Question

A bag contains 2 white and 4 black balls while another bag contains 6 white and 4 black balls. A bag is selected at random and a ball is drawn

Answer 1

Answer:

Step-by-step explanation:

Lets say bag1 has 2 white balls and 4 black balls while bag 2 has 6 white balls and 4 black balls.  

P (bag1=white ball)= 2/6;  

P (bag1=black ball)=4/6;  

P (bag2=white ball)= 6/8;  

P (bag2=black ball)= 4/8;  

Now the condition of one white and one black ball is met if we get one from either of the bag and the other from the remaining bag.  

Case1:  

Lets say we get white ball from bag 1 and black from bag 2. Then the probability of that being true is 2/6×4/8=8/48=1/6.  

Case2:  

Lets say we get white ball from bag 2 and black from bag 1. Then the probability of that being true is 6/8×4/6=24/48=1/2.  

Since both the cases are possible, the probabilities will need to be added to arrive at the total probability.  

Thus 1/6+2/6= 3/6=1/2.