JEE Main 2026 Jan 28 Shift 1, Mathematics Q9: Total Probability Theorem And Bayes' Theorem

JEE Main2026Jan 28, Shift 1Mathematics

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
$\frac{14}{55}$

Solution

This is a Bayes-theorem question with prior uniform over $k \in {0, 1, \dots, 10}$ (11 equally likely compositions).

$P (3 black ∣ k red) = \frac{( 3 10 - k )}{( 3 10 )}$ . For $k = 1$ (i.e. 9 black, 1 red): $P = \frac{( 3 9 )}{( 3 10 )}$ .

By Bayes:

$P (k = 1 ∣ 3 black) = \frac{( 3 9 )}{\sum _{k = 0}^{10} ( 3 10 - k )} = \frac{( 3 9 )}{( 3 10 ) + ( 3 9 ) + ( 3 8 ) + \dots + ( 3 3 )}$ .

Use the hockey-stick identity: $\sum_{j = 3}^{10} (3 j) = (4 11)$ . So the denominator equals $(4 11)$ .

$P = \frac{( 3 9 )}{( 4 11 )} = \frac{84}{330} = \frac{14}{55}$ .

Option (4).

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JEE Main 2026 Jan 28 Shift 1, Mathematics Q9: Total Probability Theorem And Bayes' Theorem

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