A bag contains 4 white balls and 5 black balls Another bag contains 3 white and 4 blak balls - Maths - Probability

Question

A bag contains 4 white balls and 5 black balls Another bag
contains 3 white and 4 blak balls A ball is taken out from the
first bag and without seeing its colour is put in the second bag A
ball is taken out from the latter Find the probability that the
ball drawn is white

Aakash Sharma · Accepted Answer

A white colour ball can be drawn from the second bag in two mutually exclusive ways:

(1) By transferring a white ball from first beg to the second beg and the drawing a white ball from it

(2) By transferring a black ball from the first beg to the second beg and then drawing a white ball from it.

Let E₁ , E₂ and A be the events defined as follows

E₁ = a white ball is transferred from first beg to the second

E₂ = a black ball is transferred from first beg to the second.

A = a white ball is drawn from the second beg

Since first beg contains 4 white and 5 black balls, we have

$P(E_{1}) = \frac{4}{9} and P (E_{2}) = \frac{5}{9}$

If E₁ has already occurred, i.e. a white ball has already been transferred from first beg to the second beg then the second beg contains 4 white and 4 black balls so

$P (\frac{A}{E_{1}}) = \frac{4}{8}$

If E₂ has already occurred i.e. a lack ball has been transferred from first beg to the second beg then second beg contains 3 white and 5 black balls so,

$P (\frac{A}{E_{2}}) = \frac{3}{8}$

By the law of probability we have

P( Getting a white ball) = P (A) = P (E₁) / P (A/E₁) + P (E₂) P (A/E₂)

$= \frac{4}{9} \times \frac{4}{8} + \frac{5}{9} \times \frac{3}{8} = \frac{31}{72}$

A bag contains 4 white balls and 5 black balls. Another bag contains 3 white and 4 blak balls. A ball is taken out from the first bag and without seeing its colour is put in the second bag. A ball is taken out from the latter. Find the probability that the ball drawn is white.