Question

A coin is tossed 10 times. The probability of getting exactly six heads is
(a) $\frac{512}{513}$

(b) $\frac{105}{512}$

(c) $\frac{100}{153}$

(d) ${}^{10}C_6$

Answer 1

(b) $\frac{105}{512}$

$$\begin{gathered} \mathrm{Let}X\mathit{}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{heads}\mathrm{obtained}\mathrm{in}10\mathrm{tosses}\mathrm{of}\mathrm{a}\mathrm{coin}. \\ \mathrm{Then},\mathrm{X}\mathrm{follows}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}n=6,p=\frac{1}{2}=q \\ T\mathrm{he}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{10}\mathrm{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{10-r} \\ \therefore \mathit{}P(X=6)=\frac{{}^{10}C_6}{2^{10}} \\ =\frac{105}{2^9} \\ =\frac{105}{512} \end{gathered}$$