The urn contains 30 white and 10 red balls. Find the probability that the first and second lucky

The urn contains 30 white and 10 red balls. Find the probability that the first and second lucky balls chosen will be white, and the third red.

Total balls in the urn:
30 + 10 = 40.
The probability of being the first to pull out the white ball at random is:
p1 = 30/40 = 3/4.
There will remain 39 balls, 29 of which are white.
Probability of drawing the second white ball:
p2 = 29/39.
38 balls remain, of which 10 are red.
The probability of drawing the third red ball is equal to:
p3 = 10/38.
These are independent events, their probability is equal to the product of probabilities:
P = p1 p2 p3 = 3/4 29/39 10/38 = 0.1468.
Answer: The probability that the first and second randomly chosen balls will be white, and the third will be red 0.1468.