0
$\begingroup$

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.

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

(ii) Compute the probability that no two bad apples are in the same basket.

(iii) A farmer comes and chooses a basket at random. Compute the probability that there is exactly one bad apple in the basket that the farmer chooses.

(iv) Compute the expected number of baskets that contain exactly one bad apple.

I believe for (i), I could work out the total number of permutations of $3$ apples all in the same basket (maybe $60 \cdot 14 \cdot 13$) and divide by permutations of $3$ apples in $60$ positions ($60 \cdot 59 \cdot 58$).

For (ii), I know that there would be one empty basket, so there would be ($4C1$) ways of picking the basket of all good apples, but unsure of what to do from there.

$\endgroup$
9
  • 2
    $\begingroup$ Welcome to Mathematics SE. Take a tour: math.stackexchange.com/tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context by stating what you understand about the problem, what you've tried so far, etc.; both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance: math.meta.stackexchange.com/q/9959 . $\endgroup$
    – Alan
    Aug 21 at 22:51
  • 1
    $\begingroup$ To add to Alan's excellent advice, there's the simple fact that there are many parts to this question. If you can attempt even the first couple of questions (I would say the first part at least is not too difficult), then that takes some of the burden off would be answerers. Basically, the community likes to precisely target where you're struggling, so that they don't have to type up long answers in the hope that it addresses what's causing confusion. $\endgroup$ Aug 21 at 23:00
  • 1
    $\begingroup$ Sorry, I believe for (i), I could work out the total number of permutations of 3 apples all in the same basket (maybe 60*14*13) and divide by permutations of 3 apples in 60 positions (60*59*58). For (ii), I know that there would be one empty basket, so there would be (4C1) ways of picking the basket of all good apples, but unsure of what to do from there. $\endgroup$
    – user960518
    Aug 21 at 23:06
  • 1
    $\begingroup$ At least copy the attempts you have written here after the body of the question, and edit question (iii) properly, then it may get a response. $\endgroup$ Aug 22 at 3:29
  • 2
    $\begingroup$ @true blue anil: Is the answer to (ii) $\frac{675}{1711}$? $\endgroup$ Aug 22 at 4:27
1
$\begingroup$

i) Let us calculate the opposite probability. Imagine each spot within each basket as being uniquely labeled.

Imagine each apple as being uniquely labeled and sortable.

Let the "first" of the bad apples (first according to the sorting alluded to above) be placed into one of the available spots within the available baskets. It doesn't matter which.

Place the "next" bad apple next. There remain $14$ other open positions within the basket that the first bad apple was placed out of $59$ remaining available positions. The probability it is placed in the same basket as the first is then $\dfrac{14}{59}$. (Note, $\frac{14}{59}\neq \frac{1}{4}$)

Now, place the "last" bad apple. There remain $13$ open positions within the basket that the first and second bad apples were placed in out of $58$ remaining available positions, which happens with probability $\frac{13}{58}$.

This gives $\frac{14\cdot 13}{59\cdot 58}$ probability of them all being in the same basket and so for them to not all be in the same is with probability

$$1-\dfrac{14\cdot 13}{59\cdot 58} = 1-\dfrac{91}{1711} = \dfrac{1620}{1711}\approx 0.9468\dots$$

Your working for this was perfectly correct.

ii) The same logic can apply here as well. The first bad apple goes wherever. The second bad apple goes in a different box with probability $\frac{45}{59}$ and the third in a different one yet with probability $\frac{30}{58}$ for a probability of:

$$\dfrac{45}{59}\cdot\dfrac{30}{58}=\dfrac{675}{1711}$$

iii) You can ignore the baskets... the farmer picked 15 apples at random and asks the probability that there is one bad apple present. This is a standard hypergeometric distribution.

$$\dfrac{\binom{57}{14}\binom{3}{1}}{\binom{60}{15}}$$

iv) By linearity of expectation, this is simply $4$ times the answer to part (iii).

$\endgroup$
-1
$\begingroup$

You've said, "I believe for (i), I could work out the total number of permutations of 3 apples all in the same basket (maybe 60⋅14⋅13) and divide by permutations of 3 apples in 60 positions (60⋅59⋅58)." However, this is not a permutational condition. You don't care about the place of bad apples while including them into subgroups of 15. It would be best if you worked with combinations in here.

So my attempt for $i$ this would be:

$1-\frac{4.{57 \choose 12}.{45 \choose 15}.{30 \choose 15}}{{60 \choose 15}.{40 \choose 15}.{30 \choose 15}}=1-\frac{4.{57 \choose 12}}{{60 \choose 15}}$

4 comes from the number of possible baskets for 3 bad apples among 4 baskets, and the other part is just getting the number of different choices. After finding the number of possible choices for all bad apples in the same basket, you can just subtract from 1 and find otherwise.

For $ii$, "Compute the probability that no two bad apples are in the same basket." means the same as "Compute the probability that all bad apples are in different baskets."

$\frac{4.{57 \choose 14}.{43 \choose 14}.{29 \choose 14}}{{60 \choose 15}.{40 \choose 15}.{30 \choose 15}}$ 4 comes from the number of possible ways to place 3 bad apples in different baskets(4)-$4!/3!$.After placing these bad apples, we'll have 57 apples and 14,14,14,15 gaps in each basket. Then we can choose normal apples and fill these gaps with combinations. This is my solution, but if one can also validate it, I'd be glad.

$\endgroup$
2
  • $\begingroup$ For ii) if you want to approach as you are, pick where the "first" bad apple goes, where the "second", and then where the "third" in $4!$ ways. Then, pick $14$ good apples to go with the first bad apple, $14$ more to go with the second, etc... Note that $57-14=43\neq 42$. This gives a corrected answer of $\dfrac{\color{red}{4!}\binom{57}{14}\binom{\color{red}{43}}{14}\binom{29}{14}}{\binom{60}{15}\binom{45}{15}\binom{30}{15}}=\dfrac{675}{1711}$. I find it easier to avoid binomial coefficients here personally. Compare that large expression to $\dfrac{45\cdot 30}{59\cdot 58}$ $\endgroup$
    – JMoravitz
    Aug 22 at 17:06
  • $\begingroup$ You're right; I've mistakenly written 42. Thanks for the warning. I guess you're right. Your approach seems easier thanks for sharing. $\endgroup$ Aug 22 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy