A bag contains white, black and …

Let probability of black ball, red ball and white ball be P(B), P(R) and P(W) respectively.

According to question,

P(B) + P(R) + P(W) = 1

=>          2/5 + P(R) + 3/10 = 1

=>          P(R) = 1 - (2/5 + 3/10)

=>          P(R) = 1 - 7/10 = 3/10

Now, P(B) = No. of black balls / Total balls

=>          Total balls = No. black balls / P(B)

                                = 20 / (2/5)

                                = 50

Therefore, No. of White balls = Total balls x P(W) = 50 x 3/10 = 15 balls

And           No. of Red balls = Total balls x P(R) = 50 x 3/10 = 15 balls

Ans. Total Number of Balls = 20

Number of Red Balls = R

Number of White Balls = W

Number of Black Balls = B

Probability of Getting White Ball = 3/10

Probability of Getting Black Ball = 2/5

We Know,

sum of Probability of all event = 1

=> 3/10 + 2/5 + Probability of Red Ball = 1

=> Probability of Red Ball = 1 - 3/10 - 2/5 

=> probability of Red Ball = 3/10

Number of White Balls = (Probability of whtite ball)* Total Balls

=> Number of White Balls = (3/10)*20 = 6

Number of Red Balls = (3/10)*20 = 6