An urn contains 2 red balls and 3 blue balls. A ball is drawn at random and its color noted. If it is red, then it is put back into the urn; otherwise, it is not put back into the urn. Then a second ball is drawn at random from the urn and its color noted. What's the probability that both balls are blue given that at least one is blue?
the anwser is 5/14 why????
my solution is :
both blue=3/5*2/5=6/25
at least one blue= 1- 2/5*1/4=9/10
and (6/25) / (9/10) and it's wrong
the anwser is 5/14 why????
my solution is :
both blue=3/5*2/5=6/25
at least one blue= 1- 2/5*1/4=9/10
and (6/25) / (9/10) and it's wrong
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The possibilities are drawing RR, RB, BR, BB
R denotes Red and B denotes blue ball.
P(RR) = P(R)*P(R) = 2/5*2/5 = 4/25
P(RB) = P(R)*P(B) = 2/5*3/5 = 6/25
P(BR) = P(B)*P(R) = 3/5*2/4 = 6/20
P(BB) = P(B)*P(B) = 3/5*2/4 = 6/20
P(drawing at leastt one blue) = 1 - P(RR) = 1 - 4/25 = 21/25
Required probability = (6/20) / (21/25)
= 6/20 * 25/21
= 5/14 or 0.3571
R denotes Red and B denotes blue ball.
P(RR) = P(R)*P(R) = 2/5*2/5 = 4/25
P(RB) = P(R)*P(B) = 2/5*3/5 = 6/25
P(BR) = P(B)*P(R) = 3/5*2/4 = 6/20
P(BB) = P(B)*P(B) = 3/5*2/4 = 6/20
P(drawing at leastt one blue) = 1 - P(RR) = 1 - 4/25 = 21/25
Required probability = (6/20) / (21/25)
= 6/20 * 25/21
= 5/14 or 0.3571