Question

The probability that a certain kind of component will survive a given shock test is $\frac{3}{4}.$ Find the probability that among 5 components tested
(i) exactly 2 will survive
(ii) at most 3 will survive

Answer 1

Let X denote the number of components that survive shock.
Then, X follows a binomial distribution with n = 5.
Let p be the probability that a certain kind of component will survive a given shock test.
$$\begin{gathered} \therefore p=\frac{3}{4}\mathrm{and}q=\frac{1}{4} \\ \mathrm{Hence},\mathrm{the}\mathrm{disrtibution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{5}C_r{\left(\frac{\mathit{3}}{\mathit{4}}\right)}^r{\left(\frac{1}{4}\right)}^{5-r},r=0,1,2,3,4,5 \\ \left(\mathrm{i}\right)P(\mathrm{exactly}2\mathrm{will}\mathrm{survive})=P(X=2) \\ ={}^{5}C_2{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^{5-2} \\ =\frac{10\times 9}{1024} \\ =0.0879 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)P(\mathrm{atmost}3\mathrm{will}\mathrm{survive})=\mathit{}P(X\le 3) \\ =P(X=0)+P(X=1)+P(X=2)+P(X=3) \\ ={}^{5}C_0{\left(\frac{3}{4}\right)}^0{\left(\frac{1}{4}\right)}^{5-0}+{}^{5}C_1{\left(\frac{3}{4}\right)}^1{\left(\frac{1}{4}\right)}^{5-1}+{}^{5}C_2{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^{5-2}+{}^{5}C_3{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^{5-3} \\ ={\left(\frac{1}{4}\right)}^5+5\left(\frac{3}{4}\right){\left(\frac{1}{4}\right)}^4+10{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^3+10{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^2 \\ =\frac{1+15+90+270}{1024} \\ =\frac{376}{1024} \\ =0.3672 \end{gathered}$$