A fair coin is tossed 2n times. What is the p | vedclass

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(C) The total number of outcomes when a coin is tossed $2n$ times is $2^{2n} = 4^n$.Let $A$ be the event that the number of heads and tails are not eq

Vedclass · Accepted Answer

$1 - \frac{(2n)!}{(n!)^2 	imes 4^n}$

Vedclass · Answer

(C) The total number of outcomes when a coin is tossed $2n$ times is $2^{2n} = 4^n$.Let $A$ be the event that the number of heads and tails are not equal.Let $A'$ be the event that the number of heads and tails are equal.For $A'$,we must have exactly $n$ heads and $n$ tails.The number of ways to arrange $n$ heads and $n$ tails in $2n$ tosses is given by the binomial coefficient $\binom{2n}{n} = \frac{(2n)!}{n!n!} = \frac{(2n)!}{(n!)^2}$.The probability of event $A'$ is $P(A') = \frac{\binom{2n}{n}}{2^{2n}} = \frac{(2n)!}{(n!)^2 	imes 4^n}$.The probability of event $A$ is $P(A) = 1 - P(A') = 1 - \frac{(2n)!}{(n!)^2 	imes 4^n}$.

$A$ fair coin is tossed $2n$ times. What is the probability that the number of heads and tails obtained in these $2n$ trials are not equal?

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