2

You have 20 olives. Five of them have a pit in the middle and 15 of them do not have a pit in the middle. Person B swallows five of the olives whole. Then person A takes one of the remaining olives. If the olive person A selected has a pit, what is the probability that Person B has at least one pit in his stomach?

This is what I've tried:

Let $A=\{Person\ A\ selected\ olive\ with\ a\ pit\}$ and $B=\{Person\ B\ has\ at\ least\ one\ pit\ in\ his\ stomach\}$. We are looking for $P(B|A)=\frac{P(A\cap B)}{P(A)}$.

I know that $P(A)=\frac{5}{20}$ because the order in which they choose the olives doesn't matter, and there are $5$ olives with pit out of $20$. Then I write: $$ B=\bigcup_{i=1}^{5}\{Person\ B\ has\ exactly\ i\ pits\ in\ his\ stomach\} := \bigcup_{i=1}^{5}B_i $$

Then $$ A\cap B=A\cap (\bigcup_{i=1}^{5}B_i) = \bigcup_{i=1}^{5}(A\cap B_i) $$

Then from the additivity of $P$, we get: $$ P(A\cap B)=\sum_{i=1}^{5}P(A\cap B_i) $$

Now I'm having a problem finding $P(A\cap B_i)$. I've tried to look at $P(B_i)P(A|B)$ but it didn't promote me much.

Note that this problem should be possible to solve without combinatorial knowledge.

Staltus
  • 323
  • For what its worth... it is far easier to look at the probability if the events occurred in the other order. First person $A$ goes and gets an olive with a pit. Then person $B$ goes and swallows five remaining olives whole. – JMoravitz Jan 25 '24 at 17:42
  • As for "at least 1", you can make this much easier on yourself by looking at "none" instead. Sure, you can insist on breaking into cases as you have, but it will require more effort and arithmetic. As for all calculations necessary in either interpretation/approach, this is covered by the hypergeometric distribution, the answer being a clean $1 - \dfrac{\binom{15}{5}}{\binom{19}{5}}$ – JMoravitz Jan 25 '24 at 17:45
  • @JMoravitz I've tried to use the fact that $P(B)=1-P(B^C)$ but it also didn't promote me to the answer. Also, I couldn't find a way to solve it without changing the order from $P(B|A)$ to $P(A|B)$, even though I agree it might be easier to think of. – Staltus Jan 25 '24 at 17:51
  • 1
    Note that $P(B|A)=1-P(B^C|A)$, too. If you like, you can prove this from the definition of conditional expectation together with the fact that $P(A)=P(A\cap B) + P(A\cap B^C)$ (probability of a disjoint union). – K. A. Buhr Jan 25 '24 at 18:16
  • To be clear, I never "changed the order from $P(B\mid A)$ to $P(A\mid B)$" $A$ remains $A$ and $B$ remains $B$, and in both the original phrasing and in the modified phrasing we continue to ask about $\Pr(B\mid A)$. None of that has changed. All I suggested changing was the point in time in which each event has occurred, which is perfectly legal to do... following the same logic as in If you draw two cards what is the probability that the second card is a queen? – JMoravitz Jan 25 '24 at 18:35
  • 1
    Have you realized that solving this is going to require the Bayes' theorem? – Krrish Gupta Jan 25 '24 at 19:16

2 Answers2

1

Here's how I would approach this problem.

Let event $Y$ be the event in which B swallows at least $1$ olive with a pit.

Also let

$P(X|Y) = \,$ Probability of A consuming an olive with a pit given Y has already happened

$P(X|Y^c) = \,$ Probability of A consuming an olive with a pit given $Y^c$ has already happened [B has swolled $5$ olives, none of which has a pit]

We'll start with calculating $P(Y^c)$

$$P(Y^c) = \frac{^{15}C_5}{^{20}C_5}$$

Subtracting it from $1$, we can get P(Y).

$$P(Y) = 1 - P(Y^c)$$

Next, to calculate $P(X|Y)$, we have to consider different cases.

  1. Probability of A consuming a pit when B had consumed only $1$ olive with a pit and $4$ without any. Let's label this $P_1$

$$P_1 = \frac{^4C_1}{^{19}C_1}$$

  1. Probability of A consuming a pit when B had consumed exactly $2$ olives with pit and $3$ without any. Let's label this $P_2$

$$P_2 = \frac{^3C_1}{^{19}C_1}$$

  1. Similarly, we can find $P_3$, $P_4$, and $P_5$

Now,

$$P(X|Y) = P_1 + P_2 + P_3 + P_4 + P_5$$

Also,

$$P(X|Y^c) = \frac{^5C_1}{^{19}C_1}$$

Now, according to the Bayes' theorem,

$$P(Y|X) = \frac{P(Y) \cdot P(X|Y)}{P(Y) \cdot P(X|Y) + P(Y^c) \cdot P(X|Y^c)}$$

We have already calculated everything we need on the right. so it just about substitution and calculation from here.

Please let me know if there are any errors in the solution.

You mentioned it should be possible to solve this question without the knowledge of combinatorics. One could do that, but combinatorics makes it much easier.

Haris
  • 3,409
1

It is for sure that person B did not swallow the olive that was swallowed by person A.

This leaves $19$ equiprobable candidates for being swallowed by person B.

Among these candidates are exactly $15$ that contain no pit.

Person B swallows $5$ of these candidates so the probability that afterwards he has no pit in his stomach is:$$P(B^c|A)=\frac{\binom{15}5}{\binom{19}5}$$ Then the probability that person B has at least one pit in his stomach equals:$$P(B|A)=1-\frac{\binom{15}5}{\binom{19}5}$$

Using this approach you don't need to find $P(A\cap B_i)$ but if you still want that then you can make use of this approach.

Note that for $i=1,2,3,4$ we have:$$P(B_i|A)=\frac{\binom{4}{i}\binom{15}{5-i}}{\binom{19}5}$$ Further it is evident that $P(B_5|A)=0$.

Then: $$P(A\cap B_i)=P(A)P(B_i|A)=\frac14P(B_i|A)$$

drhab
  • 151,093