Toss a Coin Six Times

Date: 02/07/98 at 16:59:43
From: Ruth Beldon
Subject: Coin tossing probabilities

A. Suppose a coin is tossed 6 times. What is the probability that 6  
    heads will occur? (Answer: 1/64)

B. What is the probablity that 3 heads will occur? (Book answer: 5/16)

   6/3 x 1/2 to 3rd power x 1/2 to 3rd power = 20x1/8x1/8 = 5/16

C.  X = 2   6/2x 1/2 squared x 1/2 to 4th = 15x1/4x1/16 = 15/64

My question is: where did the 20 come from in part B and the 15 in 
part C?   How was this answer arrived at?

Thank you,
R. Beldon

Date: 02/07/98 at 18:29:05
From: Doctor Mitteldorf
Subject: Re: Coin tossing probabilities

Dear Ruth,

The way you calculate probabilities for n coin tosses is to count the
different ways (different combinations) that the event you're looking 
at could happen.  

Say there are 6 tosses. The first toss can be either heads or tails.
The second can be either heads or tails.  2*2 = 4.  The third can be 
either heads or tails... so you end up with 2^6 = 64 possibilities. 
 
Only one of these has all heads. But there are more ways that you 
could get 3 heads. It could be the first, second, and third, or the 
first, second and fourth that are heads. Or maybe the first, second 
and fifth.

Here's a complete list:  

   123,124,125,126,134,135,136,145,146,156,234,235,
   236,245,246,256,345,346,356,456.  

That's 20 possibilities out of 64, or 20/64 = 5/16.

The answer is related to Pascal's triangle. The 6th row is 

   1 6 15 20 15 6 1

The numbers add up to 64, and the middle one is 20. There is a formula 
for these numbers, which your book is referring to:

  rth number in nth row of Pascal Triangle (counting from zero):
                  
               n! 
        ---------------
           (n-r)! r!

In your case, 6*5*4*3*2*1 in the numerator, 3*2*1 and 3*2*1 again in 
the denominator.  

-Doctor Mitteldorf,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/