_X\) is a probability density function, we must show that

(i)

\(f_X(x)\ge 0\) for all \(x\), and

(ii)

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

From part (a), \[ f_X(x)= \begin {cases} \dfrac {x}{2}, & 0<x<1,\\[2mm] \dfrac 12, & 1<x<2,\\[2mm] \dfrac {3-x}{2}, & 2<x<3,\\[2mm] 0, & \text {otherwise}. \end {cases} \]

Clearly, \[ f_X(x)\ge 0 \] for every \(x\).

Also, \begin {align*} \int _{-\infty }^{\infty }f_X(x)\,dx &= \int _0^1\frac {x}{2}\,dx +\int _1^2\frac 12\,dx +\int _2^3\frac {3-x}{2}\,dx\\ &= \frac 12\left [\frac {x^2}{2}\right ]_0^1 +\frac 12[x]_1^2 +\frac 12\left [3x-\frac {x^2}{2}\right ]_2^3\\ &= \frac 14+\frac 12+\frac 14\\ &=1. \end {align*}

Hence, \(f_X\) satisfies both conditions of a probability density function.

(c) Because \(X\) is continuous, \begin {align*} P\left (\frac 12<X\leq \frac 52\right ) &=F_X\left (\frac 52\right )-F_X\left (\frac 12\right )\\ &=\left (1-\frac {(3-5/2)^2}{4}\right )-\frac {(1/2)^2}{4}\\ &=\left (1-\frac 1{16}\right )-\frac 1{16} =\frac 78. \end {align*}

(d) Using the density, \begin {align*} \mathbb {E}[X] &=\int _0^1 x\frac {x}{2}\,dx +\int _1^2 x\frac 12\,dx +\int _2^3 x\frac {3-x}{2}\,dx\\ &=\frac 12\int _0^1x^2\,dx +\frac 12\int _1^2x\,dx +\frac 12\int _2^3(3x-x^2)\,dx\\ &=\frac 16+\frac 34+\frac 7{12} =\frac 32. \end {align*}

The symmetry of the density around \(x=3/2\) also gives the same result.

(a)

Each of the \(8000\) rides independently results in a serious mechanical failure with probability \(0.0005\). Therefore, \[ X\sim \operatorname {Bin}(8000,0.0005). \]

For the Normal approximation, we would need both \[ np \qquad \text {and}\qquad n(1-p) \] to be sufficiently large. However, \[ np=8000(0.0005)=4, \] which is less than \(10\). Thus, the usual Normal approximation is not appropriate.

For a Poisson approximation, \(n\) should be large and \(p\) should be small. Here, \[ n=8000 \] is large and \[ p=0.0005 \] is very small. Therefore, the Poisson approximation is appropriate.

Its parameter is \[ \lambda =np=8000(0.0005)=4. \]

Thus, \[ X\approx W, \qquad W\sim \operatorname {Pois}(4). \]

(b)

We want \[ P(X\geq 3). \]

Using the Poisson approximation, \[ P(X\geq 3) \approx P(W\geq 3). \]

It is easier to use the complement: \[ P(W\geq 3) = 1-P(W\leq 2). \]

Therefore, \begin {align*} P(W\geq 3) &= 1-\left [ P(W=0)+P(W=1)+P(W=2) \right ]\\ &= 1-\left [ e^{-4} + e^{-4}\frac {4}{1!} + e^{-4}\frac {4^2}{2!} \right ]\\ &= 1-e^{-4}(1+4+8)\\ &= 1-13e^{-4}\\ &\approx {0.7619}. \end {align*}

(c)

Each of the \(300\) bikes is independently returned late with probability \(0.35\). Therefore, \[ Y\sim \operatorname {Bin}(300,0.35). \]

A Poisson approximation is not appropriate because \[ p=0.35 \] is not small.

For the Normal approximation, we compute \[ np=300(0.35)=105 \] and \[ n(1-p)=300(0.65)=195. \]

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

The mean is \[ \mu =np=105, \] and the variance is \[ \sigma ^2=np(1-p) = 300(0.35)(0.65) = 68.25. \]

Thus, \[ Y\approx V, \qquad V\sim N(105,68.25), \] where \[ \sigma =\sqrt {68.25}\approx 8.26. \]

(d)

We want to approximate \[ P(95\leq Y\leq 115). \]

Using the continuity correction, \[ P(95\leq Y\leq 115) \approx P(94.5<V<115.5). \]

Standardizing, \begin {align*} P(94.5<V<115.5) &= P\left ( \frac {94.5-105}{\sqrt {68.25}} < Z < \frac {115.5-105}{\sqrt {68.25}} \right )\\ &\approx P(-1.27<Z<1.27). \end {align*}

Using the standard Normal table, \[ \Phi (1.27)=0.8980. \]

By symmetry, \[ \Phi (-1.27)=1-\Phi (1.27)=0.1020. \]

Therefore, \begin {align*} P(95\leq Y\leq 115) &\approx \Phi (1.27)-\Phi (-1.27)\\ &= 0.8980-0.1020\\ &= {0.7960}. \end {align*}