Page 34 - Schaum's Outlines - Probability, Random Variables And Random Processes

CHAP. 1) PROBABILITY

(a) Let A denote the event that the first one selected is defective. Then, by Eq. (1.38),
P(A) = = 0.2
(b) Let B denote the event that the second one selected is defective. After the first one selected is defective,
there are 99 chips left in the lot with 19 chips that are defective. Thus, the probability that the second
one selected is defective given that the first one was defective is

(c) By Eq. (l.41), the probability that both are defective is

1.43. A number is selected at random from (1, 2, . . . , 100). Given that the number selected is divisible
by 2, find the probability that it is divisible by 3 or 5.
Let A, = event that the number is divisible by 2
A, = event that the number is divisible by 3
A, = event that the number is divisible by 5
Then the desired probability is

- P(A3 n A,) + P(A, n A,) - P(A3 n As n A,) CE~. (1.29)1
-
P(A 2)
Now A, n A, = event that the number is divisible by 6
A, n A, = event that the number is divisible by 10
A, n A, n A, = event that the number is divisible by 30
and P(A, n A,) = AS n A21 = 7% P(A, n As n A,) = &,
- - 23
ZO + Ah -hi
Thus, P(A3 u As I A21 = - - 0.46
5 0
loo 50
1.44. Let A,,A ,,..., A,beeventsinasamplespaceS. Show that
1
P(A1 n A, n . n A,) = P(A,)P(A, A,)P(A, I A, n A,) . P(A, ( A, n A, n . . n A,- ,)
(1.81)

We prove Eq. (1.81) by induction. Suppose Eq. (1.81) is true for n = k:
I
P(Al n A, n . . n A,) = P(Al)P(A2 A,)P(A, I A, n A:,) . -. P(A, I A, n A, n - - n A,- ,)
Multiplying both sides by P(A,+, I A, n A, n . . . n A,), we have
P(Al n A, n --- n A,)P(A,+,IA, n A, n n A,) = P(Al n A, n -.. n A,,,)
1
and P(A, n A, n -. . n A,, ,) = P(A,)P(A, 1 A,)P(A3 A, rl A,) - -. P(A,+, 1 A, n A, n -. . n A,)
Thus, Eq. (1.81) is also true for n = k + 1. By Eq. (1 Al), Eq. (1.81) is true for n = 2. Thus Eq. (1.81) is true
for n 2 2.

1.45. Two cards are drawn at random from a deck. Find the probability that both are aces.
Let A be the event that the first card is an ace, and B be the event that the second card is an ace. The
desired probability is P(B n A). Since a card is drawn at random, P(A) = A. Now if the first card is an ace,
then there will be 3 aces left in the deck of 51 cards. Thus P(B I A) = A. By Eq. (1 .dl),