PROBLEM 16E

Bowl A contains three red and two white chips, and bowl B contains four red and three white chips. A chip is drawn at random from bowl A and transferred to bowl B. Compute the probability of then drawing a red chip from bowl B.

Solution:

Step 1 of 3 :

Here bowl A contains 3 red and 2 white chips bowl B contains 4 red and three white chips . If chip is drawn at random from bowl A and transferred to bowl B. we have to Compute the probability of drawing a red chip from bowl B.

Step 2 of 3:

It is given that

Bowl A Bowl B

No of red chips = 3 No of red chips= 4

No of white chips = 2 No of white chips = 3

It is also given that a chip is drawn randomly from bowl A and transferred to bowl B.

So then there will be two chances that drawing a red chip from Bowl B.

- Drawing a white chip from bowl A and transferred to bowl B and then drawing a red chip from bowl B.
- Drawing a red chip from bowl A and transferred to bowl B and then drawing a red chip from bowl B.

By bayes theorem probability of drawing a red chip from bowl B will be the addition of these two probabilities.

Let define the events

A = drawing a red chip from Bowl B after transferring a chip from bowl A to B which is randomly chosen.

W= a white chip transferred from bowl A to B.

R = a red chip transferred from bowl A to B.

So

P(R)= 3/5

P(W) = 2/5

P(A) = P(W) P(A/W) +P(R)P(A/R)

P(A/W)= P(that a red chip is drawn from bowl B given that a white bowl is transferred from A to B .

= 4/8

P(A/R) = P(that a red chip is drawn from bowl B given that a red bowl is transferred from A to B

= 5/8

So the probability that a red chip is drawn from bowl B given that a red bowl is transferred from A to B is 5/8.

Step 3 of 3:

Then

Probability of drawing a red chip from bowl B

P(A) = (⅗ 5/8) + (2/5 4/8)

= 15/40 + 8/40

= 23/40

= 0.575

So the probability that drawing a red chip from bowl B is 0.57.