Posted by Anonymous on Thursday, September 9, 2010 at 12:30pm.
A container has three white balls and two red balls. A first ball is drawn at random and not replaces. Then a second ball is drawn. give the following conditions. What is the probability that the second ball was red? The first ball was white? The first ball was red?

math  MathMate, Thursday, September 9, 2010 at 1:27pm
1. First ball was red
No. of red balls: 2
Total number of balls:5
P(R)=2/5
2. First ball was white
No. of white balls: 3
Total number of balls: 5
P(W)=3/5
3. Second ball was red:
The probability is given by
P(R,R)+P(W,R)
Now we'll calculate P(R,R)
P(R)=2/5
After the first red, there is one more red left, and four balls altogether, therefore
P(RR)=(2/5)*(1/4)
=1/10
Similarly,
P(WR)=(3/5)*(2/4)
=3/10
Probability of the second ball being red is
(1/10)+(3/10)
=2/5
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