Question

Vessel a contains 6 green and 4 red balls and vessel b contains 4 green and 6 red balls. 1 ball is drawn at random from vessel a and placed in vessel b.Then 1 ball is transferred at random from vessel b to vessel a. If 1 ball is now drawn at random from vessel a, the probability that it is green is__

Answer 1

Solution:

Number of balls in vessel A= 6 green + 4 red= 10 balls

Number of balls in vessel B= 4 green + 6 red= 10 balls

Case 1: If  1 ball is drawn at random from vessel A and placed in vessel B.

So, Number of balls in Vessel B= 11 Balls[Either (5 green + 6 red) or (4 green + 7 red)]

Number of balls in vessel A = 9 Balls[Either (5 green + 4 red) or (6 green + 3 red)]

Case 2 : Now, if 1 ball is transferred at random from vessel B to vessel A.

So, Number of balls in Vessel B= 10 Balls[Either (5 green + 5 red) or (4 green + 6 red) or (3 green + 7 red) ]

Number of balls in vessel A = 10 Balls[Either (5 green + 5 red) or (6 green + 4 red) or ( 7 green + 3 red) ]

Now, coming to the question

1 Ball is drawn at random from Vessel A, probability that it is green ball is:

      = $\frac{_{1}^{5}\textrm{C}}{_{1}^{10}\textrm{C}} {\text{or}}\frac{_{1}^{6}\textrm{C}}{_{1}^{10}\textrm{C}} {\text{or}}\frac{_{1}^{7}\textrm{C}}{_{1}^{10}\textrm{C}} \\\\ \frac{1}{2}{\text{or}}\frac{3}{5}{\text{or}}\frac{7}{10}$