Question

EX57
. A bag contains one black ball and two white balls. A drawing from the bag consists

of taking a ball from the bag and keeping it out of it if it is white, but putting it back if
black. Calculate the probabilities that
(1) the first drawing is a white ball,
(ii) the second drawing is a white ball,
(ii) the third drawing is a white ball.
(Type ISC 2016​

Answer 1

Answer:

I am giving below the method of solving such problems in a systematic manner.

A ball is drawn three times. What are the different routes available? They are as under:

This is similar to tossing a coin three times, with which we are all familiar.

In the table below, the different columns represent the following:

Col 1 - The route.

Col 2 - The initial mix i.e. the number of black balls and the number of white balls.

Col 3 - The probability of drawing the first ball.

Col 4 - The mix after the first ball has been drawn. If the ball drawn is black, the mix does not change since the black ball is replaced. If the ball dawn is white, the number of white balls in the mix reduces by 1.

Col 5 - The probability of drawing the second ball based on the new mix.

Col 6 - The mix after the second ball is drawn.

Col 7 - The probability of drawing the third ball based on the new mix.

Col 8 - The net probability for each route. This is the product of the three probabilities for each route determined in columns 3, 5 and 7.

We want the probability that the third ball drawn is white. This is the sum of the net probabilities for each route where the third ball drawn is white i.e for routes BBW, BWW, WBW and WWW.

Since we are interested only in these four routes, you may, if you so desire, put only these four rows in the table and not include the rows where the last ball drawn is black.

Therefore, the required probability is  227+19+16+0=1954≈35%