Question

Two bags A and B contain 4 white balls and 3 black balls ,and 2 white balls and 2 black balls respectively .From bag A two balls are transferred to bag B.find the probability of drawing

1.2 white ball from bag B

2. 2 black balls from bag B

3.1 white ball and 1 black ball from bag B

Answer 1

i have doubt in 3 answer because prob can't greater than 1 ... please correct me if other 2 answers are wrong

Answer 2

Use Law of Total Probability

Step-by-step explanation:

A white ball & a black ball can be drawn from the second bag in the following mutually exclusive ways-

(A) By transferring 2 black balls from bag A to the bag B and then drawing a white and a black ball from it

(B) By transferring 2 white balls from bag A to the bag B and then drawing a white and a black ball from it

(C) By transferring 1 white and 1 black ball from bag A to the bag B and then drawing a white and a black ball from it

Let $E_1. E_2, E_3$ and A be the events as defined below-

$E_1$ = two black balls are drawn from Bag A

$E_2$ = two white balls are drawn from Bag A

$E_3$ = one white and one black ball is drawn from Bag A

A = two balls drawn from the bag B are white and black

We have,  

$P(E_1)= \frac {^3c_{2}}{^7c_{2}} = \frac {3}{21} = \frac {1}{7}$

$P(E_2)= \frac {^4c_{2}}{^7c_{2}} = \frac {6}{21} = \frac {2}{7}$

and $P(E_3)= \frac {^3c_{1} \times ^4c_{1} }{^7c_{2}} = \frac {12}{21} = \frac {4}{7}$

If $E_1$ has already occurred, that is, if two black balls have been transferred from bag A to bag B, then the second bag will contain 2 white and 4 black balls, therefore the probability of drawing a white and a black ball from the second bag is  

$\frac {^2c_{1} \times ^4c_{1}}{^6c_{2}}$

Hence, $P(\frac {A}{E_1}) = \frac {^2c_{1} \times ^4c_{1}}{^6c_{2}} = \frac {8}{15}$

Similarly, we have ,

$P(\frac {A}{E_2}) = \frac {^4c_{1} \times ^2c_{1}}{^6c_{2}} = \frac {8}{15}$

and $P(\frac {A}{E_3}) = \frac {^3c_{1} \times ^3c_{1}}{^6c_{2}} = \frac {3}{5}$

By the law of total probability, we have -

$P(A) = P(E_1)P(\frac {A}{E_1}) + P(E_2)P(\frac {A}{E_2}) + P(E_3)P(\frac {A}{E_3})$

= $\frac {1}{7} \times \frac {8}{15} + \frac {2}{7} \times \frac {8}{15} + \frac {4}{7}\times\frac {3}{5}$

= $\frac {60}{105}$