Question

The contents of three urns are as the contents of three urns are as follows: urn 1 : 7 white, 3 black balls, urn 2 : 4 white, 6 black balls, and urn 3 : 2 white, 8 black balls. One of these urns is chosen at random with probabilities 0.20, 0.60 and 0.20 respectively. From the chosen urn two balls are drawn at random without replacement. If both these balls are white, what is the probability that these came from urn 3?

Answer 1

➡️Answer:0.025

➡️Solution:

✔️urn 1 : 7 white, 3 black balls

✔️urn 2 : 4 white, 6 black balls

✔️urn 3 : 2 white, 8 black balls

let the event of choosing urn 1 is E1,urn 2 is E2

and urn 3 is E3

it is given that

p(E1) =0.20

p(E2) =0.60

p(E3) =0.20

From the chosen urn two balls are drawn at random without replacement.

Probability of choosing first White ball,from urn 1

p(W1/E1) =7/10

Probability of choosing second White ball,from urn 1, without replacement

p(W2/E1) = 6/9

Probability of choosing first White ball,from urn 2

p(W1/E2) =4/10

Probability of choosing second White ball,from urn 2, without replacement

p(W2/E2) = 3/9

Probability of choosing first White ball,from urn 3

p(W1/E3) =2/10

Probability of choosing second White ball,from urn 3, without replacement

p(W2/E3) = 1/9

If both these balls are white, what is the probability that these came from urn 3:

Apply Bay's theorem:

$\frac{p(E3)p(W1)p(W2)}{p(E1)p(W1)p(W2) + p(E2)p(W1)p(W2) + p(E3)p(W1)p(W2)} \\ \\ = \frac{0.20 \times \frac{2}{10} \times \frac{1}{9} }{0.20 \times \frac{7}{10} \times \frac{6}{9} + 0.60 \times \frac{4}{10} \times \frac{3}{9} + 0.20 \times \frac{2}{10} \times \frac{1}{9}} \\ \\ = \frac{0.40}{42 \times 0.2 + 12 \times 0.60 + 0.20 \times 2} \\ \\ = \frac{0.40}{8.4 + 7.2 + 0.4} \\ \\ = \frac{0.4}{16} \\ \\ = 0.025$

Probability of 2 white balls choosing from urn3 is p(E) = 0.025

Hope it helps you.