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.

Step-by-step 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)

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

step-by-step explanation:

the first 5 right here the next 5 i will upload as separate question

well we know there are 4 clubs in 52 cards so it’s replaced so it should be 3/51

step-by-step explanation: