Question

Three bags contain 3 red, 7 black; 8 red, 2 black, and 4 red & 6 black balls

respectively. 1 of the bags is selected at random and a ball is drawn from it. If the ball
drawn is red, find the probability that it is drawn from the third bag.

Answer 1

4/15

Explanation:

Solution of conditional probability problem:

Given:

Bags (3R,7B), (8R,2B), (4R,6B)

Let

P(R,i) = probability of drawing a red AND from bag i

P(R, 1) = 3/10 * (1/3) = 3/30

P(R, 2) = 8/10 * (1/3) = 8/30

P(R, 3) = 4/10 * (1/3) = 4/30

Let

Let P(R) = probability of drawing a red from any bag

P(R) = sum P(R,i) for i = 1 to 3 using the addition rule

= 3/30 + 8/30 + 4/30

= 15/30

= 1 / 2

Conditional Probability of drawing from the third bag GIVEN that it is a red

= P(3 | R)

= P(R, 3) / P(R)

= 4/30 / (1/2)

= 8/30

= 4 / 15

(Since all bags contain 10 balls, by intuition, 4 red from third / 15 total red = 4/15)

Answer 2

`Probability = (4)/(15)`

Explanation:

B1 = first bag

B2= second bag

B3 = third bag

Let A = ball drawn is red

Since, there are three bags.

Probability of choosing one bag= P(B1) = P(B2) = P(B3) = 1/3.

From B1: Total balls = 10

3 red + 7 black balls.

Probability of drawing 1 red ball from it , P(A) = 3/10.

From B2: Total balls = 10

8 red + 2 black

Probability of drawing 1 red ball is, P(A) = 8/10

From B3 : Total Balls = 10

4 red + 6 black

Probability of drawing 1 red ball, P(A) = 4/10 .

To find Probability given that the ball drawn is red, that the ball is drawn from the third bag by Bayes' rule.

That is , P(B3|A)

`=\frac{(1)/(3) * (4)/(10)} { (1)/(3) * (3)/(10) + (1)/(3) *(8)/(10) + (1)/(3) * (4)/(10)}`

`=(4)/(30) * (30)/(15)\\\\=(4)/(15)`

Therefore, the probability that it is drawn from the third bag is 4/15.