Assume that on an average, one telephone out of 5 is busy. Ten telephone numbers are randomly selected and called. Find the probability that atleast three of them will be busy.

$1-\frac{4^8×101}{5^{10}}$

The correct answer is Option (1) → $1-\frac{4^8×101}{5^{10}}$

Let A be the event 'telephone is dialled and it is busy', then

$p = P(A)=\frac{1}{5}$, so $q = 1-\frac{1}{5}=\frac{4}{5}$ and $n = 10$.

Thus, we have a binomial distribution with $n =10,p=\frac{1}{5}$ and $q=\frac{4}{5}$.

Required probability = $P(X ≥ 3) = P(3) + P(4) + P(5) + ... + P(10)$

$= 1 - (P(0) + P(1) + P(2))$

$= 1 - ({^{10}C}_0q^{10} + {^{10}C}_1pq^9 + {^{10}C}_2p^2q^8)$

$= 1 - (1 × q^2 + 10pq + 45p^2) q^8$

$=1-\left(1×\left(\frac{4}{5}\right)^2+10×\frac{1}{5}×\frac{4}{5}+45\left(\frac{1}{5}\right)^2\right)\left(\frac{4}{5}\right)^8$

$= 1 - (16+40 + 45) ×\frac{4^8}{5^{10}}=1-\frac{4^8×101}{5^{10}}$