Among 60 apples collected from an apple tree, there are three bad apples. The apples are randomly placed into four baskets so that each basket contains 15 apples.

Compute the probability that the three bad apples are not all in the same basket.

Attempt: $1-\frac{(4C1)(15C3)12!}{(60C15)(45C15)(30C15)(15C15)}$

Can anyone confirm if my method is right?

I then need to find the probability that no two bad apples are in the same basket, in other words each bad apple is in a different basket. How do I do this one?

  • $\begingroup$ No your answer is not correct. Can you explain the reasoning behind your working? $\endgroup$
    – Math Lover
    May 27 at 9:36
  • $\begingroup$ Hint: For the first part, (a) determine numerator and denominator as if the $3$ bad apples were in basket $1$ (b) multiply by $4$ (c) take the complement $\endgroup$ May 27 at 13:24

For the first question: the idea is good but the result is not. You try to compute the probability of the contrary and that's good. Then, you have to count how many ways there are to place the three bad apples in the same basket and all the other apples (there are only 57 left...) to complete the baskets. For the second question: ou may first begin by answering the following question: how many ways are there to place 3 apples in 4 baskets so that no two apples are in the same basket?

  • $\begingroup$ I don't really get what you mean. Can you help me further? $\endgroup$ May 27 at 9:49

For part (a), you started well, but strayed. The formula you sought is

$1- \dfrac{\binom41\binom33\binom{57}{12}}{\binom{60}{15}}$

You should now try (b) on your own along similar lines. Here, though, is another way. Imagine $15$ slots in each basket for apples.

The first bad apple can be placed anywhere$\;$The next can be place in any of $45$ slots in other baskets out of $59$ free, and so on,

Thus $Pr = \dfrac{45}{59}\cdot\dfrac{30}{58}$

The good apples automatically fill all the remaining places.
You can countercheck using a hypergeometric formula akin to (a)


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