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1.45. Find the probability that n people (n
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365) selected at random will have n different birthdays.
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We assume that there are only 365 days in a year and that all birthdays are equally probable, assumptions which are not quite met in reality. The first of the n people has of course some birthday with probability 365 > 365 1. Then, if the second is to have a different birthday, it must occur on one of the other 364 days. Therefore, the probability that the second person has a birthday different from the first is 364 > 365. Similarly the probability that the third person has a birthday different from the first two is 363 > 365. Finally, the probability that the nth person has a birthday different from the others is (365 n l) > 365. We therefore have P(all n birthdays are different) 365 364 363 c 365 n 365 365 365 365 a1 1 b a1 365 2 b c a1 365 1 n 1 b 365
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1.46. Determine how many people are required in Problem 1.45 to make the probability of distinct birthdays less than 1 > 2.
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Denoting the given probability by p and taking natural logarithms, we find (1) ln p ln a1 1 b 365 ln a1 2 b 365 c ln a1 n 1 b 365
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But we know from calculus (Appendix A, formula 7) that (2) ln (1 x) x x2 2 x3 3 c
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CHAPTER 1 Basic Probability
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so that (1) can be written (3) ln p c 1 2 c 365 (n 1) d 1 12 c 2 22 c (365)2 (n 1)2 d c
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Using the facts that for n (4) 1 2 c
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2, 3, . . . (Appendix A, formulas 1 and 2) (n 1) n(n 2 1) , 12 22 c (n 1)2 n(n 1)(2n 6 1)
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we obtain for (3) (5) ln p n(n 1) 730 n(n 1)(2n 12(365)2 1) c
For n small compared to 365, say, n 30, the second and higher terms on the right of (5) are negligible compared to the first term, so that a good approximation in this case is (6) In p n(n 1) 730
[&!ln!p *frac*{n(n-1)}{730}&]
For p
1 2,
ln p
ln 2
0.693. Therefore, we have
n(n 1) 730
23)(n
so that n 23. Our conclusion therefore is that, if n is larger than 23, we can give better than even odds that at least 2 people will have the same birthday.
SUPPLEMENTARY PROBLEMS
Calculation of probabilities
1.47. Determine the probability p, or an estimate of it, for each of the following events: (a) A king, ace, jack of clubs, or queen of diamonds appears in drawing a single card from a well-shuffled ordinary deck of cards. (b) The sum 8 appears in a single toss of a pair of fair dice. (c) A nondefective bolt will be found next if out of 600 bolts already examined, 12 were defective. (d ) A 7 or 11 comes up in a single toss of a pair of fair dice. (e) At least 1 head appears in 3 tosses of a fair coin. 1.48. An experiment consists of drawing 3 cards in succession from a well-shuffled ordinary deck of cards. Let A1 be the event king on first draw, A2 the event king on second draw, and A3 the event king on third draw. State in words the meaning of each of the following: (a) P(A1 > Ar2 ), (b) P(A1 < A2), (c) P(Ar1 < Ar2 ), (d) P(Ar1 > Ar2 > Ar3), (e) P[(A1 > A2) < (Ar2 > A3)].
1.49. A marble is drawn at random from a box containing 10 red, 30 white, 20 blue, and 15 orange marbles. Find the probability that it is (a) orange or red, (b) not red or blue, (c) not blue, (d) white, (e) red, white, or blue. 1.50. Two marbles are drawn in succession from the box of Problem 1.49, replacement being made after each drawing. Find the probability that (a) both are white, (b) the first is red and the second is white, (c) neither is orange, (d) they are either red or white or both (red and white), (e) the second is not blue, (f) the first is orange, (g) at least one is blue, (h) at most one is red, (i) the first is white but the second is not, ( j) only one is red.
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