JEE Mathematics 2026 Question with Solution

Given: A bag contains $10$ balls with $k$ red and $(10-k)$ black, where $0 \le k \le 10$. Three balls are drawn without replacement and all are black.

Find: The posterior probability that the bag contains $1$ red and $9$ black balls.

Concept: This is a Bayes' theorem problem.

Let

$$E_k = \text{Event that the bag contains } k \text{ red balls}$$

and

$$A = \text{Event that all three drawn balls are black}$$

Since no prior information is given, all values $k = 0,1,2,\ldots,10$ are equally likely, so

$$P(E_k) = \frac{1}{11}$$

Number of black balls when there are $k$ red balls is $(10-k)$. Therefore,

$$P(A\mid E_k) = \frac{\binom{10-k}{3}}{\binom{10}{3}}$$

Now apply Bayes' theorem:

$$P(E_1\mid A) = \frac{P(A\mid E_1)P(E_1)}{\sum_{k=0}^{7} P(A\mid E_k)P(E_k)}$$

The upper limit is $7$ because for $k>7$, fewer than $3$ black balls are available. Thus,

$$P(E_1\mid A) = \frac{\binom{9}{3}}{\sum_{k=0}^{7} \binom{10-k}{3}}$$

So,

$$P(E_1\mid A) = \frac{84}{\binom{10}{3} + \binom{9}{3} + \binom{8}{3} + \binom{7}{3} + \binom{6}{3} + \binom{5}{3} + \binom{4}{3} + \binom{3}{3}}$$ $$= \frac{84}{120 + 84 + 56 + 35 + 20 + 10 + 4 + 1}$$ $$= \frac{84}{330} = \frac{7}{55}$$

Therefore, the required probability is $\dfrac{7}{55}$, so the correct option is B.

Given: The event observed is that all three drawn balls are black.

Find: $P(E_1\mid A)$.

Bayes' theorem gives

$$P(E_1\mid A) = \frac{P(A\mid E_1)P(E_1)}{P(A)}$$

Since each possible value of $k$ from $0$ to $10$ is equally likely,

$$P(E_k)=\frac{1}{11}$$

For a fixed $k$, the probability that all three drawn balls are black is

$$P(A\mid E_k)=\frac{\binom{10-k}{3}}{\binom{10}{3}}$$

Hence,

$$P(A)=\sum_{k=0}^{7} P(A\mid E_k)P(E_k)$$

Substituting in Bayes' theorem, the common factor $\frac{1}{11\binom{10}{3}}$ cancels, giving

$$P(E_1\mid A)=\frac{\binom{9}{3}}{\sum_{k=0}^{7}\binom{10-k}{3}}$$

Now evaluate:

$$\binom{9}{3}=84$$

and

$$\sum_{k=0}^{7}\binom{10-k}{3}=\binom{10}{3}+\binom{9}{3}+\binom{8}{3}+\binom{7}{3}+\binom{6}{3}+\binom{5}{3}+\binom{4}{3}+\binom{3}{3}$$ $$=120+84+56+35+20+10+4+1=330$$

Therefore,

$$P(E_1\mid A)=\frac{84}{330}=\frac{7}{55}$$

Thus the correct option is B.