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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/   
    
Associated Topics:
High School Probability

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