Question

Mark the correct alternative in the following question:

The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons 4 are swimmers is

$\left(\mathrm{a}\right){}^5\mathrm{C}_4{\left(0.7\right)}^4\left(0.3\right)\left(\mathrm{b}\right){}^5\mathrm{C}_1\left(0.7\right){\left(0.3\right)}^4\left(\mathrm{c}\right){}^5\mathrm{C}_4\left(0.7\right){\left(0.3\right)}^4\left(\mathrm{d}\right){\left(0.7\right)}^4\left(0.3\right)$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ q=\mathrm{probability}\mathrm{that}\mathrm{a}\mathrm{person}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{swimmer}=0.3\mathrm{and} \\ p=\mathrm{probability}\mathrm{that}\mathrm{a}\mathrm{person}\mathrm{is}\mathrm{a}\mathrm{swimmer}=1-q=1-0.3=0.7 \\ \mathrm{Let}X\mathrm{denote}\mathrm{a}\mathrm{success}\mathrm{that}\mathrm{a}\mathrm{person}\mathrm{selected}\mathrm{is}\mathrm{a}\mathrm{swimmer}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=5\mathrm{and}p=0.7 \\ \therefore \mathrm{P}\left(X=r\right)={}^5\mathrm{C}_r{p}^r{q}^{\left(5-r\right)}={}^5\mathrm{C}_r{\left(0.7\right)}^r{\left(0.3\right)}^{\left(5-r\right)} \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{required}\mathrm{probability}=\mathrm{P}\left(X=4\right)={}^5\mathrm{C}_4{\left(0.7\right)}^4{\left(0.3\right)}^{\left(5-4\right)}={}^5\mathrm{C}_4{\left(0.7\right)}^4\left(0.3\right) \end{gathered}$$

Hence, the correct alternative is option (a).