A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is nm, gcd (m,n)=1, then m + n is equal to :
Solutions · 1
Official solution
30 points can be reached by: 6 tails; 4 tails + 1 head; 2 tails + 2 heads; 3 heads.
\[P = \left(\frac12\right)^6+\frac{5!}{1!4!}\left(\frac12\right)^5+\frac{4!}{2!2!}\left(\frac12\right)^4+\left(\frac12\right)^3\]
\[P=\frac{1}{64}+\frac{5}{32}+\frac{3}{8}+\frac{1}{8}=\frac{1+10+24+8}{64}=\frac{43}{64}\]
So m+n=43+64=107.