A fair coin is tossed a fixed number of times | vedclass

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Question

(A) According to the Binomial probability distribution, let $n$ be the number of times the coin is tossed.Probability of getting $5$ heads is $P(X=5)

Vedclass · Accepted Answer

$\frac{495}{4096}$

Vedclass · Answer

(A) According to the Binomial probability distribution, let $n$ be the number of times the coin is tossed.Probability of getting $5$ heads is $P(X=5) = {}^{n}C_{5} (\frac{1}{2})^{n-5} (\frac{1}{2})^{5} = {}^{n}C_{5} (\frac{1}{2})^{n}$.Probability of getting $7$ heads is $P(X=7) = {}^{n}C_{7} (\frac{1}{2})^{n}$.Given that $P(X=5) = P(X=7)$, we have ${}^{n}C_{5} = {}^{n}C_{7}$.Using the property ${}^{n}C_{x} = {}^{n}C_{y} \Rightarrow x+y=n$, we get $n = 5+7 = 12$.Now, the probability of getting $4$ heads is $P(X=4) = {}^{12}C_{4} (\frac{1}{2})^{12}$.$P(X=4) = \frac{12 	imes 11 	imes 10 	imes 9}{4 	imes 3 	imes 2 	imes 1} 	imes \frac{1}{4096} = 495 	imes \frac{1}{4096} = \frac{495}{4096}$.Thus, option $A$ is correct.

$A$ fair coin is tossed a fixed number of times. If the probability of getting five heads is equal to that of getting seven heads, then the probability of getting four heads is

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