A coin is tossed 6 times. The probability of getting at least 4 heads…

The correct answer is Option (3) → $\frac{11}{32}$

The formula for getting binomial distribution,

$P(X=k)=\left({^nC}_k\right)p^kq^{n-k}$

for 4 Heads

$P(X=4)=\left({^6C}_4\right)(0.5)^4(0.5)^2=\frac{15}{64}$

for 5 Heads

$P(X=5)=\left({^6C}_5\right)(0.5)^5(0.5)^1=\frac{6}{64}$

for 6 Heads

$P(X=6)=1×\frac{1}{64}=\frac{1}{64}$

P (at lest 4 heads) = $\frac{15}{64}+\frac{6}{64}+\frac{1}{64}=\frac{22}{64}=\frac{11}{32}$