A coin is tossed 2n times. The chance that th | vedclass

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(C) The total number of outcomes when a coin is tossed $2n$ times is $2^{2n}$.The number of ways to get exactly $n$ heads and $n$ tails is given by th

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

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(C) The total number of outcomes when a coin is tossed $2n$ times is $2^{2n}$.The number of ways to get exactly $n$ heads and $n$ tails is given by the binomial coefficient $\binom{2n}{n} = \frac{(2n)!}{n!n!}$.The probability of getting exactly $n$ heads and $n$ tails is $P(	ext{equal}) = \frac{\binom{2n}{n}}{2^{2n}} = \frac{(2n)!}{(n!)^2} \cdot \frac{1}{2^{2n}} = \frac{(2n)!}{(n!)^2} \cdot \frac{1}{4^n}$.The probability that the number of heads is not equal to the number of tails is $1 - P(	ext{equal})$.Therefore,the required probability is $1 - \frac{(2n)!}{(n!)^2} \cdot \frac{1}{4^n}$.

$A$ coin is tossed $2n$ times. The chance that the number of times one gets head is not equal to the number of times one gets tail is

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