Question

One bag contains 5 white and 4 black balls. Aniother bag contains 7 white and 9 black balls. A ball is transferred from the first bag to the second and then a ball is drawn from the second. Find the probability of getting a white ball

Answer 1

Let white is transferred from 1 st bag

Now P( white ball from 1 st bag) = 5/9

Now after transferring , there is 8 white and 9 black

So P( white ball from second) = 8/ 17

Now We have to find

P( white ball from second and white or black from first)

Let's take first P( white ball from second and white is transferred from first)

That is intersection

As P( A/B) = P( A^B)/ P( B)

P( A^B) = P( A/B) × P( B)

So P ( white ball from second and white is transferred from first) = 5/9 × 8/17 = 40/ 153


Now P( white ball drawn when black is transferred from first) = 7/17

P ( white black drawn and black from first) = 7/17 × 4/9 = 28/153


Now P( white ball is drawn) = 40/153 + 28/153

= 68/153

= 4/9

Answer 2

Solutions :-

Given :
One bag contains 5 white and 4 black balls.
Another bag contains 7 white and 9 black balls.
A ball is transferred from the first bag to the second and then a ball is drawn from the second.

P(a white ball transferred from one bag to another bag) = 5/9

P(a black ball transferred from one bag to another bag) = 4/9

P(getting a white ball from another bag when a white ball is transferred) = 8/17

P(getting a black ball from another bag when a black ball is transferred) = 7/17

Now,
P(getting a white ball) = (5/9 × 8/17) + (4/9 × 7/17)
= (40/153) + (28/153)
= (40 + 28)/153
= 68/153 = 4/9

Hence,
Probability of getting a white ball = 4/9