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(D) The probability of getting $x$ heads in $n=8$ tosses is given by the binomial distribution formula: $P(X=x) = {^nC_x} (p)^x (q)^{n-x}$,where $p =

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$37/256$

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(D) The probability of getting $x$ heads in $n=8$ tosses is given by the binomial distribution formula: $P(X=x) = {^nC_x} (p)^x (q)^{n-x}$,where $p = 1/2$ and $q = 1/2$.For at least $6$ heads,we need to calculate $P(X \ge 6) = P(X=6) + P(X=7) + P(X=8)$.$P(X=6) = {^8C_6} (1/2)^8 = 28 	imes (1/256) = 28/256$.$P(X=7) = {^8C_7} (1/2)^8 = 8 	imes (1/256) = 8/256$.$P(X=8) = {^8C_8} (1/2)^8 = 1 	imes (1/256) = 1/256$.Summing these probabilities: $P(X \ge 6) = (28 + 8 + 1) / 256 = 37/256$.

If $8$ coins are tossed simultaneously,what is the probability of getting at least $6$ heads?

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