A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each

Question

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 $\frac{m}{n},gcd(m,n) = 1$, then $m + n$ is equal to :

Accepted Answer

Sol. [Key Concept: List head/tail counts giving exactly 30 points; add the binomial probabilities.]$\ P = (6$ tail $) + (4$ tail +1 head $) + (2$ tail +2 head $) + (3$ head $)$$${P = \left( \frac{1}{2} \right)^{6} + \frac{5!}{1!4!}\left( \frac{1}{2} \right)^{5} + \frac{4!}{2!2!}\left( \frac{1}{2} \right)^{4} + \left( \frac{1}{2} \right)^{3} }{P = \frac{1}{64} + \frac{5}{32} + \frac{3}{8} + \frac{1}{8} }{P = \frac{1 + 10 + 24 + 8}{64} = \frac{43}{64} = \frac{m}{n} }{m + n = 107 }$$

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