Question

Two bags contain 5 white, 3 black balls and 10 white, 3 black balls respectively.
Two balls are transferred at random from first bag to second bag and then a ball
is drawn at random from latter. The ball so drawn is found to be white in colour.
Find the probability that the transferred balls were both white

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

↝ Bag 1 contains 5 white and 3 black balls.

↝ Two balls are transferred at random from first bag to second bag and then a ball is drawn at random.

So, three cases arises.

↝ Case 1 :- Both balls are white in color

↝ Case 2 :- One is white and other is black

↝ Case :- 3 Both balls are black in color.

Let assume that,

$\rm :\longmapsto\:E_1 : transfer \: balls \: are \: white \: in \: color.$

$\rm :\longmapsto\:E_2 : one \: ball \: is \: white \: and \: other \:black \: in \: color.$

$\rm :\longmapsto\:E_3 : transfer \: balls \: are \: black \: in \: color.$

and

$\rm :\longmapsto\:E :getting \: white \: ball \: from \: second \: bag$

So,

$\red{\rm :\longmapsto\:P(E_1) = \dfrac{^5C_2}{^8C_2}}$

$\red{\rm :\longmapsto\:P(E_2) = \dfrac{^5C_1 \times \: ^3C_1}{^8C_2}}$

$\red{\rm :\longmapsto\:P(E_3) = \dfrac{^3C_2}{^8C_2}}$

Now, Two balls are transferred at random from first bag to second bag and a white ball is drawn at random from Bag 2,

So,

$\red{\rm :\longmapsto\:P(E | E_1) = \dfrac{12}{15} }$

$\red{\rm :\longmapsto\:P(E | E_2) = \dfrac{11}{15} }$

$\red{\rm :\longmapsto\:P(E | E_3) = \dfrac{10}{15} }$

Now, using Bayes, theorem, The probability of getting white ball from second bag when transferred balls from Bag is white in color is

$\red{\boxed{P(E_1|E) = \rm{ \: \dfrac{P(E_1) . P(E | E_1)}{P(E_1) . P(E | E_1) + P(E_2) . P(E | E_2) + P(E_3) . P(E | E_3)}}}}$

So, on substituting all the values evaluated above, we get

$\rm \: = \: \dfrac{\dfrac{^5C_2}{^8C_2} \times \dfrac{12}{15}}{\dfrac{^5C_2}{^8C_2} \times \dfrac{12}{15} + \dfrac{^5C_1 \times ^3C_1}{^8C_2} \times \dfrac{11}{15} + \dfrac{^3C_2}{^8C_2} \times \dfrac{10}{15} }$

$\rm \: = \: \dfrac{12 \times 5 \times 4}{12 \times 5 \times 4 + 11 \times 2 \times 5 \times 3 + 10 \times 3 \times 2}$

$\rm \: = \: \dfrac{240}{240 + 330 + 60}$

$\rm \: = \: \dfrac{240}{630}$

$\rm \: = \: \dfrac{24}{63}$

$\rm \: = \: \dfrac{8}{21}$