6 coins are tossed 320 times. The probability | vedclass

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Question

(A) This is a binomial distribution problem where $n = 6$ (number of coins) and $p = 1/2$ (probability of getting a head).The probability of getting $

Vedclass · Accepted Answer

$30^2 	imes \frac{e^{-30}}{2}$

Vedclass · Answer

(A) This is a binomial distribution problem where $n = 6$ (number of coins) and $p = 1/2$ (probability of getting a head).The probability of getting $5$ heads in one trial is $P(X=5) = \binom{6}{5} (1/2)^5 (1/2)^1 = 6 	imes (1/2)^6 = 6/64 = 3/32$.We are performing $N = 320$ trials. Let $Y$ be the number of times we get $5$ heads in $320$ trials.Then $Y$ follows a binomial distribution with $N = 320$ and $p' = 3/32$.The mean $\lambda = Np' = 320 	imes (3/32) = 30$.Using the Poisson approximation for the binomial distribution,$P(Y=k) = \frac{\lambda^k e^{-\lambda}}{k!}$.For $k = 2$,$P(Y=2) = \frac{30^2 	imes e^{-30}}{2!} = \frac{900 	imes e^{-30}}{2} = 450 	imes e^{-30}$.Since the options are provided in a specific format,the closest match based on the Poisson formula is $30^2 	imes \frac{e^{-30}}{2}$.

$6$ coins are tossed $320$ times. The probability of getting $5$ heads $2$ times is

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