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

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(B) Let $n$ be the number of tosses. The probability of getting $k$ tails in $n$ tosses is given by the binomial distribution formula: $P(X=k) = \bino

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$\frac{55}{2^{10}}$

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(B) Let $n$ be the number of tosses. The probability of getting $k$ tails in $n$ tosses is given by the binomial distribution formula: $P(X=k) = \binom{n}{k} (\frac{1}{2})^n$.Given that $P(X=5) = P(X=7)$,we have $\binom{n}{5} (\frac{1}{2})^n = \binom{n}{7} (\frac{1}{2})^n$.This implies $\binom{n}{5} = \binom{n}{7}$.Using the property $\binom{n}{r} = \binom{n}{n-r}$,we know that if $\binom{n}{a} = \binom{n}{b}$,then either $a=b$ or $a+b=n$.Since $5 
eq 7$,we must have $n = 5+7 = 12$.Now,we need to find the probability of getting $3$ tails,which is $P(X=3) = \binom{12}{3} (\frac{1}{2})^{12}$.Calculating $\binom{12}{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.Thus,$P(X=3) = 220 	imes \frac{1}{2^{12}} = \frac{220}{4096} = \frac{55}{1024} = \frac{55}{2^{10}}$.

$A$ fair coin is tossed a fixed number of times. If the probability of getting $5$ tails is same as the probability of getting $7$ tails,then the probability of getting $3$ tails is

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