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(C) Let $h$ be the number of heads and $t$ be the number of tails. The total points are given by $10h + 5t = 30$, which simplifies to $2h + t = 6$.Sin

Vedclass · Accepted Answer

$107$

Vedclass · Answer

(C) Let $h$ be the number of heads and $t$ be the number of tails. The total points are given by $10h + 5t = 30$, which simplifies to $2h + t = 6$.Since $h$ and $t$ must be non-negative integers, the possible pairs $(h, t)$ are $(0, 6), (1, 4), (2, 2),$ and $(3, 0)$.The total number of tosses $N = h + t$ varies for each case: $6, 5, 4,$ and $3$ respectively.The probability of getting $h$ heads and $t$ tails in $N$ tosses is given by $\binom{N}{h} (\frac{1}{2})^N$.For $(h, t) = (0, 6)$, $N=6$, $P_1 = \binom{6}{0} (\frac{1}{2})^6 = \frac{1}{64}$.For $(h, t) = (1, 4)$, $N=5$, $P_2 = \binom{5}{1} (\frac{1}{2})^5 = \frac{5}{32} = \frac{10}{64}$.For $(h, t) = (2, 2)$, $N=4$, $P_3 = \binom{4}{2} (\frac{1}{2})^4 = \frac{6}{16} = \frac{24}{64}$.For $(h, t) = (3, 0)$, $N=3$, $P_4 = \binom{3}{3} (\frac{1}{2})^3 = \frac{1}{8} = \frac{8}{64}$.The total probability is $P = P_1 + P_2 + P_3 + P_4 = \frac{1 + 10 + 24 + 8}{64} = \frac{43}{64}$.Thus, $m = 43$ and $n = 64$. Since $	ext{gcd}(43, 64) = 1$, $m + n = 43 + 64 = 107$.

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

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