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Mathematics · Probability

JEE Advanced 2019 — Paper 1 — Question 40

There are three bags B1,B2B_{1}, B_{2} and B3B_{3}. The bag B1B_{1} contains 5 red and 5 green balls, B2B_{2} contains 3 red and 5 green balls and B3B_{3} contains 5 red and 3 green balls. Bags B1,B2B_{1}, B_{2} and B3B_{3} have probabilities 310,310\frac{3}{10}, \frac{3}{10} and 410\frac{4}{10} respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?

  1. Option A:

    Probability that the chosen ball is green equals 3980\frac{39}{80}

    Correct
  2. Option B:

    Probability that the chosen ball is green, given that the selected bag is B3B_{3}, equals 38\frac{3}{8}

    Correct
  3. Option C:

    Probability that the selected bag is B3B_{3} and the chosen ball is green equals 310\frac{3}{10}

  4. Option D:

    Probability that the selected bag is B3B_{3}, given that the chosen ball is green, equals 513\frac{5}{13}

Answer: A, B

Step-by-step solution

P(B1)=310,P(B2)=310,P(B3)=410\mathrm{P}\left(\mathrm{B}_{1}\right)=\frac{3}{10}, \mathrm{P}\left(\mathrm{B}_{2}\right)=\frac{3}{10}, \mathrm{P}\left(\mathrm{B}_{3}\right)=\frac{4}{10}

(A) P(G)=P(B1)×P(GB1)+P(B2)×P(GB2)+P(B3)×P(GB3)\mathrm{P}(\mathrm{G})=\mathrm{P}\left(\mathrm{B}_{1}\right) \times \mathrm{P}\left(\frac{\mathrm{G}}{\mathrm{B}_{1}}\right)+\mathrm{P}\left(\mathrm{B}_{2}\right) \times \mathrm{P}\left(\frac{\mathrm{G}}{\mathrm{B}_{2}}\right)+\mathrm{P}\left(\mathrm{B}_{3}\right) \times \mathrm{P}\left(\frac{\mathrm{G}}{\mathrm{B}_{3}}\right)

=310×510+310×58+410×38=\frac{3}{10} \times \frac{5}{10}+\frac{3}{10} \times \frac{5}{8}+\frac{4}{10} \times \frac{3}{8} =60+75+60400=195400=3980=\frac{60+75+60}{400}=\frac{195}{400}=\frac{39}{80}

(B) P(GB3)=38\quad \mathrm{P}\left(\frac{\mathrm{G}}{\mathrm{B}_{3}}\right)=\frac{3}{8}

(C) P(B3G)=P(B3)×P(GB3)P(G)=410×383980=413\mathrm{P}\left(\frac{\mathrm{B}_{3}}{\mathrm{G}}\right)=\frac{\mathrm{P}\left(\mathrm{B}_{3}\right) \times \mathrm{P}\left(\frac{\mathrm{G}}{\mathrm{B}_{3}}\right)}{\mathrm{P}(\mathrm{G})}=\frac{\frac{4}{10} \times \frac{3}{8}}{\frac{39}{80}}=\frac{4}{13}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 1
Subject
Mathematics
Chapter
Probability
Topic
Conditional Probability and Multiplication Theorem

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