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Required probability
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number of selections of 3 out of 8 red balls number of selections of 3 out of 20 balls
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8C3 20C3
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(b) Using the second method indicated in part (a), P(all 3 are white)
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The first method indicated in part (a) can also be used. (c) P(2 are red and 1 is white) (selections of 2 out of 8 red balls)(selections of 1 out of 3 white balls) number of selections of 3 out of 20 balls (8C2)(3C1) 7 95 20C3 (d) P(none is white)
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34 . Then 57 P(at least 1 is white) 1 34 57 23 57
(e) P(l of each color is drawn)
(8C1)(3C1)(9C1) 20C3
18 95
(f) P(balls drawn in order red, white, blue)
Another method
P(R1 > W2 > B3)
1 18 6 95
1 P(l of each color is drawn) 3! 3 , using (e) 95
P(R1) P(W2 u R1) P(B3 u R1 > W2) 8 3 9 20 19 18 3 95
CHAPTER 1 Basic Probability
1.36. In the game of poker 5 cards are drawn from a pack of 52 well-shuffled cards. Find the probability that (a) 4 are aces, (b) 4 are aces and 1 is a king, (c) 3 are tens and 2 are jacks, (d) a nine, ten, jack, queen, king are obtained in any order, (e) 3 are of any one suit and 2 are of another, (f) at least 1 ace is obtained.
(a) P(4 aces) (b) (c) (d) (e) 1 . 54,145 (4C4)(4C1) 1 . P(4 aces and 1 king) 649,740 52C5 (4C3)(4C2) 1 . P(3 are tens and 2 are jacks) 108,290 52C5 (4C1)(4C1)(4C1)(4C1)(4C1) 64 . P(nine, ten, jack, queen, king in any order) 162,435 52C5 (4 13C3)(3 13C2) 429 , P(3 of any one suit, 2 of another) 4165 52C5 since there are 4 ways of choosing the first suit and 3 ways of choosing the second suit. 35,673 18,472 35,673 48C5 . Then P(at least one ace) 1 . P(no ace) 54,145 54,145 54,145 52C5 (4C4)(48C1) 52C5
(f )
1.37. Determine the probability of three 6s in 5 tosses of a fair die.
Let the tosses of the die be represented by the 5 spaces . In each space we will have the events 6 or not 6 (6r). For example, three 6s and two not 6s can occur as 6 6 6r6 6r or 6 6r6 6r6, etc. Now the probability of the outcome 6 6 6r 6 6r is P(6 6 6r 6 6r) P(6) P(6) P(6r) P(6) P(6r) 1 5 6 6
1 1 5 1 5 6 6 6 6 6
since we assume independence. Similarly, P
5 1 6 6
for all other outcomes in which three 6s and two not 6s occur. But there are 5C3 are mutually exclusive. Hence, the required probability is P(6 6 6r6 6r or 6 6r6 6r6 or c)
5C3 6
10 such outcomes, and these
In general, if p P(A) and q 1 p P(Ar), then by using the same reasoning as given above, the probability of getting exactly x A s in n independent trials is
nCx p x qn x
1.38. A shelf has 6 mathematics books and 4 physics books. Find the probability that 3 particular mathematics books will be together.
All the books can be arranged among themselves in 10P10 10! ways. Let us assume that the 3 particular mathematics books actually are replaced by 1 book. Then we have a total of 8 books that can be arranged among themselves in 8P8 8! ways. But the 3 mathematics books themselves can be arranged in 3P3 3! ways. The required probability is thus given by 8! 3! 10! 1 15
n px qn x
5 6
5! 1 5 3!2! 6 6
125 3888
Miscellaneous problems 1.39. A and B play 12 games of chess of which 6 are won by A, 4 are won by B, and 2 end in a draw. They agree to play a tournament consisting of 3 games. Find the probability that (a) A wins all 3 games, (b) 2 games end in a draw, (c) A and B win alternately, (d ) B wins at least 1 game.
Let A1, A2, A3 denote the events A wins in 1st, 2nd, and 3rd games, respectively, B1, B2, B3 denote the events B wins in 1st, 2nd, and 3rd games, respectively. On the basis of their past performance (empirical probability),
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