Question

The first bag contains 3 white balls, 2 red balls and 4 black balls. Second bag contains 2 white. 3 red and 5 black balls and third bag contains 3 white. 4 red and 2 black balls. One bag is chosen at random and from it 3 balls are drawn. Out of three balls two balls are white and one red. What are the probabilities that they were taken from first bag, second bag and third bag.​

Answer 1

Answer:

5/28

Step-by-step explanation:

Probability of not getting red ball in first draw =  8/6

Probability of not getting red ball in second draw =  5/7

And probability of getting red ball in third draw =  6/2

Hence required probability =  

8/6  ×  7/5×  6/2  =  5/28

Answer 2

Answer:

5/28

Step-by-step explanation:

Bag 1

White balls = 3

Red balls= 2

Black balls= 4

Bag 2

White balls= 2

Red balls = 3

Black balls= 5

So now,

The probability of not getting red ball in first draw =  8/6

The probability of not getting red ball in second draw =  5/7

And probability of getting red ball in third draw =  6/2

Hence the possibility of taking the balls from first bag are :-    

8/6  ×  7/5×  6/2  =  5/28

So, the final answer is 5/28.