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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$\frac{53}{64}$

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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 $P(X=x) = ^{10}C_x (\frac{1}{2})^{10}$.We need to find $P(X \leq 6) = 1 - P(X > 6) = 1 - [P(X=7) + P(X=8) + P(X=9) + P(X=10)]$.$P(X=7) = ^{10}C_7 (\frac{1}{2})^{10} = \frac{120}{1024}$.$P(X=8) = ^{10}C_8 (\frac{1}{2})^{10} = \frac{45}{1024}$.$P(X=9) = ^{10}C_9 (\frac{1}{2})^{10} = \frac{10}{1024}$.$P(X=10) = ^{10}C_{10} (\frac{1}{2})^{10} = \frac{1}{1024}$.Sum $= \frac{120+45+10+1}{1024} = \frac{176}{1024} = \frac{11}{64}$.Therefore,$P(X \leq 6) = 1 - \frac{11}{64} = \frac{53}{64}$.

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

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