The urn contains 3 white 4 yellow and 2 black balls. From it at random take out (without returning)

The urn contains 3 white 4 yellow and 2 black balls. From it at random take out (without returning) one by one, one ball at a time. What is the probability that the white ball will appear before the yellow one?

It will be possible to remove the white ball before the yellow one (we will designate this as event A) in three cases: the white ball was taken out first, the white one was taken out after the first black one, the white one was taken out after the first and second black ones. In all other cases, the yellow was taken out earlier.
Let’s accept hypotheses:
H1 – the white ball was taken out first;
H2 – the black ball was taken out first;
H3 – the first and second to take out the black balls.
Probabilities of hypotheses:
P (H1) = 3/9 = 1/3;
P (H2) = 2/9;
P (H3) = 2/9 * 1/8 = 1/36.
The conditional probability that event A will occur when these hypotheses are fulfilled:
P (A | H1) = 1;
P (A | H2) = 3/8;
P (A | H3) = 3/7;
Applying the formula for total probability, we get:
P (A) = P (H1) P (A | H1) + P (H2) P (A | H2) + P (H3) P (A | H3) = 1/3 * 1 + 2/9 * 3/8 + 1/36 * 3/7 = 1/3 + 1/12 + 1/84 = 28/84 + 7/84 + 1/84 = 36/84 = 3/7 = 0.42857.
Answer: 0.42857.