SOLUTION: Use the pattern in Pascal's triangle to find the probability of getting 2 heads when four fair coins are tossed. (type an integer or a simplified fraction)

Algebra ->  Probability-and-statistics -> SOLUTION: Use the pattern in Pascal's triangle to find the probability of getting 2 heads when four fair coins are tossed. (type an integer or a simplified fraction)      Log On


   

Question 1026474: Use the pattern in Pascal's triangle to find the probability of getting 2 heads when four fair coins are tossed. (type an integer or a simplified fraction)
Found 2 solutions by ikleyn, Timnewman:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
Use the pattern in Pascal's triangle to find the probability of getting 2 heads
when four fair coins are tossed. (type an integer or a simplified fraction)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

It is C%5B4%5D%5E2%2A%281%2F2%29%5E4 = 6%2A%281%2F16%29 = 3%2F8.

Here  C%5B4%5D%5E2  is  the binomial coefficient equal to %284%2A3%29%2F%281%2A2%29 = 6.


Answer by Timnewman(323) About Me  (Show Source):
You can put this solution on YOUR website!
lets follow your question up by expansion as suggested by pascal using pascals triangle.
In a coin,we have tail (T) and head (H).
Therefore a coin =(H+T)
since there are four coin here,we raise (H+T) to power of 4
That is,(H+T)^4
Expand the above,we have
(H+T)^4=H^4+4H^3T+6H²T²
+HT^3+H^4
Two head lies on 6H²T²
but p(H)=1/2 also, p(T)=1/2
Therefore,
p(two head)=p(6H²T²)
=6(1/2)²(1/2)²
=6(1/4)(1/4)
=6(1/16)
=6/16
=3/8
Hope this helps too?