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

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(A) Let the coin be tossed $n$ times.Since the coin is fair,the probability of getting a head is $p = \frac{1}{2}$ and the probability of getting a ta

Vedclass · Accepted Answer

$\frac{15}{2^{13}}$

Vedclass · Answer

(A) Let the coin be tossed $n$ times.Since the coin is fair,the probability of getting a head is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.The probability of getting $r$ heads in $n$ tosses is given by the binomial distribution formula: $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r} = {}^{n}C_{r} (\frac{1}{2})^{n}$.Given that $P(X=7) = P(X=9)$,we have:${}^{n}C_{7} (\frac{1}{2})^{n} = {}^{n}C_{9} (\frac{1}{2})^{n}$${}^{n}C_{7} = {}^{n}C_{9}$Using the property ${}^{n}C_{x} = {}^{n}C_{y} \implies x+y=n$ (if $x 
eq y$),we get $n = 7 + 9 = 16$.Now,we need to find the probability of getting $2$ heads,which is $P(X=2)$:$P(X=2) = {}^{16}C_{2} (\frac{1}{2})^{16}$$P(X=2) = \frac{16 	imes 15}{2 	imes 1} 	imes \frac{1}{2^{16}}$$P(X=2) = 8 	imes 15 	imes \frac{1}{2^{16}} = \frac{15}{2^{3} 	imes 2^{16-3}} = \frac{15}{2^{13}}$.

$A$ fair coin is tossed a fixed number of times. If the probability of getting $7$ heads is equal to the probability of getting $9$ heads,then the probability of getting $2$ heads is:

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