Question

From a group of $10$ persons consisting of $5$ lawyers, $3$ doctors and $2$ engineers, four persons are selected at random. The probability that the selection contains at least one of each category is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}$

Let's represent lawyers by $L$, doctors by $D$ and engineers by $E$.

Thus, we have $L_1,L_2,L_3,L_4,L_5,D_1,D_2,D_3$ and $E_1,E_2$

Total number of ways in which $4$ people can be selected ${}^{10}{C}_4$

Now according to question we need to select $4$ people such that at least one person from each category gets selected.

One from each category can be selected in ${}^5{C}_1\times {}^3{C}_1\times {}^4{C}_1$ ways.

Now, the fourth person can be selected from any category in $({}^4{C}_1+{}^2{C}_1+{}^1{C}_1)$

$\therefore$ Total number of favourable outcome $=\frac{{}^5{C}_1{\times }^3{C}_1{\times }^2{C}_1\times \left({}^4{C}_1{+}^2{C}_1{+}^1{C}_1\right)}{2}$ We are dividing by $2$ because $L_1{,D}_1,E_1{E}_2$ and $L_1{,D}_1,E_2{E}_1$ will get repeated.

$\therefore$ Probability $=\frac{\text{Total number of favourable outcome}}{\text{Total number of ways in which}4\text{people can be selected}}$

$=\frac{{}^5{C}_1{\times }^3{C}_1{\times }^2{C}_1\times \left({}^4{C}_1{+}^2{C}_1{+}^1{C}_1\right)}{2{\times }^{10}{C}_4}$

$\Rightarrow$ Probability $=\frac{1}{2}$