Question

Two urns contains 10 white,6 red , 9 black balls and 3 white,7 red and 15 black balls. One ball

is drawn from each urn. Find the probability that both balls are red.​

Answer 1

Ways to draw 2 balls, without replacement, from 11 balls = 11!/(9!)(2!) = 55.

Ways to draw 2 red balls from 5 = 5!/(3!)(2!) = 10.

Ways to draw 2 white balls from 6 = 6!/(4!)(2!) = 15.

Ways to draw 1 red and 1 white ball = 55-(10+15)= 30.

Probability of 2 red balls to 2nd urn = 10/55 = 2/11.

Probability of 2 white balls to 2nd urn = 15/55 = 3/11.

Probability of 1 red and 1 white ball to 2nd urn = 30/55 = 6/11 .

The 2nd urn now has 12 balls with 3 possible cases of balls distribution, each with a probability of occurrence as follows:

Case 1… 5 red and 7 white balls, with a probability of occurrence = 2/11;

Case 2… 3 red and 9 white balls with a probability of occurrence = 3/11;

Case 3… 4 red and 8 white balls with a probability of occurrence =6/11.

Ways to draw 2 balls from 12, without replacement = 12!/(2!)(10!) = 66.

Case 1… probability of 2 red balls being drawn = {[5!/(3!)(2!)]/66}*(2/11) = 20/726 = 10/363 =.027548….~ 2.75%.

Case 2… probability of 2 red balls being drawn = {[3!/(1!)(2!)]/66}*(3/11) = 9/726 = 3/242= .0123966….~ 1.24% .

Case 3… probability of 2 red balls being drawn = {[4!/(2!)(2!)]/66}*(6/11) = 36/726 = 6/121= .049586….~ 4.96% .

Therefore, in response to the question, the probability of randomly drawing 2 red balls from the 2nd urn subsequent to receiving 2 balls randomly drawn from the 1st urn is at least ~ 1.24% and at most ~4.96%.

Edit: The probabilities as shown are based on the outcome of a single draw of 2 red balls. If the routine as described above regarding 2 urns with certain red and white balls distributions, including drawing 2 balls from the 2nd urn, were to be repeated many times over and over, the expectation would be to see the outcome of 2 red balls approach the sum of the above probabilities = .027548…+ .0123966…+ .049586… = .0895316…~ 8.95% of the time and the sum of outcome probabilities respective to 2 white balls and 1 white plus 1 red ball approach (1 - .0895306…) = .9104694…~ 91.05% of the time