One box contains 7 red and 4 green balls, and the second box contains 3 red and 5 green balls. one ball is drawn from the first box and placed unseen in the second box. what is the probability that a ball now drawn from the second box is red

The probability that the ball drawn from the second box is red is 40/99

Step-by-step explanation:

We can divide the event that the second ball drawn is red into 2 disjoint events, and obtain its probability by summing both probabilities for the disjoint cases.

For the first case, we have 11 balls in the first box, 7 of them being red. Thus, the probability of drawing a red ball is 7/11. After that, we will put the red ball drawn in the second box, so that it will have 9 balls in total, 4 of them being red. As a consecuence, the probability of drawing a red ball in this instance is 4/9. We need to multiply that probability by the probability that a red ball was drawn first, so we have a probability of 7/11 * 4/9 = 28/99 that the balls drawn are both red.

Now, lets go to the second case. The probability of drawing a green ball from the first box is 4/11, and after placing that ball in the second box, the probability that the new ball drawn is red is 3/9 = 1/3; this is because we will have 3 red balls out of a total of 9 (6 are green). Thus, the probability of this second event is 4/11 * 1/3 = 4/33.

Combining both probabilities, the probability that the ball drawn in the second box is red is 28/99 +4/33 = 40/99 = 0.40404040404...

answer: c is correct, a regular pentagon

answer: 590

step-by-step explanation: