Question

On dialling certain telephone numbers, assume that on an average, one telephone number out of five is busy, ten telephone numbers are randomly selected and dialled. Find the probability that at least three of them will be busy.

Answer 1

$P$(Probability of being busy) $=\frac{1}{5}$, $Q$(Probability of not busy) $=1-\frac{1}{5}$ and $n=10$
$\therefore$ Required probability (at least three phones are busy)
$=1-$ (Probability maximum two phones are busy)

$=1-\left[{}^{10}{C}_0{\left(\frac{1}{5}\right)}^0{\left(\frac{4}{5}\right)}^{10}{+}^{10}{C}_1{\left(\frac{1}{5}\right)}^1{\left(\frac{4}{5}\right)}^9{+}^{10}{C}_2{\left(\frac{1}{5}\right)}^2{\left(\frac{4}{5}\right)}^8\right]$$=1-[1\times 1\times {\left(\frac{4}{5}\right)}^{10}+10\times \frac{1}{5}\times {\left(\frac{4}{5}\right)}^9+45\times {\left(\frac{1}{5}\right)}^2{\left(\frac{4}{5}\right)}^8]$

$=1-{\left(\frac{4}{5}\right)}^8[\frac{16}{5}+\frac{8}{5}+\frac{9}{5}]$

$=1-{\left(\frac{4}{5}\right)}^8\left(\frac{16+40+45}{25}\right)$

$=1-0.1678\left(\frac{101}{25}\right)$

$=1-0.1678\times 4.04$

$=1=0.68=0.32$