A man tosses a coin 10 times,scoring 1 point | vedclass

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(B) Let $H$ be the number of heads and $T$ be the number of tails. Since the coin is tossed $10$ times,$H + T = 10$.The total score $S$ is given by $S

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Let $H$ be the number of heads and $T$ be the number of tails. Since the coin is tossed $10$ times,$H + T = 10$.The total score $S$ is given by $S = 1 	imes H + 2 	imes T = H + 2(10 - H) = 20 - H$.We want to find the largest $K$ such that $P(S \geq K) > \frac{1}{2}$.Substituting $S = 20 - H$,we get $P(20 - H \geq K) = P(H \leq 20 - K) > \frac{1}{2}$.Let $m = 20 - K$. We need $P(H \leq m) > \frac{1}{2}$.The probability distribution of $H$ is binomial with $n = 10$ and $p = \frac{1}{2}$.$P(H \leq m) = \frac{1}{2^{10}} \sum_{i=0}^{m} \binom{10}{i}$.We know that $\sum_{i=0}^{10} \binom{10}{i} = 2^{10} = 1024$. Since the distribution is symmetric,$P(H \leq 4) = \sum_{i=0}^{4} \binom{10}{i} / 1024 = (1 + 10 + 45 + 120 + 210) / 1024 = 386 / 1024 $P(H \leq 5) = (386 + \binom{10}{5}) / 1024 = (386 + 252) / 1024 = 638 / 1024 > \frac{1}{2}$.Thus,the largest $m$ is $5$.Since $m = 20 - K$,we have $5 = 20 - K$,which gives $K = 15$.

$A$ man tosses a coin $10$ times,scoring $1$ point for each head and $2$ points for each tail. Let $P(K)$ be the probability of scoring at least $K$ points. The largest value of $K$ such that $P(K) > \frac{1}{2}$ is

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