Here's the problem I have:

The probability that an apple is red is 80%. The probability that a red apple is rotten is 10%. If 10 apples are picked at random, what are the odds that half are red and rotten?

Approach 1:

The odds of an apple being both red and rotten is 8%. Therefore, we have a binomial distribution with 10 trials, 5 expected successes, and probability of success 0.08.

This gives us a probability of 0.054%.

Approach 2:

We would expect, on average, to get 8 red apples when we pick 10. This allows us to skip the 80% and compute a binomial distribution with the following: 8 trials, 5 expected successes, and probability 0.1.

Here, we get a probability of 0.041%

I believe the right answer comes from Approach 1, but I'm not sure. Which one is correct and why do we get different numbers?

  • 1
    $\begingroup$ In approach 1, there is a chance of drawing 9 or 10 red ones, and in approach 2 you are fixing the draws to exactly 8. $\endgroup$
    – Dave2e
    Sep 24, 2023 at 3:46
  • 4
    $\begingroup$ You simply cannot replace a random variable ("number of reds") by its expectation at one stage of the derivation of the distribution. $\endgroup$
    – Xi'an
    Sep 24, 2023 at 6:03

1 Answer 1


I hope I understand the phrasing correctly. So we have $P(``\textrm{Red}")=0.8$ and $P(``\textrm{Rotten}"\mid ``\textrm{Red}")=0.1$. So let's just get the probability of a single apple being red and rotten and work with that. \begin{align*} P(``\textrm{Rotten}"\cap~``\textrm{Red}")&=P(``\textrm{Rotten}"\mid ``\textrm{Red}")P(``\textrm{Red}")\\ &=0.1\cdot0.8\\ &=0.08 \end{align*} Plugging in to binomial, \begin{align*} P(X=k)&=\binom nkp^k(1-p)^{n-k}\\ &=\binom{10}{5}(0.08)^5(1-0.08)^{10-5} \\ &=0.0005442389 \end{align*} And that is technically your approach 1.

Approach 2 seems a little suspicious to me since you're working with the expected value which kinda excludes the chances of stuff like having 10 red apples but only having 5 rotten ones where as approach 1 covers all possible outcomes.


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