One Person of Seven Born on Monday

Date: 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