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(C) The probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.For $n = 15$ tr

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$\frac{3003}{32768}$

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(C) The probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.For $n = 15$ trials,the probability of getting $r = 10$ heads is given by the binomial distribution formula: $P(X = r) = {}^{n}C_{r} p^{r} q^{n-r}$.Substituting the values,we get: $P(X = 10) = {}^{15}C_{10} \left(\frac{1}{2}ight)^{10} \left(\frac{1}{2}ight)^{5}$.Since ${}^{15}C_{10} = {}^{15}C_{5}$,we have: $P(X = 10) = {}^{15}C_{5} 	imes \left(\frac{1}{2}ight)^{15}$.Calculating the combination: ${}^{15}C_{5} = \frac{15 	imes 14 	imes 13 	imes 12 	imes 11}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 3003$.Thus,the probability is: $P(X = 10) = \frac{3003}{2^{15}} = \frac{3003}{32768}$.

If $15$ coins are tossed,then the probability of getting $10$ heads is:

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