Finding $P(B|A)$ when $B$ involves a "at least" term

Question

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.

Answer 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:$$P(B|A)=1-\frac{\binom{15}5}{\binom{19}5}$$

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

Answer 2

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 $P_1$

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

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

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

3. 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.

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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.