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Practice Problems
Contents
1 Discrete Random Variables, PMFs, and CDFs 1 1.1 Problems...................................... 1 1.2 Solutions...................................... 2
2 Common Discrete Random Variables 3 2.1 Problems...................................... 3 2.2 Solutions...................................... 4
3 Expectation 6 3.1 Problems...................................... 6 3.2 Solutions...................................... 6
1 Discrete Random Variables, PMFs, and CDFs
1.1 Problems
Problem 1. Let the random variable X take values { 1 , 2 , 3 , 4 , 5 }, with the following prob- ability mass function with non-zero values:
x 1 2 3 4 5 P(X = x) (^171411432727)
- Compute P(X ≤ 3).
- Compute P(X < 3).
- Compute P(X < 4. 12 |X > 1 .638).
Problem 2. Suppose a discrete random variable X takes values { 1 , 2 , 3 , 4 } and has a prob- ability mass function of the form:
pX (x) = cx for x = 1, 2 , 3 , 4 ,
for some constant c.
- Find c.
- Find the probability that X is odd.
1.2 Solutions
Solution to Problem 1. 1. We calculate the probability that X takes a value less than or equal to 3 by summing the probabilities of X = 1, X = 2, and X = 3:
P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)
P(X ≤ 3) =
- To calculate P(X < 3), we sum the probabilities of X = 1 and X = 2:
P(X < 3) = P(X = 1) + P(X = 2)
P(X < 3) =
- The conditional probability P(X < 4. 12 | X > 1 .638) can be simplified. Since X ∈ { 1 , 2 , 3 , 4 , 5 }, the event X > 1 .638 is equivalent to X ∈ { 2 , 3 , 4 , 5 } and the event X < 4 .12 is equivalent to X ∈ { 2 , 3 , 4 }. Hence, we calculate:
P(X < 4. 12 | X > 1 .638) =
P(X ∈ { 2 , 3 , 4 })
P(X ∈ { 2 , 3 , 4 , 5 })
First, we calculate the numerator:
P(X ∈ { 2 , 3 , 4 }) = P(X = 2) + P(X = 3) + P(X = 4)
P(X ∈ { 2 , 3 , 4 }) =
Next, the denominator:
P(X ∈ { 2 , 3 , 4 , 5 }) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
P(X ∈ { 2 , 3 , 4 , 5 }) =
Finally, the conditional probability is:
P(X < 4. 12 | X > 1 .638) =
4 7 6 7
Problem 6. Tom and Sally play a game of basketball. The game consists of them alter- natively throwing a ball, aiming for it to go through the hoop. The game starts with Sally taking the first shot. In a given turn, Tom gets the ball through the hoop with probability p and Sally with probability q. The game ends as soon as any player scores. Let T be the duration of the game (number of times the basketball is thrown). What is the distribution of T? If it is a named distribution, identify its parameters. Otherwise, specify its PMF.
Problem 7. Tom and Sally play a sequence of games, where Sally has a probability p of winning. The game ends when one person has won two games more than the other. What is the distribution of T , the duration of the game? If this is a named distribution, identify the distribution and its parameters. Otherwise, specify its probability mass function.
2.2 Solutions
Solution to Problem 3. 1. The number of women employees with a bonus is Bin(30, 0 .1). To see this, label the 30 women employees with IDs 1, 2 ,... , 30 and let:
Xi =
1 Woman i receives a bonus, 0 Woman i does not receive a bonus.
i = 1, 2 ,... , 30.
The number of women employees that receive a bonus is X 1 + X 2 + · · · + X 30 and X1: are independent Bern(0.1) random variables. Hence the number of women employees who receive a bonus is Bin(30, 0 .1).
- The number of women employees with a bonus is HG(100, 30 , 10). This is because we have a population of N = 100 employees out of which k = 30 employees have a special feature (they are women). We draw a sample of n = 10 employees to award the bonus, and we are interested in the number of female employees in our drawn sample. This matches with the definition of HG(100, 30 , 10) distribution.
Solution to Problem 4. X ∼ HG(ℓ, m, q). Here is an explanation: in this problem we have a population of ℓ dolphins, out of which m dolphins have the special feature that they were tagged in the first phase. In the second phase, the marine biologist draws a sample of q dolphins and counts the number of dolphins with the special feature (that they were tagged in the first phase). This matches the definition of HG(ℓ, m, q) distribution.
Solution to Problem 5. Label the n spelling errors are 1, 2 ,... , n. Define:
Xi =
1 Tom and Sally both miss error i, 0 Either Tom or Sally catch error i.
i = 1, 2 ,... , n.
Notice that the number of uncaught errors can be written as X 1 +X 2 +· · ·+Xn. Furthermore:
- The random variables X1:n are Bernoulli (they can either be 0 or 1).
- They are independent (since Tom and Sally catch/miss each error independently of the other errors).
- Let Ti be the event that Tom misses error i and Si be event that Sally misses error i. Notice that P(Xi = 1) = P(Si ∩ Ti). Since Si, Ti are independent, P(Xi = 1) = P(Si)P(Ti) = (1 − p 1 )(1 − p 2 ). Hence, Xi ∼ Bern(1 − p 1 − p 2 + p 1 p 2 ).
By the definition of the Binomial distribution, the number of uncaught errors follows Bin(n, 1 − p 1 − p 2 + p 1 p 2 ).
Solution for Problem 6. The duration of the game can be any natural number. For k ∈ N, we need to compute P(T = k).
Case 1: k is odd. This means that Sally scores first. In the first k − 1 throws, Tom threw the ball (k − 1)/2 times and failed to score and Sally threw the ball (k − 1)/2 times and also failed to score. On the kth throw, Sally succeeds in scoring. Hence,
P(T = k) = (1 − p)
k− 1 (^2) · (1 − q) k− 1 (^2) · q if k = 1, 3 , 5 ,....
Case 2: k is odd. This means that Tom scores first. In the first k − 1 throws, Tom threw the ball k/ 2 − 1 times and failed to score and Sally threw the ball k/2 times and also failed to score. On the kth throw, Tom succeeds in scoring. Hence,
P(T = k) = (1 − p)
k 2 −^1 · (1 − q) k (^2) · p if k = 2, 4 , 6 ,....
To summarize,
P(T = k) =
(1 − p)(k−1)/^2 (1 − q)(k−1)/^2 q k = 1, 3 , 5 , · · · (1 − q)k/^2 (1 − p)k/^2 −^1 p k = 2, 4 , 6 , · · · 0 otherwise.
Notice that the distribution of T is not Geometric.
Solution for Problem 7. First, observe that the game cannot end after an odd number of trials. This is because when the game ends, the winner has won exactly two more games than the loser. If the loser has won t games, then the winner has won t + 2 games, which means the total number of games is 2t + 2, which is even.
Hence, we only need to compute P(T = 2k) where k ∈ N. To compute this, we consider the games in pairs, game 1 and game 2 are pair 1, game 3 and game 4 are pair 2, and so on.
Define the random variables Y 1 , Y 2 , Y 3 ,... as follows: for any i ∈ N, Yi is 0 if pair i ends in a draw (that is Sally wins one game in the pair and Tom wins the other) and 1 if a single player wins both the games in the pair. Observe that:
P(T = 2k) = P(Y 1 = 0, Y 2 = 0,... , Yk− 1 = 0, Yk = 1) = P(Y 1 = 0) × P(Y 2 = 0) × · · · P(Yk = 1).
Observe that P(Yi = 1) = p^2 + (1 − p)^2 (either Tom wins both the Games or Sally wins both the games). And P(Yi = 0) = 2p(1 − p) (Sally wins the first game, Tom wins the second, or vice-versa). Hence, the non-zero values of the PMF are:
P(T = 2k) = (p^2 + (1 − p)^2 ) · (2p(1 − p))k−^1 ∀ k ∈ N.
Notice that the distribution of T is not Geometric.
We are interested in the expected value of the total number of guests who will receive prizes, which is X 1 + X 2 + · · · + Xn. By linearity,
E
X^ n
i=
Xi
X^ n
i=
E(Xi).
Since Xi’s are Bernoulli, E[Xi] = P(Xi = 1). The probability that guest i does not share a birthday with anyone else is the probability that all other n − 1 guests have birthdays different from guest i’s birthday. For any specific birthday of guest i, the probability that each of the remaining n − 1 guests does not have that same birthday is 364365. Thus, we have:
P(Xi = 0) =
n− 1 .
Therefore, the probability that guest i shares a birthday with at least one other guest is:
P(Xi = 1) = 1 −
n− 1 .
Hence, the expected total number of prizes is
X^ n
i=
E(Xi) = n 1 −
n− 1! .
Solution of Problem 10. Let X be the total number of occurrences of the pattern tht in the sequence of n tosses. Define indicator variables Xi for i = 1, 2 ,... , n − 2 as follows:
Xi =
1 if an occurrence of the pattern tht appears starting at position i, 0 otherwise.
Then, X =
Pn− 2 i=1 Xi^ is the total number of times^ tht^ appears. By the linearity of expecta- tion, we have:
E(X) = E
X^ n−^2
i=
Xi
X^ n−^2
i=
E(Xi).
Since Xi’s are Bernoulli, E[Xi] = P(Xi = 1). Notice that Xi is 1 if we observe the pattern tht occurs starting at toss i. That is,
- The i-th toss is tails.
- The (i + 1)-th toss is heads.
- The (i + 2)-th toss is tails.
The probability of observing tht in positions i, i + 1, i + 2 is therefore:
P(Xi = 1) = (1 − p) · p · (1 − p) = p(1 − p)^2.
Thus, E(Xi) = P(Xi = 1) = p(1 − p)^2.
Summing over all n − 2 possible starting positions for the pattern, we get:
E(X) =
X^ n−^2
i=
E(Xi) = (n − 2) · p(1 − p)^2.
Hence, the expected number of times the pattern tht appears in n tosses is:
E(X) = (n − 2) · p(1 − p)^2.
