A bag contains 10 balls out of which � are red and � are black, where �. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

A bag contains 10 balls out of which k are red and (10−k) are black – JEE Main Probability Question

Q. A bag contains 10 balls out of which k are red and (10 − k) are black, where 0 ≤ k ≤ 10. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

(A) 7/110

(B) 7/11

(D) 14/55

Correct Answer: 14/55

Explanation

This is a problem based on Bayes’ Theorem.

Let the event A_k denote that the bag contains k red balls and (10 − k) black balls.

Since no information about k is given, all values of k from 0 to 10 are equally likely.

P(A_k) = 1/11, for k = 0, 1, 2, …, 10

Let B be the event that all three drawn balls are black.

If the bag contains k red balls, then it contains (10 − k) black balls.

So,

P(B | A_k) = C(10 − k, 3) / C(10, 3)

Using Bayes’ theorem:

P(A₁ | B) = [ P(B | A₁) · P(A₁) ] / [ Σ P(B | A_k) · P(A_k) ]

Substitute values:

P(A₁ | B) = [ C(9, 3) × (1/11) ] / [ (1/11) Σ C(10 − k, 3) ]

Cancel 1/11 from numerator and denominator:

= C(9, 3) / Σ C(10 − k, 3)

Now,

C(9, 3) = 84

And,

Σ C(10 − k, 3) = C(10,3)+C(9,3)+…+C(0,3)

Using identity:

C(10,3)+C(9,3)+…+C(3,3) = C(11,4) = 330

Hence,

P = 84 / 330 = 14 / 55

Therefore, the required probability is

14 / 55

Explanation

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