In one urn there are 6 white and 4 black balls, in the second – 8 white and 2 black. Choose an urn at random

In one urn there are 6 white and 4 black balls, in the second – 8 white and 2 black. Choose an urn at random, choose one ball at random from the urn. It turns out to be white. What is the probability that this ball is removed from the first urn?

The first urn has a total of 6 + 4 = 10 balls.
The second urn has a total of 8 + 2 = 10 balls.
Let the hypotheses be accepted:
Hypothesis H1 – the ball is removed from the first urn.
Hypothesis H2 – the ball is taken out of the second urn.
Probabilities of hypotheses:
P (H1) = 0.5 = 1/2;
P (H2) = 0.5 = 1/2;
A – an event such that the taken out ball is white.
Conditional probability of event A when hypothesis H1 is fulfilled:
P (A | H1) = 6/10 = 3/5;
Conditional probability of event A when hypothesis H2 is fulfilled:
P (A | H2) = 8/10 = 4/5;
According to Bayes’ formula, the probability that the ball is removed from the first urn is:
P (H1 | A) = P (H1) P (A | H1) / (P (H1) P (A | H1) + P (H2) P (A | H2)) =
= 1/2 3/5 / (1/2 3/5 + 1/2 4/5) = (3/10) / (3/10 + 4/10) = 3/7 = 0.429.
Answer: The probability of hypothesis H1 that the ball was removed from the first urn is 0.429.