Question

A bag contains 6 white and 4 red balls. One ball is drawn at random and put aside without
noticing its colour. Now another ball is drawn from the bag at random. What is the
probability that the second ball is white

Answer 1

Solution. Suppose event A is a second ball is white; B1 is the first ball is white; B2 is the first ball is red. Using the Law of Total Probability for our case get

$$P(A)=P(A/B_1)P(B_1)+P(A/B_2)P(B_2)$$

According to the condition of the problem

$$P(B_1)=\frac {6} {10}$$

$$P(B_2)=\frac {4} {10}$$

$$P(A/B_1)=\frac {5} {9}$$ $$P(A/B_2)=\frac {6} {9}$$

As a result, we get

$$P(A)=\frac {6} {10} *\frac {5} {9}+\frac {4} {10} *\frac {6} {9}=\frac{30} {90}+\frac{24} {90}=\frac{54} {90}=0.6$$

Answer. 0.6