A coin is tossed 8 times. If the probability | vedclass

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(D) Let the outcomes of the $8$ tosses be $X_1, X_2, ..., X_8$. Each toss is independent with $P(H) = P(T) = 1/2$.Let $k$ be the number of heads in th

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Let the outcomes of the $8$ tosses be $X_1, X_2, ..., X_8$. Each toss is independent with $P(H) = P(T) = 1/2$.Let $k$ be the number of heads in the overlapping tosses $X_4, X_5, X_6$. The first $6$ tosses $(X_1, ..., X_6)$ have $4$ heads, so $X_1, X_2, X_3$ must have $4-k$ heads.The last $5$ tosses $(X_4, ..., X_8)$ have $3$ heads, so $X_7, X_8$ must have $3-k$ heads.The constraints on $k$ are $0 \le 4-k \le 3$, $0 \le k \le 3$, and $0 \le 3-k \le 2$. This implies $k \in \{1, 2, 3\}$.The number of favorable outcomes is $\sum_{k=1}^3 \binom{3}{4-k} \binom{3}{k} \binom{2}{3-k}$.For $k=1$: $\binom{3}{3} \binom{3}{1} \binom{2}{2} = 1 	imes 3 	imes 1 = 3$.For $k=2$: $\binom{3}{2} \binom{3}{2} \binom{2}{1} = 3 	imes 3 	imes 2 = 18$.For $k=3$: $\binom{3}{1} \binom{3}{3} \binom{2}{0} = 3 	imes 1 	imes 1 = 3$.Total favorable outcomes $= 3 + 18 + 3 = 24$.The total number of possible outcomes for $8$ tosses is $2^8 = 256$.Thus, $p = 24/256 = 3/32$.Therefore, $96p = 96 	imes (3/32) = 3 	imes 3 = 9$.

$A$ coin is tossed $8$ times. If the probability that exactly $4$ heads appear in the first six tosses and exactly $3$ heads appear in the last five tosses is $p$, then $96p$ is equal to ————

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