Mathematics · Probability

JEE Advanced 2024 — Paper 2 — Question 9

A bag contains NN balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For i=1,2,3i=1,2,3, let Wi,GiW_{i}, G_{i}, and BiB_{i} denote the events that the ball drawn in the ith \mathrm{i}^{\text {th }} draw is a white ball, green ball, and blue ball, respectively. If the probability P(W1∩G2∩B3)=25NP\left(W_{1} \cap G_{2} \cap B_{3}\right)=\frac{2}{5 N} and the conditional probability P(B3∣W1∩G2)=29P\left(B_{3} \mid W_{1} \cap G_{2}\right)=\frac{2}{9}, then NN equals \qquad

Answer: 11

Numerical answer — enter this value.

Step-by-step solution

P(B3∣W1∩G2)=P(B3∩W1∩G2)P(W1∩G2)=29P\left(B_{3} \mid W_{1} \cap G_{2}\right)=\frac {P\left(B_{3} \cap W_{1} \cap G_{2}\right)}{P\left(W_{1} \cap G_{2}\right)}=\frac{2}{9}

⇒3N×N−9N−1×6N−23N×6N−1=29\Rightarrow \frac{\frac{3}{N} \times \frac{N-9}{N-1} \times \frac{6}{N-2}}{\frac{3}{N} \times \frac{6}{N-1}}=\frac{2}{9}

⇒3×(N−9)×6( N−2)×18=29\Rightarrow \frac{3 \times(\mathrm{N}-9) \times 6}{(\mathrm{~N}-2) \times 18}=\frac{2}{9}

9 N−81=2 N−4;7 N=779 \mathrm{~N}-81=2 \mathrm{~N}-4 ; 7 \mathrm{~N}=77 ∴ N=11\therefore \mathrm{~N}=11

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 2
Subject
Mathematics
Chapter
Probability
Topic
Conditional Probability and Multiplication Theorem