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 in a single toss is $p = \frac{1}{2}$ and the probabili

Vedclass · Accepted Answer

$\frac{35}{2^{12}}$

Vedclass · Answer

(A) Let the coin be tossed $n$ times. Since the coin is fair,the probability of getting a head in a single toss 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: $P(X = r) = {}^nC_r p^r q^{n-r} = {}^nC_r (\frac{1}{2})^n$. Given that $P(X = 7) = P(X = 9)$,we have: ${}^nC_7 (\frac{1}{2})^n = {}^nC_9 (\frac{1}{2})^n$. This implies ${}^nC_7 = {}^nC_9$. Using the property ${}^nC_a = {}^nC_b \Rightarrow a + b = n$ (if $a 
eq b$),we get $n = 7 + 9 = 16$. Now,we need to find the probability of getting $3$ heads: $P(X = 3) = {}^{16}C_3 (\frac{1}{2})^{16}$. Calculating ${}^{16}C_3 = \frac{16 	imes 15 	imes 14}{3 	imes 2 	imes 1} = 16 	imes 5 	imes 7 = 560$. So,$P(X = 3) = 560 	imes (\frac{1}{2})^{16} = \frac{560}{65536} = \frac{35}{4096} = \frac{35}{2^{12}}$. Thus,the correct option is $A$.

$A$ fair coin is tossed a fixed number of times. If the probability of getting $7$ heads is equal to that of getting $9$ heads,then the probability of getting $3$ heads is

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