(25 points) Answer True or False to each of the following statements.
(5 points) Suppose that the events \(A_1\), \(A_2\), and \(B_1\) are mutually independent, and that the events \(A_1\), \(A_2\), and \(B_2\) are also mutually independent. Assume further that \(B_1\) and \(B_2\) are disjoint, i.e., \(B_1 \cap B_2 = \emptyset\). Then the three events \(A_1\), \(A_2\), and \(B_1 \cup B_2\) are mutually independent. (True)
(5 points) A box contains 100 balls, of which \(r\) are red. Suppose that the balls are drawn from the box one at a time, at random, without replacement. Then, the probability that the third ball drawn will be red is \((r-2)/98\). (False)
(5 points) Let \(P(X=x)=(x+1)(1-p)^{x} p^2\), \(x=0,1,2,\ldots,\infty\). Then \(\mathrm{E}(X) = 2/p-1\). (False)
(5 points) Let \(P(X=x)={1000 \choose x} 0.001\ ^{x} (1-0.0001)^{1000-x}\), \(x=0,1,2,\ldots,n\). Then \(P (X \geq 1)\) is approximately the same as \(1-e^{-2}\). (False)
(5 points) Suppose that a biased coin is tossed repeatedly. Each time there is a probability \(p\) of a head turning up. Let \(p_n\) be the probability that an even number of heads has occurred after \(n\) tosses. Then, \(p_n - p_{n-1} = p_{n+1} - p_n\). (False)
(25 points) Suppose that you roll a die and your score is the number shown on the die. On the other hand, suppose that your friend rolls five dice and his score is the number of 6’s shown out of five rollings. Compute the probability
(10 points) Compute the probability of event \(A\) that the two scores are equal.
Solution: Let \(A\) be the event that your score and your friend’s score are equal, and let \(F_i\) denote the event that your friend’s score equals \(i\), for \(i = 0,1,2,3,4,5\). Then \[ P(A) = P\!\left(A \cap \left(\cup_{i=1}^{5} F_i \right) \right) = \sum_{i=1}^{5} P(A \mid F_i) P(F_i). \] Since your die roll is independent of your friend’s five rolls, and your own score equals \(i\) with probability \(1/6\), we have \[ P(A \mid F_i) = P(\text{your score} = i) = \frac{1}{6}. \] Therefore, \(P(A) = \sum_{i=1}^{5} \frac{1}{6} P(F_i) = \frac{1}{6} \sum_{i=1}^{5} P(F_i) = \frac{1}{6} \left(1 - P(F_0)\right).\) Since \(F_0\) is the event that no 6’s appear in your friend’s five rolls, \(P(F_0) = \left(\frac{5}{6}\right)^5\). Hence, \(P(A) = \frac{1}{6} \left( 1 - \left( \frac{5}{6} \right)^5 \right)\).
Alternative solution: Let \(X\) be your score and let \(Y\) be your friend’s score. Note that \(P(X=i)=\frac{1}{6}\), \(i=1,\ldots,6\) and \(P(Y=j)=\binom{5}{j}\left(\frac{1}{6}\right)^{j}\left(\frac{5}{6}\right)^{5-j}\),
\(j=0,\ldots,5\), and that \(X\) and \(Y\) are independent. Thus,
\[
\begin{aligned}
P(X=Y)&=\sum_{i=1}^{5}P(X=Y=i)=\sum_{i=1}^{5}\left(P(X=i)\cdot
P(Y=i)\right)\\
&=\sum_{i=1}^{5}\frac{1}{6}\binom{5}{i}\left(\frac{1}{6}\right)^{i}\left(\frac{5}{6}\right)^{5-i}\\
&=\frac{1}{6}\sum_{i=1}^{5}\binom{5}{i}\left(\frac{1}{6}\right)^{i}\left(\frac{5}{6}\right)^{5-i}\\
&=\frac{1}{6}(1-P(Y=0))=\frac{1}{6}\left(1-\left(\frac{5}{6}\right)^{5}\right).
\end{aligned}
\]
(15 points) Compute the probability of event \(B\) that your friend’s score is strictly smaller than yours.
Solution: Let \(B\) be the event that your score is strictly larger than your friend’s. Note that \(P(B|F_{i})=(6-i)/6\). Then, \[ \begin{aligned} P(B)&=P\left(B\cap \left(\cup_{i=0}^{6}F_{i}\right)\right)=\sum_{i=0}^{6}P(B|F_{i})P(F_{i})\\ &=\sum_{i=0}^{5}P(F_{i})\cdot\left(\frac{6-i}{6}\right)\\ &=\frac{1}{6}\cdot 6\cdot \sum_{i=0}^{5}P(F_{i})-\frac{1}{6}\sum_{i=0}^{5}i\cdot P(F_{i})\\ &=1-\frac{1}{6}\sum_{i=0}^{5}i\cdot P(F_{i})\\ &=1-\frac{1}{6}\sum_{i=0}^{5}i\cdot\binom{5}{i}\left(\frac{1}{6}\right)^{i}\left(\frac{5}{6}\right)^{5-i} =1-\frac{5}{36}=\frac{31}{36}. \end{aligned} \]
Alternative solution: To derive \(P(X>Y)\), \[ \begin{aligned} P(X>Y)&=\sum_{i=1}^{6}P(X=i)\cdot P(Y=0,\ldots (i-1))\\ &=\sum_{i=1}^{6}\sum_{j=0}^{i-1}P(X=i)\cdot P(Y=j)\\ &=\sum_{i=1}^{6}\sum_{j=0}^{i-1}\frac{1}{6}\cdot \binom{5}{j}\left(\frac{1}{6}\right)^{j}\left(\frac{5}{6}\right)^{5-j}\\ &=\frac{1}{6}\sum_{i=0}^{5}(6-i)\binom{5}{i}\left(\frac{1}{6}\right)^{i}\left(\frac{5}{6}\right)^{5-i}\\ &=\frac{1}{6}\cdot \mathrm{E}(6-Y)=\frac{1}{6}(6-\mathrm{E}(Y)) =\frac{1}{6}\left(6-\frac{5}{6}\right)=\frac{31}{36}. \end{aligned} \]
(25 points) You roll two fair dice. If the sum of the numbers shown is 7, you win. If it is 8, you lose. If it is any other number \(j\), you continue to roll until the sum is 7 or 8.
(10 points) What is the probability that you win before the \(n(>1)\)th roll?\
Solution: Let \(W_i\) denote the event that the game ends on the \(i\)th roll, and let \(A\) denote the event that you win (i.e., the sum of the dice is 7). Then \(A \cap W_i\) represents the event that you win exactly on the \(i\)th roll. Hence, the probability that you win before the \(n\)th roll is \[ P(A \text{ before } n\text{th roll}) = \sum_{i=1}^{n-1} P(A \cap W_i) = \sum_{i=1}^{n-1} P(A \mid W_i)\, P(W_i). \] Given that the game ends on the \(i\)th roll, the conditional probability of winning is \[ P(A \mid W_i) = \frac{P(\text{sum}=7)}{P(\text{sum}=7)+P(\text{sum}=8)} = \frac{6}{6+5} = \frac{6}{11}. \] The probability that the game ends on the \(i\)th roll is \[ P(W_i) = \left(\frac{25}{36}\right)^{i-1} \left(\frac{11}{36}\right), \] since the first \((i-1)\) rolls must produce neither 7 nor 8 (probability \(25/36\)), and the \(i\)th roll must produce either 7 or 8 (probability \(11/36\)).
Combining these, we obtain \[ \begin{aligned} P(A \text{ before } n\text{th roll}) &= \sum_{i=1}^{n-1} \frac{6}{11} \cdot \left(\frac{25}{36}\right)^{i-1} \cdot \frac{11}{36} \\ &= \sum_{i=1}^{n-1} \frac{1}{6} \left(\frac{25}{36}\right)^{i-1} = \frac{\frac{1}{6}\left(1 - \left(\frac{25}{36}\right)^{n-1}\right)}{1 - \frac{25}{36}} \\ &= \frac{6}{11}\left(1 - \left(\frac{25}{36}\right)^{n-1}\right). \end{aligned} \]
(15 points) Let \(X\) be an indicator random variable for your winning. That is, if you win, \(X=1\). If you lose, \(X=0\). Find the variance of \(X\), \(\mathrm{Var}(X)\).
Solution: Let \(X\) be an indicator random variable for your winning: \[ X = \begin{cases} 1, & \text{if you win (sum of 7 appears before 8)},\\ 0, & \text{if you lose (sum of 8 appears before 7)}. \end{cases} \] Then \(X\) follows a Bernoulli distribution with parameter \[ X \sim \mathrm{Bernoulli}(p),~p = P(X=1) = P(\text{win}) = \frac{6}{11}. \] For a Bernoulli random variable \(X \sim \text{Bernoulli}(p)\), \(\mathrm{Var}(X) = p(1-p)\). Hence, \(\mathrm{Var}(X) = \frac{6}{11}\left(1 - \frac{6}{11}\right) = \frac{6}{11} \cdot \frac{5}{11} = \frac{30}{121}\).
(25 points) An elevator opens on the ground floor and \(N\) passengers enter it. Each passenger selects one floor (above the ground floor), and the elevator proceeds upward, stopping at each selected floor. Assume the following :
There are 10 floors above the ground floor.
Each passenger selects one floor above the ground floor.
Each passenger is equally likely to choose any of the 10 floors.
All passenger choices are independent.
If no passenger enter, then the elevator makes zero stops.
Find the expected number of stops that the elevator makes (not counting the ground floor).
Solution: First, define the random variables as follows:
Let \(X\) denote the number of stops the elevator makes (not counting the ground floor).
Let \(W_1, W_2, \ldots, W_{10}\)
be indicator random variables such that \[
W_i = \begin{cases}
1, & \text{if the elevator stops at the $i$th floor above the
ground,}\\
0, & \text{otherwise.}
\end{cases}
\]
Our goal is to find \(\mathrm{E}[X]\).
Since the total number of stops is the sum of the individual floor
indicators, \(X = \sum_{i=1}^{10}
W_i\). By linearity of expectation, \(\mathrm{E}[X] = \sum_{i=1}^{10}
\mathrm{E}[W_i]\).
For any given floor (say, the first floor above ground), \[ \begin{aligned} \mathrm{E}[W_1] &= P(W_1 = 1) = 1 - P(W_1 = 0) \\ &= 1 - P(\text{no passenger chooses floor 1}) \\ &= 1 - (0.9)^N. \end{aligned} \] By symmetry, all floors have the same expectation: \[ \mathrm{E}[W_1] = \mathrm{E}[W_2] = \cdots = \mathrm{E}[W_{10}] = 1 - (0.9)^N. \] Therefore, \(\mathrm{E}[X] = \sum_{i=1}^{10} \mathrm{E}[W_i] = 10\bigl(1 - (0.9)^N\bigr)\).