PROBABILITY EXAM 1
(1) Suppose there are three horses, A, B, and C, in a race. A and B are equally likely to win, but C is twice as
likely to win as A. What is the probability that B or C wins the race?
ANS. Let p the probabiity that A wins. The probability that C wins is 2p. So p + p + 2p = 1 and p = .25.
Hence the probabiity that B or C wins is .25 + .5 or 3/4.
(2) Suppose you flip a fair coin until you obtain Heads. What is the probability that you flip the coin at least
three times?
ANS. Let E be the event that you flip the coin at least three times. Then Prob (E) = 1 − Prob(Ẽ). But
Prob(Ẽ) = .5 + .25 = .75. So Prob (E) = .25.
(3) For a bill to come before the president it must first pass the House and the Senate. Suppose that 60% of all
bills pass the House, 30% pass the Senate, and 70% pass at least one of the House or Senate. What percentage
of bills come before the president?
ANS. Let H be the event that the bill comes before the House, and let S be the event that the bill comes
before the Senate. We want Prob (H ∩ S). By inclusion-exclusion Prob (H ∩ S) = Prob (H) + Prob (S) −
Prob (H ∪ S) = .6 + .3 − .7 = .2.
(4) A coin lands Heads with probability p. Suppose you know that the probability of getting at least one Head in
two flips of the coin is 9/25? What is the value of p?
ANS. Let E be the event of getting at least one H in two flips. Then Prob Ẽ = (1 − p)2 = 1 − 9/25 = 16/25.
Hence 1 − p = 4/5 and p = 1/5 = .2.
(5) An urn contains 5 red and 10 black balls. Suppose you select two balls at random without replacement. What
is the probabiity that the two balls are of the same color?
ANS. Let E be the event that the two balls are of the same color. The Prob(E) = 1 − Prob(Ẽ) = 1 −
(5 ∗ 10) /C(15, 2) = 1 − 10/21 = 11/21.
(6) An urn contains 7 red and 12 balls. You select two balls at random without replacement. What is the
probability that the two balls are of the same color given that one of the balls is red?
ANS. Let E be the event that the two balls are of the same color. Let R be the event that one of the balls is red.
Let RR be the event that both balls are red, and let RB be the event that one is red and one is black. Then
Prob(E|R) = Prob(RR|R) = Prob(RR)/Prob(R). Prob(RR) = C(7, 2)/C(19, 2). Prob(R) = Prob(RR) +
Prob(RB) = C(7, 2)/C(19, 2) + (7 ∗ 12)/C(19, 2). Hence Prob(E|R) = C(7, 2)/[C(7, 2) + 7 ∗ 12] = 1/5.
(7) Suppose the pair of random variables (X, Y ) is uniformly distributed on the unit square, [0, 1] × [0, 1]. What
is the probability that X + Y is less that 1/2 given that X is greater than Y ?
ANS. Prob(X + Y < 1/2|X > Y ) = Prob(X + Y < 1/2 and X > Y )/Prob(X > Y ) . Prob(X > Y ) = 1/2 .
Prob(X + Y < 1/2 and X > Y ) is the area of the triangle with vertices at (0, 0), (1/2, 0), and (1/4, 1/4). The
area of this triangle is 1/16 so Prob(X + Y < 1/2|X > Y ) = (1/16)/(2/16) = 2/16.
(8) The random variable W takes values in [−1, 1], and W has density function f (x) proportional to 1 − |x|. What
is the probability that W > 1/3?
ANS. First find the proportionality constant by graphing the function 1 − |x| from −1 to 1. Since the area
under the graph is 1 ( two triangles), the proportionality constant is 1. So Prob(W > 1/3) is the area under
the graph of 1 − |x| for 1/3 ≤ x ≤ 1. This area is (2/3)(2/3)(1/2) = 2/9.
(9) Suppose 10% of a large population has a certain genetic trait. A test will detect the trait when it is present
80% of the time. The test results in a false positive 5% of the time (The test result is positive but the individual
does not have the trait.) If a randomly selected individual is tested for the trait, what is the probability that
the test result will be positive?
ANS. Let P be event that the test is positive, T the event that the trait is present, and N the event that the
trait is not present. Then Prob(P ) = Prob(P ∩T )+Prob(P ∩N ) = Prob(P |T )Prob(T )+Prob(P |N )Prob(N ) =
.8 ∗ .1 + .05 ∗ .9 = .125
(10) Suppose X and Y are jointly distributed and take values in the set {0, 1}. Suppose the probability that X = Y
is 3/10, the probability that X = 0 is 1/2, and the probability that Y = 0 given that X = 1 is 3/5? What is
the probability that Y = 0 and X = 1?
ANS. Prob(Y = 0 and X = 1) = Prob(Y = 0|X = 1)Prob(X = 1) = (3/5)(1/2) . Note Prob(X = 1) =
1 − Prob(X = 0) = 1 − (1/2).
1
(11) In the poker game “Texas Hold’em each player is dealt two cards from a standard 52 card deck. After a round
of betting, three cards are dealt face up on the table (called the “flop”). Suppose there are 8 players in the
game, and you are one of the players. Suppose exactly one of your cards is an Ace and you see exactly one
Ace in the “flop”. (You do not see the cards held by the other players.) What is the probabilty that at least
one other player holds an Ace? You must give a numerical expression for your answer, but you do not have
to evaluate that numerical expression. (For example an answer that was expressed in terms of factorials or
combinations is acceptable even if you do not express the value of the expression in decimal form.)
ANS.There are 52-5 or 47 cards other than the cards in your hand or the flop, and two of them are aces.
The other 7 players receive 14 of these cards. The probability tht none of them receives an Ace is therefore
C(45, 14)/C(47, 14). Therefore the probability that at least one of them has an ace is 1 − C(45, 14)/C(47, 14).
2