In the first urn there are 2 white and 4 blue balls, in the second – 5 white, 1 blue.

In the first urn there are 2 white and 4 blue balls, in the second – 5 white, 1 blue. What is the probability of drawing a white ball from a randomly chosen urn.

1. Hypotheses and corresponding prior probabilities:

A1 – the first urn is selected;
A2 – second urn selected;
P (A1) = 1/2;
P (A2) = 1/2.
2. The conditional probabilities of event B are that a white ball is drawn from a randomly chosen urn:

P (B | A1) = 2 / (2 + 4) = 2/6;
P (B | A2) = 5 / (5 + 1) = 5/6.
3. Events A1 and A2 constitute a complete group of incompatible events, therefore, the total probability of event B is equal to:

P (B) = P (A1) * P (B | A1) + P (A2) * P (B | A2);
P (B) = 1/2 * 2/6 + 1/2 * 5/6 = 1/2 * (2/6 + 5/6) = 1/2 * 7/6 = 7/12.
Answer: 7/12.