Question

6. There are 5 red balls and 7 blue balls in an urn. Two balls are drawn consecutively without replacement. What is the probability that the first ball drawn is red given that the second ball drawn is also red?

Answer 1

Solution:

P(first ball is red given that second ball is also red) = 0.3636

Explanation:

Given that,

Number of red balls = 5

Number of blue balls = 7

Total balls = 12

Two balls are drawn consecutively without replacement.

Now the probability that first ball is red given that second ball is also red is given by,

Probability that second ball is red = $\frac{5}{12}\times \frac{4}{11}+\frac{7}{12}\times \frac{5}{11}$

Probability that second ball is red = $\frac{55}{132}$

And,

Probability that first ball is red and second ball is red = $\frac{5}{12}\times \frac{4}{11}=\frac{20}{132}$

So,

The probability that first ball is red given that second ball is also red is given by,

P(first ball is red given that second ball is also red) = P(Probability that first ball is red and second ball is red) / P(Probability that second ball is red)

P(first ball is red given that second ball is also red) = $\frac{\frac{20}{132}}{ \frac{55}{132}}$

P(first ball is red given that second ball is also red) =

20 55 - 0.3636

P(first ball is red given that second ball is also red) = 0.3636

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