从urn中取球的概率问题：求5次有放回抽取全白球的概率及适用模型

Alright, let's work through your probability problem step by step — I'll make sure to break down both questions clearly.

We need to use the law of total probability here, since the outcome of the first draw (from the first urn) changes the composition of the second urn, which directly impacts the subsequent draws. Here's the breakdown:

• First, calculate the probability of drawing a white ball from the first urn:
  P(white from urn1) = 6 / (6 + 7) = 6/13
  If we add this white ball to the second urn, the urn now holds 11 white balls and 5 black balls. With replacement, the probability of drawing a white ball in one trial is 11/16. For 5 consecutive white draws, this probability becomes (11/16)^5.

• Next, calculate the probability of drawing a black ball from the first urn:
  P(black from urn1) = 7 / (6 + 7) = 7/13
  Adding this black ball to the second urn gives us 10 white balls and 6 black balls. The probability of drawing a white ball here is 10/16 = 5/8, so 5 consecutive white draws is (5/8)^5.

• Combine these two scenarios using total probability:
  Total Probability = (6/13)*(11/16)^5 + (7/13)*(5/8)^5

For a numerical approximation, this works out to roughly 0.123 (or ~12.3%).

This is not a Poisson model — let's break down why and what model does fit:

The correct model: Binomial distribution (with total probability weighting) #

For the 5 draws from the second urn (once we know what ball was added), this is a classic binomial distribution scenario. The binomial model applies when:

• We have a fixed number of independent trials (here, 5 draws with replacement, so each trial is independent of the last)
• Each trial has two mutually exclusive outcomes (white or black)
• The probability of "success" (drawing white) stays constant for every trial (since we replace the ball after each draw)

Since we have two possible initial scenarios (white added vs. black added), the overall probability is a weighted combination of two separate binomial outcomes (specifically, the probability of 5 successes in each binomial case).

Why Poisson doesn't apply here #

The Poisson distribution is built to model rare events happening in a fixed interval (e.g., number of customer complaints per day, number of defects in a batch). Its core assumptions are:

• Events are extremely rare (the probability p of success is very small)
• The number of trials n is very large
• The product n*p remains a constant (the expected number of events)

In our problem, n=5 is small, and the probability of drawing white isn't a rare event (it's either 11/16 or 5/8, both fairly high). So Poisson is not the right fit here.

内容的提问来源于stack exchange，提问作者Lola