Question

An urn contains 10 red and 3 black balls. Another urn contains 3 red and 5 black
balls. Two balls are transferred from the 1st urn to the second urn without noticing
the color. One ball is drawn from the second urn and it is found to be red. What is
the probability that 1 red and 1 black ball were transferred?

Answer 1

There are three mutually exclusive and exhaustive ways in which 2 balls can be transferred from first bag to second bag.

First way: Two white balls are transferred from first bag to second bag, probability of which is 13C210C2

In the second bag we now have 5 white and 5 black balls and probability of getting a white ball is 105,

∴ Required probability=13C210C2×105=7845×105=780225

Second way: two black balls are transferred from first bag to the second bag

probability of which is 13C23C2

Now we have 3 white and 7 black balls in second bag and probability of getting a white ball is 103

∴ Required probability13C23

Answer 2

$\huge{\mathfrak\green{♡Answer}}$

Before transferring, the odds are 3:5 against

grabbing 2 whites from the first has a 10\13 * 9/12 chance which is 90/156 or 15/26

If 2 whites are picked, the ratio is 5:5 or 1:1 or 1/2 chance

If 2 reds, it is 3:7 or 3/10.

The range is 3/10 to 5/10(1/2), and we are 15/26 in between.

1/26 of 2/10 (the difference of 5/10&3/10) is 1/130

15 /26 of 2/10 is 3/26

3/26 + 3/10 or (3/2.6)/(26/2.6) + 3/10

(15/13)/10 + 3/10 = (4&2/13)/10 or (54/13)/10

(54/13)/(130/13)=54/130

130–54=76, ratio is 54:76 or 27:38

Khadijah21