A fair coin is tossed n times. If the probabi | vedclass

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(B) For a fair coin,the probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \fra

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$14$

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(B) For a fair coin,the probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.The probability of getting $k$ heads in $n$ tosses is given by the binomial distribution formula: $P(X = k) = {}^nC_k p^k q^{n-k} = {}^nC_k (\frac{1}{2})^k (\frac{1}{2})^{n-k} = {}^nC_k (\frac{1}{2})^n$.Given that the probability of $6$ heads is equal to the probability of $8$ heads:$P(X = 6) = P(X = 8)$${}^nC_6 (\frac{1}{2})^n = {}^nC_8 (\frac{1}{2})^n$${}^nC_6 = {}^nC_8$Using the property of combinations,if ${}^nC_a = {}^nC_b$,then either $a = b$ or $a + b = n$.Since $6 
eq 8$,we must have $6 + 8 = n$.Therefore,$n = 14$.

$A$ fair coin is tossed $n$ times. If the probability that head occurs $6$ times is equal to the probability that head occurs $8$ times,then $n$ is equal to:

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