Probability of Being Born on Monday
Date: 8/26/96 at 12:1:19 From: Robert Ferra Subject: Probability of being born on Monday Dear Dr. Math: Could I please get your take on the following problem: To the nearest percent, the probability that any one person selected at random was born on a Monday is 14 percent. What is the probability, to the nearest percent, that of any seven persons chosen at random, exactly one was born on a Monday? Thanks so much for any help, Bob Ferra, Ph.D.
Date: 10/24/96 at 11:17:52 From: Doctor Jaime Subject: Re: Probability of being born on Monday Let's begin solving a simpler problem. Let's suppose you just want the first person to be born on a Monday and the other six to be born on any other day. Then the probability of this happening is equal to the probability of being born on a Monday times the probability of *not* being born on a Monday (six times for the other six persons): 0.14 x (1-0.14)^6. But if you want any one of the seven persons (not just the first one) to be born on a Monday, then you want also the second, or third or ... person to be born on a Monday (and the others on any other day). So you have a total of seven possibilities each with the same probability just calculated. So the total probability is 7 x 0.14 x (1-0.14)^6 = 0.39647... and the answer is 40 percent. -Doctor Jaime, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 10/24/96 at 12:18:49 From: Doctor Bill Subject: Re: Probability of being born on Monday Dear Bob, Here's another way to think about it. Think of the problem as rolling 7 dice (one for each person), each die having 7 sides (one for each day of the week). Thus, we have a binomial probability, i.e. getting exactly one "Monday" in 7 rolls of the dice. The probability of getting a Monday is 1/7 and the probability of getting 6 "non-Mondays" is (6/7)^6. The number of ways this can happen is a combination of "7 picking 1," i.e. 7C1=7. Therefore, P(exactly 1 Monday out of 7 people)=7C1 * (1/7)^1 *(6/7)^6 = .3966, or 40 percent to the nearest percent. -Doctor Bill, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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