JEE Main 2012 — Probability Question with Solution

Given set S = $$\left\{1,2,3,..8\right\}$$

Choosing 3 numbers from 8 numbers can be done $${}^8{C}_3$$ ways.

Choosing 3 numbers from 8 numbers while minimum no is 3 can be done $$1\times {}^5{C}_2$$ ways.

$$\therefore$$ Probablity P(min = 3) = $$\frac{1\times {}^5{C}_2}{{}^8{C}_3}$$

Choosing 3 numbers from 8 numbers while maximum no is 6 can be done $$1\times {}^5{C}_2$$ ways.

$$\therefore$$ Probablity P(max = 6) = $$\frac{1\times {}^5{C}_2}{{}^8{C}_3}$$

Choosing 3 numbers from 8 numbers while minimum number 3 and maximum no is 6 can be done $$1\times {}^2{C}_1\times 1$$ ways.

$$\therefore$$ $$P\left(\min =3\cap \max =6\right)$$ = $$\frac{1\times {}^2{C}_1\times 1}{{}^8{C}_3}$$

The probability that their minimum is $$3,$$ given that their maximum is $$6,$$ is :
$$P\left(\frac{\min =3}{\max =6}\right)$$

= $$\frac{P\left(\min =3\cap \max =6\right)}{P\left(\max =6\right)}$$

= $$\frac{\frac{{}^2{C}_1}{{}^8{C}_3}}{\frac{{}^5{C}_2}{{}^8{C}_3}}$$

= $$\frac{{}^2{C}_1}{{}^5{C}_2}$$

= $$\frac{1}{5}$$