This practice exam covers the material from chapters 1 through 8. Give yourself 120 minutes to solve the six problems, which you may assume have equal point score.
1. Recall that a full deck of cards contains 13 cards of each of the four suits (♣, ♦, ♥, ♠). Select cards from the deck at random, one by one, without replacement.
(a) Compute the probability that the first four cards selected are all hearts (♥).
(b) Compute the probability that all suits are represented among the first four cards selected.
(c) Compute the expected number of different suits among the first four cards selected.
(d) Compute the expected number of cards you have to select to get the first hearts card.
2. Eleven Scandinavians: 2 Swedes, 4 Norwegians, and 5 Finns are seated in a row of 11 chairs at random.
(a) Compute the probability that all groups sit together (i.e., the Swedes occupy adjacent seats, as do the Norwegians and Finns).
(b) Compute the probability that at least one of the groups sits together.
(c) Compute the probability that the two Swedes have exactly one person sitting between them.
3. You have two fair coins. Toss the first coin three times and let X be the number of Heads.
Then toss the second coin X times, that is, as many times as the number of Heads in the first coin toss. Let Y be the number of Heads in the second coin toss. (For example, if X = 0, Y is automatically 0; if X = 2, toss the second coin twice and count the number of Heads to get Y .) (a) Determine the joint probability mass function of X and Y , that is, write down a formula for P (X = i, Y = j) for all relevant i and j.
(b) Compute P (X ≥ 2 | Y = 1).
4. Assume that 2, 000, 000 single male tourists visit Las Vegas every year. Assume also that each of these tourists, independently, gets married while drunk with probability 1/1, 000, 000.
(a) Write down the exact probability that exactly 3 male tourists will get married while drunk next year.
(b) Compute the expected number of such drunk marriages in the next 10 years.
(c) Write down a relevant approximate expression for the probability in (a).
(d) Write down an approximate expression for the probability that there will be no such drunk marriage during at least one of the next 3 years.
5. Toss a fair coin twice. You win $2 if both tosses comes out Heads, lose $1 if no toss comes out Heads, and win or lose nothing otherwise.
(a) What is the expected number of games you need to play to win once?
(b) Assume that you play this game 500 times. What is, approximately, the probability that you win at least $135?
(c) Again, assume that you play this game 500 times. Compute (approximately) the amount of money x such that your winnings will be at least x with probability 0.5. Then, do the same with probability 0.9.
6. Two random variables X and Y are independent and have the same probability density function
g(x) =
(c(1 + x) x ∈ [0, 1],
0 otherwise.
(a) Find the value of c. Here and in (b): use R1
0 xndx = n+11 , for n > −1.
(b) Find Var(X + Y ).
(c) Find P (X + Y < 1) and P (X + Y ≤ 1). Here and in (d): when you get to a single integral involving powers, stop.
(d) Find E|X − Y |.
Solutions to Practice Final
1. Recall that a full deck of cards contains 13 cards of each of the four suits (♣, ♦, ♥, ♠).
Select cards from the deck at random, one by one, without replacement.
(a) Compute the probability that the first four cards selected are all hearts (♥).
Solution:
¡13
4
¢
¡52
4
¢
(b) Compute the probability that all suits are represented among the first four cards selected.
Solution:
134
¡52
4
¢
(c) Compute the expected number of different suits among the first four cards selected.
Solution:
If X is the number of suits represented, then X = I♥+ I♦+ I♣+ I♠, where I♥ = I{♥ is represented}, etc. Then,
EI♥ = 1 −
¡39
4
¢
¡52
4
¢ , which is the same for the other three indicators, so
EX = 4 EI♥= 4 Ã
1 −
¡39
4
¢
¡52
4
¢
! .
(d) Compute the expected number of cards you have to select to get the first hearts card.
Solution:
Label non-♥ cards 1, . . . , 39 and let Ii = I{card i selected before any ♥ card}. Then, EIi =
141 for any i. If N is the number of cards you have to select to get the first hearts card, then
EN = E (I1+ · · · + EI39) = 39 14.
2. Eleven Scandinavians: 2 Swedes, 4 Norwegians, and 5 Finns are seated in a row of 11 chairs at random.
(a) Compute the probability that all groups sit together (i.e., the Swedes occupy adjacent seats, as do the Norwegians and Finns).
Solution:
2! 4! 5! 3!
11!
(b) Compute the probability that at least one of the groups sits together.
Solution:
Define AS = {Swedes sit together} and, similarly, AN and AF. Then, P (AS∪ AN ∪ AF)
= P (AS) + P (AN) + P (AF)
− P (AS∩ AN) − P (AS∩ AF) − P (AN ∩ AF) + P (AS∩ AN ∩ AF)
= 2! 10! + 4! 8! + 5! 7! − 2! 4! 7! − 2! 5! 6! − 4! 5! 3! + 2! 4! 5! 3!
11!
(c) Compute the probability that the two Swedes have exactly one person sitting between them.
Solution:
The two Swedes may occupy chairs 1, 3; or 2, 4; or 3, 5; ...; or 9, 11. There are exactly 9 possibilities, so the answer is
¡119
2
¢ = 9 55.
3. You have two fair coins. Toss the first coin three times and let X be the number of Heads.
Then, toss the second coin X times, that is, as many times as you got Heads in the first coin toss. Let Y be the number of Heads in the second coin toss. (For example, if X = 0, Y is automatically 0; if X = 2, toss the second coin twice and count the number of Heads to get Y .)
(a) Determine the joint probability mass function of X and Y , that is, write down a formula for P (X = i, Y = j) for all relevant i and j.
Solution:
P (X = i, Y = j) = P (X = i)P (Y = j|X = i) = µ3
i
¶1 23 ·
µi j
¶1 2i, for 0 ≤ j ≤ i ≤ 3.
(b) Compute P (X ≥ 2 | Y = 1).
Solution:
This equals
P (X = 2, Y = 1) + P (X = 3, Y = 1)
P (X = 1, Y = 1) + P (X = 2, Y = 1) + P (X = 3, Y = 1) = 5 9.
4. Assume that 2, 000, 000 single male tourists visit Las Vegas every year. Assume also that each of these tourists independently gets married while drunk with probability 1/1, 000, 000.
(a) Write down the exact probability that exactly 3 male tourists will get married while drunk next year.
Solution:
With X equal to the number of such drunk marriages, X is Binomial(n, p) with p = 1/1, 000, 000 and n = 2, 000, 000, so we have
P (X = 3) = µn
3
¶
p3(1 − p)n−3
(b) Compute the expected number of such drunk marriages in the next 10 years.
Solution:
As X is binomial, its expected value is np = 2, so the answer is 10 EX = 20.
(c) Write down a relevant approximate expression for the probability in (a).
Solution:
We use that X is approximately Poisson with λ = 2, so the answer is λ3
3! e−λ = 4 3e−2.
(d) Write down an approximate expression for the probability that there will be no such drunk marriage during at least one of the next 3 years.
Solution:
This equals 1−P (at least one such marriage in each of the next 3 years), which equals 1 −¡
1 − e−2¢3 .
5. Toss a fair coin twice. You win $2 if both tosses comes out Heads, lose $1 if no toss comes out Heads, and win or lose nothing otherwise.
(a) What is the expected number of games you need to play to win once?
Solution:
The probability of winning in 14. The answer, the expectation of a Geometric¡1
4
¢ random variable, is 4.
(b) Assume that you play this game 500 times. What is, approximately, the probability that you win at least $135?
Solution:
where Z is standard Normal. Using n = 500, we get the answer
1 − Φ
(c) Again, assume that you play this game 500 times. Compute (approximately) the amount of money x such that your winnings will be at least x with probability 0.5.
Then, do the same with probability 0.9.
Solution:
For probability 0.5, the answer is exactly ESn= n · 14 = 125. For probability 0.9, we approximate
P (Sn≥ x) ≈ P
Z ≥ x − 125 q
500 ·1916
= P
Z ≤ 125 − x q
500 ·1916
= Φ
125 − xq 500 ·1916
,
where we have used that x < 125. Then, we use that Φ(z) = 0.9 at z ≈ 1.28, leading to the equation
125 − x q
500 ·1916
= 1.28
and, therefore,
x = 125 − 1.28 · r
500 ·19 16. (This gives x = 93.)
6. Two random variables X and Y are independent and have the same probability density function
g(x) =
(c(1 + x) x ∈ [0, 1],
0 otherwise.
(a) Find the value of c. Here and in (b): useR1
0 xndx = n+11 , for n > −1.
Solution:
As
1 = c · Z 1
0
(1 + x) dx = c ·3 2, we have c = 23
(b) Find Var(X + Y ).
Solution:
By independence, this equals 2Var(X) = 2(E(X2) − (EX)2). Moreover,
The two probabilities are both equal to µ2