Question

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times. [NCERT EXEMPLAR]

Answer 1

Let getting an odd number be a success in a trial.

We have,

$$\begin{gathered} p=\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{in}\mathrm{a}\mathrm{trial}=\frac{3}{6}=\frac{1}{2} \\ \mathrm{Also},q=1-p=1-\frac{1}{2}=\frac{1}{2} \\ \mathrm{Let}X\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{success}\mathrm{in}\mathrm{a}\mathrm{sample}\mathrm{of}5\mathrm{trials}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=5\mathrm{and}p=q=\frac{1}{2} \\ \therefore \mathrm{P}\left(X=r\right)={}^5\mathrm{C}_r{p}^r{q}^{\left(5-r\right)}={}^5\mathrm{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{\left(5-r\right)}={}^5\mathrm{C}_r{\left(\frac{1}{2}\right)}^5,\mathrm{where}r=0,1,2,3,4,5 \\ \mathrm{Now}, \\ \mathrm{Required}\mathrm{probability}=\mathrm{P}\left(X=3\right) \\ ={}^5\mathrm{C}_3{\left(\frac{1}{2}\right)}^5 \\ =\frac{10}{32} \\ =\frac{5}{16} \end{gathered}$$