Question

If a fair coin is tossed $5$ times. Then the probaility that we get at least $3$ tails is
$\left(a\right)\frac{1}{2}\left(b\right)\frac{1}{4}\left(c\right)\frac{1}{8}\left(d\right)\frac{3}{32}$

Answer 1

Probability of heads = $p=\frac{1}{2}$
Probability of tails = $q=\frac{1}{2}$
Probability of getting at least $3$ tails = ${}^5{C}_3{q}^3{p}^2{+}^5{C}_4{q}^4{p}^1{+}^5{C}_5{q}^5{p}^0$
$=\frac{1}{32}(10+5+1)=\frac{1}{2}$

Answer: $\left(a\right)$