Question

In a bag, there are 2 blue balls, 5 green balls and 6 red balls. If a person took two balls without replacement, what is the probability that the second ball will be red given that the first ball is red?

• A. $\frac{5}{12}$
• B. $\frac{1}{2}$
• C. $\frac{6}{13}$
• D. $\frac{30}{156}$

Answer 1

The correct option is A $\frac{5}{12}$

Let E and F denote the following events:
E: 'the first ball is red'.
F: 'the second ball is red'.
We have to find P(F|E).
Total number of balls = 2+5+6 = 13
Then,
$P\left(E\right)=\frac{6}{13}\mathrm{and}P\left(F\right)=\frac{5}{12}$
Now, since there is no replacement,
$P\left(E\cap F\right)=\frac{6}{13}\times \frac{5}{12}$
Then,
$P\left(F|E\right)=\frac{P\left(F\cap E\right)}{P\left(E\right)}\Rightarrow P\left(F|E\right)=\frac{\frac{6}{13}\times \frac{5}{12}}{\frac{6}{13}}=\frac{5}{12}$