There were r red balls and y yellow balls in a bag. Three red balls and four yellow ones were added to the bag. What is the probability that a yellow ball and then a red ball will be selected if Jerry pulls out two balls at random, and puts the first ball back before pulling the second ball?
Choose the correct choice from the following answers: A) (y+4)/(y+r+7)*(r+3)/(y+r+7) B) (y+3)/(y+r+7) + (r+4)/(y+r+6) C) (y+3)/(y+r+7)*(r+4)/(y+r+6) D) (y+3)/(y+r+7)*(r+3)/(y+r+7) E) (y+3)/(y+r+7) + (r+3)/(y+r+6)
Currently we have r red and y yellow balls.   After adding new balls, we have   red = r + 3   Yellow = y + 4.   Total = r + y + 7   Now question says we are taking one ball first putting it back and then taking another ball. So, it means both the times, the number of ways of selecting one ball will be Total C 1 or (r+y+7)C1   So, Probability of selecting 1st as yellow and then 2nd as red = [(y+4)C1 / (r+y+7)C1] * [(r +3)C1 / (r+y+7)C1]   (y+4)/(y+r+7)*(r+3)/(y+r+7)nswer : option (y+4)/(y+r+7)*(r+3)/(y+r+7).
<result>A</result>
There were r red balls and y yellow balls in a bag. Three red balls and four yellow ones were added to the bag. What is the probability that a yellow ball and then a red ball will be selected if Jerry pulls out two balls at random, and puts the first ball back before pulling the second ball?