JEE Challenger
More from Probability (Advanced)
MathMedAdv 2019 • Paper 1

Conditional and Joint Probability of Choosing Green Balls from Bags

Consider three bags B1,B2B_1, B_2, and B3B_3. 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. The probabilities of selecting bags B1,B2B_1, B_2, and B3B_3 are 310\frac{3}{10}, 310\frac{3}{10}, and 410\frac{4}{10}, respectively. A bag is chosen at random, and then a ball is drawn uniformly at random from it. Which of the following statements is/are correct?

Options

A

Probability that the chosen ball is green, given that the selected bag is B3B_3, equals 38\frac{3}{8}

Correct
B

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

Correct
C

Probability that the selected bag is B3B_3, given that the chosen ball is green, equals 513\frac{5}{13}

D

Probability that the selected bag is B3B_3 and the chosen ball is green equals 310\frac{3}{10}

Topics & Concepts

Probability (Advanced)ProbabilityConditional ProbabilityBayes' Theorem

Step-by-Step Solution

To determine the correct options, we evaluate the relevant conditional and joint probabilities using the given bag compositions and selection probabilities:

The probabilities of choosing each bag are P(B1)=310P(B_1) = \frac{3}{10}, P(B2)=310P(B_2) = \frac{3}{10}, and P(B3)=410P(B_3) = \frac{4}{10}. The conditional probabilities of drawing a green ball GG from each bag are: P(G∣B1)=510=12,P(G∣B2)=58,P(G∣B3)=38P(G \mid B_1) = \frac{5}{10} = \frac{1}{2}, \quad P(G \mid B_2) = \frac{5}{8}, \quad P(G \mid B_3) = \frac{3}{8}

  1. For Option (A): The probability of drawing a green ball given bag B3B_3 is P(G∣B3)=38P(G \mid B_3) = \frac{3}{8}. Hence, option (A) is correct.

  2. For Option (B): By the law of total probability, P(G)=P(B1)P(G∣B1)+P(B2)P(G∣B2)+P(B3)P(G∣B3)=310(12)+310(58)+410(38)=3980P(G) = P(B_1)P(G \mid B_1) + P(B_2)P(G \mid B_2) + P(B_3)P(G \mid B_3) = \frac{3}{10}\left(\frac{1}{2}\right) + \frac{3}{10}\left(\frac{5}{8}\right) + \frac{4}{10}\left(\frac{3}{8}\right) = \frac{39}{80} Hence, option (B) is correct.

  3. For Option (C): Using Bayes' theorem, P(B3∣G)=P(B3∩G)P(G)=410×383980=1239=413P(B_3 \mid G) = \frac{P(B_3 \cap G)}{P(G)} = \frac{\frac{4}{10} \times \frac{3}{8}}{\frac{39}{80}} = \frac{12}{39} = \frac{4}{13} Hence, option (C) is incorrect.

  4. For Option (D): The joint probability is P(B3∩G)=P(B3)P(G∣B3)=410×38=320P(B_3 \cap G) = P(B_3)P(G \mid B_3) = \frac{4}{10} \times \frac{3}{8} = \frac{3}{20} Hence, option (D) is incorrect.

Therefore, the correct options are (A) and (B).

PreviousNext