Question

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?

Answer 1

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