What is the probability of getting exactly 6 | vedclass

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(B) Let $X$ be the number of heads in $10$ tosses of a coin.This follows a binomial distribution with parameters $n = 10$,$p = 1/2$,and $q = 1 - p = 1

Vedclass · Accepted Answer

$105/512$

Vedclass · Answer

(B) Let $X$ be the number of heads in $10$ tosses of a coin.This follows a binomial distribution with parameters $n = 10$,$p = 1/2$,and $q = 1 - p = 1/2$.The probability of getting exactly $x$ heads is given by $P(X = x) = {}^{n}C_{x} p^{x} q^{n-x}$.For $x = 6$:$P(X = 6) = {}^{10}C_{6} \left(\frac{1}{2}ight)^{6} \left(\frac{1}{2}ight)^{10-6}$$P(X = 6) = {}^{10}C_{6} \left(\frac{1}{2}ight)^{10}$Since ${}^{10}C_{6} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$,$P(X = 6) = \frac{210}{2^{10}} = \frac{210}{1024} = \frac{105}{512}$.

What is the probability of getting exactly $6$ heads when a coin is tossed $10$ times?

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