Assume the usual birthday model with 365 equally likely birthdays and no leap day. The smallest possible value of \(X\) is 2, since one person alone cannot create a match. The largest possible value is 366, since after 365 people with distinct birthdays, the next person must match one of them.

For \(X=k\), the first \(k-1\) people must all have distinct birthdays, and the \(k\)th person must match one of those \(k-1\) birthdays. The probability that the first \(k-1\) birthdays are distinct is \[ \frac {365\cdot 364\cdots (365-k+2)}{365^{k-1}}. \] Once this has happened, the \(k\)th person has exactly \(k-1\) favorable birthdays out of 365. Therefore, for \(k=2,3,\ldots ,366\), \[ P(X=k) = \frac {365\cdot 364\cdots (365-k+2)}{365^{k-1}}\cdot \frac {k-1}{365}. \]

Let \(F_Y\) denote the CDF of \(Y\). For any real number \(y\), \[ F_Y(y)=P(Y\le y). \] Since \(Y=\mu +\sigma X\) and \(\sigma >0\), the inequality may be rearranged without reversing its direction: \[ \mu +\sigma X\le y \qquad \Longleftrightarrow \qquad X\le \frac {y-\mu }{\sigma }. \] Thus \[ F_Y(y) = P\left (X\le \frac {y-\mu }{\sigma }\right ) = F\left (\frac {y-\mu }{\sigma }\right ). \]

(a) The function \(F\) is nondecreasing because \(\lfloor x\rfloor \) is nondecreasing. It also has the correct limiting behavior. For \(x<0\), \(F(x)=0\), so \[ \lim _{x\to -\infty }F(x)=0. \] For \(x>n\), \(F(x)=1\), so \[ \lim _{x\to \infty }F(x)=1. \] The only possible jumps occur at the integers. At an integer \(j\) with \(1\le j\le n\), the left-hand limit is \[ F(j^-)=\frac {j-1}{n}, \] while \[ F(j)=\frac {j}{n}. \] Thus the function is right-continuous at each jump. Away from the integers, it is constant on intervals and therefore right-continuous there as well. Hence \(F\) has the required properties of a CDF.

(b) The PMF is obtained from the jumps of the CDF. At each integer \(j=1,2,\ldots ,n\), \[ P(X=j)=F(j)-F(j^-)=\frac {j}{n}-\frac {j-1}{n}=\frac {1}{n}. \] There are no jumps elsewhere, so the probability is zero outside \(\{1,2,\ldots ,n\}\).

(a) Each value \(p(n)\) is nonnegative. It remains to check that the probabilities add to 1. Using the geometric series, \[ \sum _{n=0}^{\infty }\left (\frac 12\right )^{n+1} = \frac 12+\frac 14+\frac 18+\cdots = 1. \] Therefore \(p(n)\) is a valid PMF on the nonnegative integers.

(b) Let \(X\) have this PMF. If \(x<0\), then no possible value of \(X\) is at most \(x\), so \(F(x)=0\). If \(x\ge 0\), then \[ F(x)=P(X\le x) = \sum _{n=0}^{\lfloor x\rfloor }\left (\frac 12\right )^{n+1}. \] The finite geometric sum gives \[ F(x)=1-\left (\frac 12\right )^{\lfloor x\rfloor +1}, \qquad x\ge 0. \]

Let \(M\) be the value of the most valuable prize among the 5 prizes selected. The sample space consists of all 5-element subsets of the 100 prizes, so there are \[ \binom {100}{5} \] equally likely selections.

For the maximum value to be \(m\), the prize worth \(m\) dollars must be selected, and the other 4 selected prizes must all come from the prizes worth \(1,2,\ldots ,m-1\). This can be done in \[ \binom {m-1}{4} \] ways. This is possible only when \(m\ge 5\).

Therefore \[ P(M=m)=\frac {\binom {m-1}{4}}{\binom {100}{5}}, \qquad m=5,6,\ldots ,100. \] \[ P(M=m)=0 \qquad \text {otherwise.} \]

(a) Since \(F_1\) and \(F_2\) are nondecreasing and \(p\) and \(1-p\) are positive weights, their weighted average is also nondecreasing. Also, \[ \lim _{x\to -\infty }F(x) = p\lim _{x\to -\infty }F_1(x)+(1-p)\lim _{x\to -\infty }F_2(x) = 0, \] and \[ \lim _{x\to \infty }F(x) = p\lim _{x\to \infty }F_1(x)+(1-p)\lim _{x\to \infty }F_2(x) = p+(1-p)=1. \] Finally, because \(F_1\) and \(F_2\) are right-continuous, for every \(x\), \[ \lim _{t\downarrow x}F(t) = p\lim _{t\downarrow x}F_1(t)+(1-p)\lim _{t\downarrow x}F_2(t) = pF_1(x)+(1-p)F_2(x) = F(x). \] Thus \(F\) is a valid CDF.

(b) Let \(C\) be the event that the coin lands Heads. The random variable is generated from \(F_1\) on \(C\) and from \(F_2\) on \(C^c\). Therefore, by the law of total probability, \[ P(X\le x) = P(X\le x\mid C)P(C)+P(X\le x\mid C^c)P(C^c). \] Since \(P(C)=p\), \(P(C^c)=1-p\), \(P(X\le x\mid C)=F_1(x)\), and \(P(X\le x\mid C^c)=F_2(x)\), we get \[ P(X\le x)=pF_1(x)+(1-p)F_2(x)=F(x). \] So the random variable constructed by this procedure has CDF \(F\).

(a) Conditional on choosing coin 1, the number of Heads has distribution \(\operatorname {Bin}(n,p_1)\). Conditional on choosing coin 2, the number of Heads has distribution \(\operatorname {Bin}(n,p_2)\). Since the two coins are chosen with equal probabilities, \[ P(X=k) = \frac 12\binom {n}{k}p_1^k(1-p_1)^{n-k} + \frac 12\binom {n}{k}p_2^k(1-p_2)^{n-k}, \] for \(k=0,1,\ldots ,n\).

(b) If \(p_1=p_2=p\), then the two possible coins have the same Heads probability. The mixture collapses to the ordinary Binomial distribution, \[ P(X=k)=\binom {n}{k}p^k(1-p)^{n-k}. \] Thus \[ X\sim \operatorname {Bin}(n,p). \]

(c) When \(p_1\ne p_2\), the flips are not independent in the unconditional sense. Seeing many Heads early makes it more likely that the chosen coin is the one with the larger Heads probability, and that changes our prediction for later flips. For large \(n\), the number of Heads tends to cluster around \(np_1\) if coin 1 was chosen and around \(np_2\) if coin 2 was chosen. This is a mixture of two Binomial patterns rather than one Binomial pattern with a single fixed success probability.

(a) Let \(W\) be the number of women promoted. Since the company chooses \(t\) employees from \(n+m\) employees without replacement, \(W\) has a Hypergeometric distribution. To have \(W=k\), the company must choose \(k\) of the \(n\) women and \(t-k\) of the \(m\) men. Thus \[ P(W=k)=\frac {\binom {n}{k}\binom {m}{t-k}}{\binom {n+m}{t}}, \] where \(k\) must satisfy \[ \max (0,t-m)\le k\le \min (n,t). \]

(b) Now each employee is promoted independently with probability \(p\).

The number of women promoted is the number of successes among \(n\) independent promotion decisions for women, so \[ W_p\sim \operatorname {Bin}(n,p). \] The number of women not promoted is the number of failures among those same \(n\) women, so \[ W_{np}\sim \operatorname {Bin}(n,1-p). \] The total number of employees promoted is the number of successes among all \(n+m\) employees, so \[ T\sim \operatorname {Bin}(n+m,p). \]

(c) Given that exactly \(t\) employees are promoted, all sets of \(t\) promoted employees are equally likely. This brings us back to the same sampling-without-replacement structure as part (a). Therefore the conditional distribution of the number of women promoted is Hypergeometric: \[ P(W=k\mid T=t)=\frac {\binom {n}{k}\binom {m}{t-k}}{\binom {n+m}{t}}, \] for the same possible values of \(k\), i.e., \(\ \max (0,t-m)\le k\le \min (n,t).\)