One bag contains 4 red balls and 3 blue balls and second bag contains 3 red and 5 blue balls . One ball is drawn at random from the first bag and put unseen in the second bag . What is the probability that a ball now drawn from the second bag is blue
Answers
Answer = 38/63
There are two ways that you can get a blue ball from the second bag
These are
Red from the first followed by blue from the second
Blue from the first followed by blue from the second
The probability that you draw a red from the first is 4/7 (7 balls in total and 4 are red)
Now assuming you have drawn a red when placed in the second bag there are now 4 red and 5 blue in the second bag. Therefore the probability that you draw blue given that a red was placed in it = 5/9, ie 5 blues out of 9 in total
Therefore P(red then blue) = 4/7 * 5/9 = 20/63
Now The probability that you draw a BLUE from the first is 3/7 (7 balls in total and 3 are blue)
Now assuming you have drawn a blue, when placed in the second bag there are now 3 red and 6 blue in the second bag. Therefore the probability that you draw blue given that a blue was placed in it = 6/9, ie 6 blues out of 9 in total
Therefore P(blue then blue) = 3/7 * 6/9 = 18/63
Therefore the probability that you draw a blue ball from the second bag
= P(red then blue) + P(blue then blue)
= 20/63 + 18/63
= 38/63
Answer = 38/63.
FYI the only other solutions are
Red from the first followed by red from the second = 4/7 * 4/9 = 16/63
Blue from the first followed by red from the second = 3/7 * 3/9 = 9/63
Check
Red + Blue = 20/63
Blue + Blue = 18/63
Red + Red = 16/63
Blue + Red = 9/63
Total = 63/63