Box 1 contains 3 blue and 5 red balls, whereas Box 2 contains 2 blue and 4 red ball. A ball is randomly chosen from Box 1 and then transferred to Box 2, and a ball is then randomly selected from Box 2. What is the conditional probability that the transferred ball was blue, given that a red ball is selected from Box 2 (round off to second decimal place)?

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MechEngineer MechEngineer · Answer 1 · 2020-03-10T08:32:13+01:00

Answer:

0.32

Step-by-step explanation:

Let A be the event that the transferred ball was blue.

B be the event that the red ball is selected from box 2

C be the event that the transferred ball was red

Then:

P(A) be the probability that the transferred ball was blue, which is 3/8

P(C) be the probability that the transferred ball was red, which is 5/8

P(B|A) is the probability that selected ball is red, given that the transferred ball was blue, which is 4/7

P(B|C) is the probability that selected ball is red, given that the transferred ball was red, which is 5/7

P(B) is the probability that the selected ball is red, which there are 2 scenarios, given that the transferred ball is blue and red

P(B) = P(B|A)P(A) + P(B|C)P(C)

[tex]P(B) = \frac{4}{7}\frac{3}{8} + \frac{5}{7}\frac{5}{8}[/tex]

[tex]P(B) = \frac{3}{14} + \frac{25}{56} = \frac{37}{56}[/tex]

P(A|B) is the probability that the transferred ball was blue, given that a red ball is selected from Box 2, which can be solved using Bayes theorem

[tex]P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{\frac{4}{7}*\frac{3}{8}}{\frac{37}{56}} = \frac{3}{14}\frac{56}{37} = \frac{12}{37} = 0.324[/tex]