JEE Mathematics 2023 Question with Solution

Given: A bag contains $6$ balls. Two drawn balls are found to be black.

Find: The probability that the bag contains at least $5$ black balls.

From the extracted working, consider possible numbers of black balls in the bag as $2,3,4,5,6$. The event “at least $5$ black balls” corresponds to $5$ or $6$ black balls.

Using the combination count shown in the solution:

$$\frac{{}^5C_2+{}^6C_2}{{}^2C_2+{}^3C_2+{}^4C_2+{}^5C_2+{}^6C_2}$$

Substitute the values:

$$\frac{10+15}{1+3+6+10+15}$$

So,

$$\frac{25}{35}$$

Hence,

$$\frac{5}{7}$$

Therefore, the probability that the bag contains at least $5$ black balls is $\frac{5}{7}$. The listed the solution marks option C, but the worked value is $\frac{5}{7}$, which matches option A.

Given: Two balls drawn from the bag are black.

Find: Probability that the bag had at least $5$ black balls.

The provided solution counts the admissible black-ball cases by the number of ways to choose $2$ black balls from a bag having $k$ black balls, for $k=2,3,4,5,6$.

For the favorable cases:

• If the bag has $5$ black balls, the count is ${}^5C_2=10$.
• If the bag has $6$ black balls, the count is ${}^6C_2=15$.

Thus favorable total:

$$10+15=25$$

For all considered cases:

$${}^2C_2+{}^3C_2+{}^4C_2+{}^5C_2+{}^6C_2$$

That is,

$$1+3+6+10+15=35$$

Therefore,

$$P(\text{at least }5\text{ black balls})=\frac{25}{35}=\frac{5}{7}$$

So the correct numerical value is $\frac{5}{7}$, and the corresponding correct option is A.