Here are all the questions and their solutions from the Week_10_Graded_Solution.pdf1:
1. Expectation of the Sum of Numbers on Cards
Question:
There are $2^n$ numbered cards in a deck, among which ${}^n C_i$ cards bear the number $i$ ($i = 0,1,2,\ldots,n$). From the deck, $m$ cards are drawn with replacement. What is the expectation of the sum of their numbers? (Enter the answer correct to one decimal accuracy.)
Answer:
$$
\frac{m n}{2}
$$
Solution:
Let $X_j$ be the random variable representing the number on the $j^\text{th}$ card drawn.
The probability that $X_j = x$ is $\frac{{}^n C_x}{2^n}$.
The expectation for one draw is:
2. Expected Value of $Y$ (Random Walk and Sign Patterns)
Question:
An unbiased die is thrown $n+2$ times. After each throw, a ‘+’ is recorded for 2 or 5 and ‘-’ is recorded for 1, 3, 4, or 6, forming an ordered sequence. To each except the first and last sign, a random variable $X_i$ ($i=1,2,\ldots,n$) is associated, which takes the value 1 if both neighboring signs differ from the one between them, and 0 otherwise. If $Y = a S + b$ where $S = \sum_{i=1}^n X_i$, find the expected value of $Y$.
Options:
a. $\left(a \times \frac{2 n}{9}\right)+b$
b. $\left(a \times \frac{2 n}{9}\right)$
c. $\frac{2 n}{9}+b$
d. $\frac{2 n}{9}$
Answer:
a. $\left(a \times \frac{2 n}{9}\right)+b$
Solution:
The pattern for $X_i = 1$ is either ‘$+ - +$’ or ‘$- + -$’.
Probability for each:
$$
P(X_i = 1) = \left(\frac{4}{6}\right)^2 \cdot \frac{2}{6} + \left(\frac{4}{6}\right) \cdot \left(\frac{2}{6}\right)^2 = \frac{2}{9}
$$$$
E(S) = \sum_{i=1}^n E(X_i) = \frac{2n}{9}
$$$$
E(Y) = a E(S) + b = a \cdot \frac{2n}{9} + b
$$
3. True Statements about $Y$
Question:
Which of the following statement(s) is/are true about $Y$ as defined above?
a. $V(Y) = a^2 V(S) + b$
b. $V(Y) = a^2 V(S)$
c. $V(Y) \neq a^2 V(S)$
d. $E(Y) = a^2 E(S) + b$
e. $E(Y) = a E(S) + b$
Answer:
b, e
Solution:
By properties of expectation and variance:
$$
V(Y) = a^2 V(S)
$$$$
E(Y) = a E(S) + b
$$
4. Expected Distance After $n$ Steps (Random Walk)
Question:
Amandeep is in the middle of a bridge of infinite length. He takes a unit step to the right with probability $p$ and to the left with probability $1-p$. Assume movements are independent. What is the expected distance between the starting point and endpoint after $n$ steps?
Options:
a. $2p-1$
b. $n(2p-1)$
c. $1-2p$
d. $n(1-2p)$
Answer:
b. $n(2p-1)$
Solution:
Let $X_i = 1$ (right) with probability $p$, $-1$ (left) with probability $1-p$.
Question:
A box contains $a$ white and $b$ black balls. $c$ balls are drawn at random without replacement. Find the expected value of the number of white balls drawn. (Enter the answer correct to 2 decimal places.)
Solution:
The expected number of white balls drawn is:
$$
E(X) = c \cdot \frac{a}{a+b}
$$
For example, with $a=7$, $b=4$, $c=2$:
$$
E(X) = 2 \cdot \frac{7}{11} \approx 1.27
$$
7. Expected Number of Trials to Open a Door
Question:
Rohit wants to open his door with 5 keys (out of which 1 will open the door) and tries the keys independently and at random. If unsuccessful keys are eliminated, find the expected number of trials required to open the door.
Options:
a. $1$
b. $9$
c. $3$
d. $2$
Answer:
c. $3$
Solution:
The probability of success at the $k^\text{th}$ trial is $\frac{1}{5}$ for all $k$ (as each permutation is equally likely).
$$
E(X) = \frac{1+2+3+4+5}{5} = 3
$$
8. Variance of $aX + bY$
Question:
$X$ and $Y$ are independent random variables with means $m_1$ and $m_2$, and variances $v_1$ and $v_2$ respectively. Find the variance of $aX + bY$.
10. Values of $a$ and $b$ for Standardized Random Variable
Question:
Suppose $X$ is a random variable for which $E(X) = m$ and $\text{Var}(X) = v$. Find the positive values of $a$ and $b$ such that $Y = aX - b$ has expectation 0 and variance 1.
Options:
a. $\frac{-1}{\sqrt{v}}, \frac{-m}{\sqrt{v}}$
b. $\frac{1}{v}, \frac{m}{\sqrt{v}}$
c. $\frac{1}{\sqrt{v}}, \frac{m}{\sqrt{v}}$
d. $1, m$
Answer:
c. $\frac{1}{\sqrt{v}}, \frac{m}{\sqrt{v}}$
Solution:
$$
E(Y) = aE(X) - b = 0 \implies m a - b = 0 \implies b = m a
$$$$
V(Y) = a^2 V(X) = 1 \implies a^2 v = 1 \implies a = \frac{1}{\sqrt{v}}
$$$$
b = m \cdot \frac{1}{\sqrt{v}}
$$
This covers all questions and solutions from the Week_10_Graded_Solution.pdf1.
