One Person of Seven Born on MondayDate: 8/30/96 at 15:41:12 From: Robert Ferra Subject: One Person of Seven 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, PhD. Date: 8/30/96 at 19:9:32 From: Doctor Tom Subject: Re: One Person of Seven Born on Monday Hi Bob, Just use the standard binomial distribution, or reason as follows: first let's calcuate the probability that person 1 was born on Monday and persons 2 through 7 were not. That's: (1/7)*(6/7)*(6/7)*(6/7)*(6/7)*(6/7)*(6/7) = .056652779... Since the one person could have been any of the seven, the result you want is seven times that = .3965694566... The binomial distribution says that the probability of getting m hits out of n trials where the probability of a hit is p is given by: (n choose m)*p^m*(1-p)^(n-m) For your problem, p = 1/7, m = 1, n = 7, (7 choose 1) = 7, giving the same result. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 9/1/96 at 12:37:31 From: Robert Ferra Subject: Re: One Person of Seven Born on Monday Hi Doctor Tom: Many, many thanks for helping me out! I was unsure about multiplying the individual probability that each of each person was born on a Monday by the number of persons in the sample (7 in this case). I kind of thought it was just (.14)*(6/7)^6, but you're right, that would be for just one of the seven! Since you did the problem two different ways, I am very confident now in your answer. After I received your solution, the thought of using trees for this problem occurred to me (which I guess is a graphical picture of Binomial trials). I did a tree of Monday-Not Monday up to a population of four persons and I found four paths of only one Monday- each with a probability along each of the four paths of (1/7)*(6/7)(6/7)*6/7). Since the graphical tree gets too hairy after six or so persons, by induction (I guess) of the four paths tree, we arrive at your solution as well. Also, I mildly wondered where the author of the problem got a probability of being born on a Monday being equal to 0.14? As soon as I read your solution, noticing that you used 1/7 in place of 0.14. That 1/7 is the reciprocal of 0.14! So, 1 out of seven is 0.14 which seems to be simply that one is equally likely to be born on any of seven days of the week! Thanks too for pointing that out as well. Thanks again, Bob Date: 9/3/96 at 17:51:48 From: Doctor Tom Subject: Re: One Person of Seven Born on Monday Exactly. If you've ever looked at Pascal's triangle, the numbers in it just count the number of paths from the apex to the particular spot. For example: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 (To get any number in Pascal's triangle, add the two numbers above it.) So if you think of starting at the apex, and going either right or left at each number as you go down, the numbers in the triangle represent the number of paths to reach that number. For example, there are 1s all the way down the right, since the only path is the one that makes all right turns. The '2' in the third row can be reached by going left, then right, or right, then left. Consider the first '10' in the bottom row that I showed above. How many ways are there to get there? Well, you could take any of the paths to get to the '4' above it (4 ways), and then go right, or you can take any of the paths to the '6' (6 of them), and then go left. You can think of each path as a sequence of 'L' and 'R', so, for example, to get to the leftmost '3' in the fourth row, you had to make one step to the right and 2 to the left. The combinations that'll do this for you are RLL, LRL, LLR. To reach the '6' in the next row, there are 2 moves to the right and 2 to the left. Here are the possibilities: LLRR, LRLR, LRRL, RLLR, RLRL, RRLL. The other way to look at this is that it counts the number of ways to put the 2 Rs in the 4 available slots. For problems like yours, the probability of a birthday on monday (moving left) is 1/7 and the probability of moving right is 6/7. You can use Pascal's triangle to work out all the other possibilities as well. For example, suppose you want to know the probability of exactly 3 Monday birthdays with 5 people. That corresponds to the '10' in the bottom row above, so the probability will be 10 times (1/7)^3 times (6/7)^2. There's a world of information in Pascal's triangle if you look at it right. I assume the probability of being born on Monday is 1/7, but now with doctors not wanting to work on weekends and having the ability to induce labor, I'd be willing to bet the probabilities for all 7 days are not equal any more :^) -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 9/4/96 at 18:34:10 From: Robert Ferra Subject: Re: One Person of Seven Born on Monday Dear Dr. Tom: Thanks for the insight into the practical meaning of Pascal's Triangle. This has been very enlightening for me; the 'Triangle' has never really been explained to me in that context. Also, your comment is very perceptive about the probability of being born on a certain day of the week may now be skewed by the five day work week of physicians in America. But the equi-probability probably might still hold in third world nations... sounds like there might be a dissertation topic in there somewhere, or at least a Master's project! Thanks again, it's always nice to learn something new, don't you think! Bob Ferra |
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