A contains some red and blue balls. There are four more red than blue balls. A ball is removed at random and replaced. A second ball is removed. The probability that the two balls are different colours is 21/50. How many balls are in the box.
Ans
Drawing a tree diag gives you that
2p(r)xp(b) = 21/50
p(r)^2 + p(b)^2 = 29/50
The ans is that there are 7 red and 3 blue but how do you get there?
2 answers
Ok the way you do that is to multiply the red times the blue balls. There are 4 red balls. 3 X 4 = 12.
r = red , b = blue , t = total = r + b
r = b + 4 ... b = r - 4
... t = 2b + 4 = 2r - 4
4 possible outcomes
... r+r , b+b , r+b , b+r
p(r+b) = p(b+r) = (r*b) / t^2
2 [r(r-4)] / (2r - 4)^2 = 21/50
100 r^2 - 400r = 84 r^2 - 336r + 336
16 r^2 - 64r - 336 = 0 = r^2 - 4r - 21
... 0 = (r - 7)(r + 3) ... r = 7
t = 2r - 4 = 10
r = b + 4 ... b = r - 4
... t = 2b + 4 = 2r - 4
4 possible outcomes
... r+r , b+b , r+b , b+r
p(r+b) = p(b+r) = (r*b) / t^2
2 [r(r-4)] / (2r - 4)^2 = 21/50
100 r^2 - 400r = 84 r^2 - 336r + 336
16 r^2 - 64r - 336 = 0 = r^2 - 4r - 21
... 0 = (r - 7)(r + 3) ... r = 7
t = 2r - 4 = 10
5 answers