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(C) The total number of possible outcomes when tossing $6$ coins is $2^6 = 64$. We need to find the probability of getting at least $4$ heads,which is

Vedclass · Accepted Answer

$\frac{11}{32}$

Vedclass · Answer

(C) The total number of possible outcomes when tossing $6$ coins is $2^6 = 64$. We need to find the probability of getting at least $4$ heads,which is $P(X \ge 4) = P(X=4) + P(X=5) + P(X=6)$. Using the binomial distribution formula $P(X=r) = \binom{n}{r} p^r q^{n-r}$,where $n=6$ and $p=q=1/2$: $P(X=4) = \binom{6}{4} (1/2)^6 = \frac{15}{64}$ $P(X=5) = \binom{6}{5} (1/2)^6 = \frac{6}{64}$ $P(X=6) = \binom{6}{6} (1/2)^6 = \frac{1}{64}$ Summing these probabilities: $P(X \ge 4) = \frac{15+6+1}{64} = \frac{22}{64} = \frac{11}{32}$. Thus,the correct option is $C$.

When six coins are tossed simultaneously,the probability of getting at least $4$ heads is

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