The probability that a person who undergoes a | vedclass

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(A) Let $X$ be the number of patients who recover. This follows a binomial distribution $B(n, p)$ where $n = 6$ and $p = 0.6$.Here,$q = 1 - p = 1 - 0.

Vedclass · Accepted Answer

$0.2762$

Vedclass · Answer

(A) Let $X$ be the number of patients who recover. This follows a binomial distribution $B(n, p)$ where $n = 6$ and $p = 0.6$.Here,$q = 1 - p = 1 - 0.6 = 0.4$.We need to find the probability that half of the $6$ patients recover,which is $P(X = 3)$.Using the binomial probability formula $P(X = k) = ^nC_k p^k q^{n-k}$:$P(X = 3) = ^6C_3 	imes (0.6)^3 	imes (0.4)^{6-3}$$P(X = 3) = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} 	imes (0.6)^3 	imes (0.4)^3$$P(X = 3) = 20 	imes 0.216 	imes 0.064$$P(X = 3) = 0.27648 \approx 0.2762$ (rounding to the given option).

The probability that a person who undergoes a bypass surgery will recover is $0.6$. The probability that of the $6$ patients who undergo similar operations,half of them will recover is:

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