JEE Mathematics 2023 Question with Solution

Given: A bag contains $6$ balls of different colours and draws are made with replacement.

Find: If $p:q = m:n$, find $m+n$.

For $p$, both drawn balls must have the same colour.

$$p = 6\left(\frac{1}{6}\times\frac{1}{6}\right) = \frac{1}{6}$$

For $q$, exactly three out of four balls must be of one colour and the fourth of a different colour.

• Choose the colour repeated three times: $6$ ways
• Choose the position of the different-coloured ball: $\binom{4}{1}=4$ ways
• Choose its colour: $5$ ways

Thus,

$$q = \frac{6\times 4\times 5}{6^4} = \frac{120}{1296} = \frac{5}{54}$$

Now,

$$\frac{p}{q} = \frac{\frac{1}{6}}{\frac{5}{54}} = \frac{1}{6}\times\frac{54}{5} = \frac{9}{5}$$

So,

$$p:q = 9:5$$

Hence $m=9$ and $n=5$, therefore

$$m+n=14$$

Therefore, the required answer is $14$.

Given: There are $6$ distinct colours, and each draw is independent because the ball is replaced.

Find: The value of $m+n$ when $p:q=m:n$.

For the event defining $p$:

• Total outcomes in two draws = $6^2$
• Favourable outcomes = choose any one colour and get it twice = $6$

Hence,

$$p = \frac{6}{6^2} = \frac{1}{6}$$

For the event defining $q$:

• Total outcomes in four draws = $6^4$
• Favourable outcomes for exactly three same-coloured balls:

$$6 \times 5 \times 4$$

where $6$ chooses the tripled colour, $5$ chooses the different colour, and $4$ chooses the position of that different ball.

Therefore,

$$q = \frac{6\times 5\times 4}{6^4} = \frac{120}{1296} = \frac{5}{54}$$

Now compare:

$$p:q = \frac{1}{6} : \frac{5}{54}$$

Multiply both terms by $54$:

$$p:q = 9:5$$

Thus $m=9$, $n=5$, and

$$m+n=14$$

Therefore, the answer is $14$.