In one urn there are 3 white and 5 black balls, and in the other 6 white and 6 black balls. 4 balls are taken

In one urn there are 3 white and 5 black balls, and in the other 6 white and 6 black balls. 4 balls are taken out of the first urn at random and lowered into the second urn. After that, one ball is also accidentally taken out of the second urn. Find the probability that all balls taken out of the second urn are white.

We have 4 hypotheses
Н1-removed 3 white and 1 black;
H2 – removed 2 white and 2 black;
H3 – removed 1 white and 3 black;
Н4-removed 0 white and 4 black.
Find the probability of each hypothesis
p (H1) = C33 C15 / C48 = 5/70;
p (H2) = C23 C25 / C48 = 30/70;
We count by the formula
Сn (m) = n! / ((n-m)! m!
A – an event meaning that a white ball is taken out of the second urn.

A / H1 – an event meaning that a white ball is removed from the second urn, provided that the event H1 has taken place
p (A / H1) = 9/16;
p (A / H2) = 8/16;
p (A / H3) = 7/16;
p (A / H4) = 6/16.
According to the formula of total probability

p (A) = p (H1) p (A / H1 + p (H2) p (A / H2) + p (H3) p (A / H3) + p (H4) p (A / H4 ) =

= (5/70) (9/16) + (30/70) (8/16) + (30/70) (7/16) + (5/70) (6/16) =

= (45 + 240 + 210 + 30) / 1120 = 525/1120 = 0.46875.
Answer: 0.46875