MATH 511, Meade HW Solutions
2.3 – 3, 5, 9, 15, 10, 16 2/6/04
3.
a. P (A1 ∩ B2) = 5/35
b. P (A1 U B1) = 26/35
c. P (A1 | B1) = 5/19
d. P (B2 | A2) = 9/23
e. Left. Given that the person is left thumb on top, there is a 74% chance that
they be right eye dominant, but there is only a 56% chance given that they
are right thumb on top. P(A2 | B1) = 14/19 > P(A2 | B2) = 9/16
5. P(A) = 0.7, P(B)= 0.5, P[(A U B)’] = 0.1
a. P[(A U B)’] = 1 – P(A U B) = 1 – P(A) – P(B) + P(A ∩ B)
0.1 = 1 – 1.2 + P(A ∩ B)
P(A ∩ B) = 0.3
b. P(A | B) = P(A ∩ B) / P(B) = 0.3 / 0.5 = 3/5
c. P(B | A) = P(A ∩ B) / P(A) = 0.3 / 0.7 = 3/7
9. NOTE: Read the problem carefully. The question does NOT say that the first
ball is orange. The given event is that at least one orange ball is selected.
An urn contains four balls: two are orange and two are blue. Two balls are
selected at random w/o replacement. You are told that at least one of them is
orange. What is the probability that the other is also orange?
Combinatorics Approach:
Let A = {at least one ball is orange}
Let B = {both balls are orange}
The question is asking what is P(B | A)?
How many ways can any two balls be selected? 4C2 = 6
How many ways can at least one be selected? Either one blue and one orange
are selected or two oranges and zero blues are selected.
(2C1) (2C1) + (2C2) (2C0) = (2)(2) + (1)(1) = 4+1 = 5, so P(A) = 5/6
How many ways can exactly two orange balls be selected? This means to
select two orange and zero blue. (2C2) (2C0) = (1) (1) = 1, so P(B) = 1/6
P(B ∩ A) = 1/6, since B is a subset of A.
Finally, P(B | A) = P(B ∩ A)/ P(A) = (1/6) / (5/6) = 1/5
