If a fair coin is tossed 10 times,find the pr | vedclass

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(A) The repeated tosses of a coin are Bernoulli trials. Let $X$ denote the number of heads in an experiment of $10$ trials.Clearly,$X$ follows a binom

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$\frac{193}{512}$

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(A) The repeated tosses of a coin are Bernoulli trials. Let $X$ denote the number of heads in an experiment of $10$ trials.Clearly,$X$ follows a binomial distribution with $n=10$ and $p=\frac{1}{2}$.The probability mass function is given by $P(X=x) = ^{n}C_{x} q^{n-x} p^{x}$,where $x = 0, 1, 2, \dots, 10$.Here,$n=10$,$p=\frac{1}{2}$,and $q=1-p=\frac{1}{2}$.Thus,$P(X=x) = ^{10}C_{x} (\frac{1}{2})^{10-x} (\frac{1}{2})^{x} = ^{10}C_{x} (\frac{1}{2})^{10}$.We need to find $P(X \geq 6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)$.$P(X \geq 6) = [^{10}C_{6} + ^{10}C_{7} + ^{10}C_{8} + ^{10}C_{9} + ^{10}C_{10}] 	imes (\frac{1}{2})^{10}$.Calculating the combinations:$^{10}C_{6} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$.$^{10}C_{7} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.$^{10}C_{8} = \frac{10 	imes 9}{2 	imes 1} = 45$.$^{10}C_{9} = 10$.$^{10}C_{10} = 1$.Sum $= 210 + 120 + 45 + 10 + 1 = 386$.$P(X \geq 6) = \frac{386}{2^{10}} = \frac{386}{1024} = \frac{193}{512}$.

If a fair coin is tossed $10$ times,find the probability of getting at least six heads.

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