Question 940217
Given:
flips (fair) coin 7 times.
rolls (fair) die 7 times.
Find 
Probability of getting 3 heads or 2 sixes or both.


Using binomial distribution, 
For coin, p=0.5, n=7, r=3, C(n,r)=n!/((n-r)!r!)
P(3H) = {{{C(7,3)*0.5^3*0.5^(7-3) = 0.27344 }}}
For die, p=1/6, n=7, r=2, 
P(2-"6")  {{{C(7,2)*(1/6)^2*(5/6)^5 = 0.23443 }}}
  
Since the two processes can be considered independent, so P(A & B)=P(A)*P(B),
Using
P(A or B) = P(A) + P(B) -P(A & B)
= 0.27344+0.23443-(0.27344*0.23443)
= 0.4438