Question

There are two bags. One bag contains six green and three red balls. The second bag contains five green and four red balls. One ball is transferred from the first bag to the second bag. Then one ball is drawn from the second bag. Find the probability that it is a red ball.

Answer 1

first bag has 6 green balls and 3 red balls
second bag 5 green 4red ball
4÷9 is the probability

Answer 2

The probability that it is a red ball is 3/10, or 0.3.

• Using conditional probability, we can determine the likelihood that the ball selected from the second bag will be red after a ball has been moved from the first bag to the second.

• Find the likelihood that the transferred ball is red first. The likelihood of choosing a red ball is 1/3 given that the first bag includes three red balls and six green balls.

• When one ball is moved from the first bag to the second bag, a total of 10 balls are in the second bag (5 green and 5 red). If a red or green ball was transferred determines the likelihood of choosing a red ball from the second bag.

• The likelihood of choosing a red ball from the second bag is 4/10, or 2/5, if a green ball was transferred, meaning that the second bag will include 4 red balls out of 10 tot

• We must weigh each of these scenarios according to their relative probabilities in order to determine the overall likelihood of drawing a red ball from the second bag. We may apply the following calculation since the likelihood of transferring a green ball is 2/3 and the likelihood of transferring a red ball is 1/3:

• P(Red) is equal to P(Green transferred) times the product of P(Red pulled from second bag | Green transferred) and P(Red transferred) times the product of P(Red drawn from second bag | Red transferred).

• P(Red) equals (1/3 x 1/2) plus (2/3 x 2/5)
• P(Red) equals 4/15 plus 1/6, 9/30, and 3/10.

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