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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