JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Probability

A bag contains 10 balls out of which k are red and

(10 - k)

are black, where

0 \leq k \leq 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) $\frac{7}{11}$
(B) $\frac{7}{55}$
(C) $\frac{7}{110}$
(D) $\frac{14}{55}$ ✓ Correct

Answer: (D)

Solution

With

k

red balls the bag has

10 - k

black, and the chance of drawing three black is

\frac{( 3 10 - k )}{( 3 10 )}

. Taking each of the eleven values

k = 0, 1, \dots, 10

as equally likely, the total probability of three black draws is

\frac{1}{11} \sum_{k = 0}^{7} \frac{( 3 10 - k )}{( 3 10 )} = \frac{1}{11} \cdot \frac{( 4 11 )}{( 3 10 )} = \frac{1}{11} \cdot \frac{330}{120} = \frac{1}{4}

. For the bag with

1

red and

9

black (

k = 1

) the likelihood is

\frac{( 3 9 )}{( 3 10 )} = \frac{7}{10}

. By Bayes' theorem the required probability is

\frac{\frac{1}{11} \cdot \frac{7}{10}}{\frac{1}{4}} = \frac{14}{55}

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Probability

Solution

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