The urn contains 6 black and white balls, 3 white balls are added to them. After that, 4 white balls are randomly
The urn contains 6 black and white balls, 3 white balls are added to them. After that, 4 white balls are randomly taken out of the urn. Find the probability that all the balls taken out are white, assuming that all possible assumptions about the original contents of the urn are equally possible. About the last sentence. As I understand, at the beginning there may be 6 white balls, or 5b and 1h, etc.
1. Hypotheses:
Ai – there were i white and 6 – i black balls in the urn.
Equal probability of placement with repetition:
p (Ai) = C (6; i) (1/2) ^ i * (1/2) ^ (6 – i) = C (6; i) * (1/2) ^ 6.
p (A0) = 1/64;
p (A1) = 6/64;
p (A2) = 15/64;
p (A3) = 20/64;
p (A4) = 15/64;
p (A5) = 6/64;
p (A6) = 1/64.
(In the case of equiprobability by combination, we get P (Ai) = 1/7).
2. The conditional probabilities of event B are that all the balls taken out are white:
P (B | Ai) = C (i + 3, 4) / C (9, 4);
P (B | A0) = 0;
P (B | A1) = C (4, 4) / C (9, 4) = 1/126;
P (B | A2) = C (5, 4) / C (9, 4) = 5/126;
P (B | A3) = C (6, 4) / C (9, 4) = 15/126;
P (B | A4) = C (7, 4) / C (9, 4) = 35/126;
P (B | A5) = C (8, 4) / C (9, 4) = 70/126;
P (B | A6) = C (9, 4) / C (9, 4) = 126/126.
3. Total probability of event B:
P (B) = Σ [0; 6] (P (Ai) * P (B | Ai));
P (B) = 1 / (64 * 126) (1 * 0 + 6 * 1 + 15 * 5 + 20 * 15 + 15 * 35 + 6 * 70 + 1 * 126);
P (B) = 1 / (64 * 126) (0 + 6 + 75 + 300 + 525 + 420 + 126);
P (B) = 1452 / (64 * 126) = 121 / (32 * 21) = 121/672 ≈ 0.18.
Answer: 0.18.
