双Urn两阶段概率实验：事件A与事件B的技术问询

Let’s walk through the key probability calculations for this experiment, starting with the basics and moving to conditional and total probabilities.

Experiment Setup Recap #

• Urn 1 (initial): 3 Yellow (Y), 3 Red (R), 3 Green (G) → Total 9 balls
• Urn 2 (initial): 1 Y, 2 G, 4 Purple (P) → Total 7 balls
• Two stages:
  1. Randomly move 3 balls from Urn 1 to Urn 2
  2. Randomly draw 1 ball from the updated Urn 2
• Events:
  • A = {Moved 2 Y and 1 G to Urn 2}
  • B = {Drew a G from the updated Urn 2}

Calculating P(A): Probability of Event A #

To find the probability of moving exactly 2 Y and 1 G, we use binomial combinations (since selection order doesn’t matter):

1. Total ways to choose 3 balls from Urn 1:
   C(9, 3) = 84 (where C(n, k) represents "n choose k")
2. Favorable ways to choose 2 Y and 1 G:
   Choose 2 Y from 3: C(3,2) = 3; Choose 1 G from 3: C(3,1) =3
   Total favorable combinations: 3 * 3 =9
3. Final probability P(A):
   P(A) = 9 / 84 = 3/28 ≈ 0.1071

Calculating P(B|A): Conditional Probability of B Given A #

If event A occurred, we add 2 Y and 1 G to Urn 2. Let’s update Urn 2’s composition:

• New Y count: 1 + 2 = 3
• New G count: 2 + 1 = 3
• P count remains 4
• Total balls in updated Urn 2: 3 + 3 + 4 = 10

The probability of drawing a G is the number of G balls divided by total balls:
P(B|A) = 3/10 = 0.3

Calculating P(A∩B): Probability of Both A and B Occurring #

Using the multiplication rule for conditional probability:
P(A∩B) = P(A) * P(B|A) = (3/28) * (3/10) = 9/280 ≈ 0.0321

Bonus: Total Probability P(B) (Overall Chance of Drawing G) #

To find the overall probability of event B, we use the law of total probability—summing the weighted probabilities of drawing G across all possible transfer scenarios.

A quick shortcut using expectation:

• The expected number of G balls moved from Urn 1 is (3/9)*3 =1 (each drawn ball has a 3/9 chance of being G)
• Expected G count in Urn 2 after transfer: 2 + 1 =3
• Total balls in Urn 2 after transfer is always 7 +3 =10
• Thus, P(B) = 3/10 =0.3

Detailed Verification #

We can confirm this by calculating contributions from all transfer cases (grouped by number of G balls moved, g):

1. g=0: Contribution = (20/84)*(2/10) =1/21 ≈0.0476
2. g=1: Contribution = (45/84)*(3/10) =9/56 ≈0.1607
3. g=2: Contribution = (18/84)*(4/10) =3/35 ≈0.0857
4. g=3: Contribution = (1/84)*(5/10) =1/168 ≈0.00595

Summing these gives 252/840 =3/10 =0.3, matching the expectation result.

内容的提问来源于stack exchange，提问作者White Mamba