Question

In an examination, the probability of a candidate solving a question is $\frac{1}{2}$. Out of given $5$ questions in the examination, what is the probability that the candidate was able to solve at least $2$ questions?

• A. $\frac{1}{64}$
• B. $\frac{3}{16}$
• C. $\frac{1}{2}$
• D. $\frac{13}{16}$

Answer 1

The correct option is D

$\frac{13}{16}$

Explanation for the correct option:

Finding the probability:

The probability of solving a question is $\frac{1}{2}$, it means the probability of not solving a question is $1-\frac{1}{2}=\frac{1}{2}$.

The required probability, $$\begin{gathered} P=P\left(\mathrm{the}\mathrm{candidate}\mathrm{can}\mathrm{solve}2\mathrm{questions}\right)+P\left(\mathrm{the}\mathrm{candidate}\mathrm{can}\mathrm{solve}3\mathrm{questions}\right) \\ +P\left(\mathrm{the}\mathrm{candidate}\mathrm{can}\mathrm{solve}4\mathrm{questions}\right)+P\left(\mathrm{the}\mathrm{candidate}\mathrm{can}\mathrm{solve}5\mathrm{questions}\right) \end{gathered}$$

It will be

$\begin{array}{rcl}P& =& {}^{5}C_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^3+{}^{5}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^2+{}^{5}C_4{\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^1+{}^{5}C_5{\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^0\\ & =& {\left(\frac{1}{2}\right)}^5\left[{}^{5}C_2+{}^{5}C_3+{}^{5}C_4+{}^{5}C_5\right]\\ & =& \frac{1}{32}\left(\frac{5\times 4\times 3!}{2\times 3!}+\frac{5\times 4\times 3!}{2\times 3!}+\frac{5\times 4!}{4!}+1\right)\mathbf{[}\mathbf{\because }\mathbf{}{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\left(n-r\right)\mathbf{!}}\mathbf{]}\\ & =& \frac{1}{32}\left(10+10+5+1\right)\\ & =& \frac{13}{16}\end{array}$

Hence, option D is correct.