Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

140 PROBABILITY [CHAP. 7

Fig. 7-6

7.14. Prove Theorem 7.4 (Addition Principle): For any eventsAandB,

P(A∪B)=P (A)+P(B)−P(A∩B)

As indicated by Fig. 7-6(c),A∪B=(A\B)∪BwhereA\BandBare disjoint sets. Thus, using Theorem 7.3(ii),

P(A∪B)=P(A\B)+P(B)=P (A)−P(A∩B)+P(B)

=P (A)+P(B)−P(A∩B)

CONDITIONAL PROBABILITY

7.15.A pair of fair dice is thrown. (See Fig. 7-1(b).) Find the probability that the sum is 10 or greater if:

(a) 5 appears on the first die; (b) 5 appears on at least one die.

(a) If a 5 appears on the first die, then the reduced sample space is

A={( 5 , 1 ), ( 5 , 2 ), ( 5 , 3 ), ( 5 , 4 ), ( 5 , 5 ), ( 5 , 6 )}

The sum is 10 or greater on two of the six outcomes: (5, 5), (5, 6). Hencep=^26 =^13.

(b) If a 5 appears on at least one of the dice, then the reduced sample space has eleven elements.

B={( 5 , 1 ), ( 5 , 2 ), ( 5 , 3 ), ( 5 , 4 ), ( 5 , 5 ), ( 5 , 6 ), ( 1 , 5 ), ( 2 , 5 ), ( 3 , 5 ), ( 4 , 5 ), ( 6 , 5 )}

The sum is 10 or greater on three of the eleven outcomes: (5, 5), (5, 6), (6, 5). Hencep= 113.

7.16. In a certain college town, 25% of the students failed mathematics(M), 15% failed chemistry(C), and

10% failed both mathematics and chemistry. A student is selected at random.

(a) If he failed chemistry, find the probability that he also failed mathematics.

(b) If he failed mathematics, find the probability that he also failed chemistry.

(c) Find the probability that he failed mathematics or chemistry.

(d) Find the probability that he failed neither mathematics nor chemistry.

(a) The probability that a student failed mathematics, given that he failed chemistry, is

P(M|C)=

P(M∩C)

P(C)

=

0. 10

0. 15

=

2

3

(b) The probability that a student failed chemistry, given that he failed mathematics is

P(C|M)=

P(C∩M)

P(M)

=

0. 10

0. 25

=

2

5

(c) By the Addition Principle (Theorem 7.4),

P(M∪C)=P(M)+P(C)−P(M∩C)= 0. 25 + 0. 15 − 0. 10 = 0. 30