# Bowl of cherries probability

1. Aug 25, 2005

### Alexsandro

Can someone help me with this interesting question ?

A bowl contains twenty cherries, exactly fifteen of which have had their stones removed. A greedy pig eats five whole cherries, picked at random, without remarking on the presence or absence of stones. Subsequently, a cherry is picked randomly from the remaining fifteen.

(a) What is the probability that this chery contains a stone ?
(b) Given that this cherry contains a stone, what is the probability that the pig consumed at least one stone ?

2. Aug 27, 2005

### EnumaElish

(a) The prob of a stone is still 5/20 = 0.25.

(b) Prob{StonEaten > 1|stone} = Prob{StonEaten > 1 and stone}/Prob{stone} = 4Prob{StonEaten = 1 or 2 or 3 or 4} = 4(Prob{Ate 1s, 4 ns} + Prob{Ate 2s, 3 ns} + Prob{Ate 3s, 2 ns} + Prob{Ate 4s, 1 ns}) = ...

The rest is combinatorics, which is not interesting but rather tedious.

Last edited: Aug 27, 2005
3. Aug 27, 2005

### CRGreathouse

For (b), I'd think it would be 15/19*14/18*13/17*12/16*11/15 -- this treats the picked cherry as if it wasn't there at all.

4. Aug 28, 2005

### EnumaElish

I don't see where these fractions are coming from. Why 15/19 through 11/15, and why are they multiplied together?

P.S. Is there something wrong with my answer to (b)?

Last edited: Aug 28, 2005
5. Aug 28, 2005

### CRGreathouse

Actually my statement is incorrect, I meant 1-15/19*14/18*13/17*12/16*11/15.

As I see the problem, there are 19 cherries of which 4 have stones. The chance of choosing at least 1 cherry with a stone is 1 - the chance of choosing only cherries without stones:

P(no stones in 5 picks) = no stone in pick 1 * no stone in pick 2 * ... * no stone in pick 5.

Since there are 15 stoneless cherries and 19 in total, the probabilities are as stated.

As to your solution, I'm not sure what's going on in it -- I see conditional probability but I'm not sure of what it says.

6. Aug 28, 2005

### EnumaElish

I think your solution is wasting information -- the fact that you picked a random cherry and it turned out to have a stone. If the pig had eaten 4 stones then the probability of this event would have been lower than if it had eaten only 1 stone. Now you can reverse this logic and say that "since I found a stone, the chances that the pig ate only 1 stone are greater than it ate 4 stones." Hence the cond. prob.

7. Aug 29, 2005

### Alexsandro

(a) The probability that the cherry with stone picked from the remaining fifteen, is still 5/20 = 0.25 ????

Let SC = cherry without stone and CC = cherry with stone.
There are six cases to pig to eat five cherries, in my opinion:
A1: pig ate 5 SC = (15/20*14/19*13/18*12/17*11*16), remaining 10SC and 5CC;
A2: pig ate 4 SC, 1 CC = (15/20*14/19*13/18*12/17*5/16), remaining 11SC and 4CC;
A3: pig ate 3 SC, 2 CC = (15/20*14/19*13/18*5/17*4/16), remaining 12SC and 3CC;
A4: pig ate 2 SC, 3 CC = (15/20*14/19*5/18*4/17*3/16), remaining 13SC and 2CC;
A5: pig ate 1 SC, 4 CC = (15/20*5/19*4/18*3/17*2/16), remaining 14SC and 1CC;
A5: pig ate 0 SC, 5 CC = (5/20*4/19*3/18*2/17*1/16), remaining 15SC and 0CC;

(a)P(next cherry has stone) = sum(i) P(eat CC|Ai) =
= P(eat CC|A1)*P(A1) + P(eat CC|A2)*P(A2) + P(eat CC|A3)*P(A3)+ P(eat CC|A4)*P(A4) + P(eat CC|A5)*P(A5).

Am I right ?

8. Aug 29, 2005

### EnumaElish

If the pig didn't care what it ate, why should the prior probability change? It shouldn't.

Example: There are 5 red and 5 blue balls in an urn. Someone removes 2 balls randomly. What is the probability of drawing a red ball before and after the 2 balls have been removed?

Example: Somebody tosses a fair coin 100 times and records the outcomes. What is your "best guess" of the proportion of heads to total? What is your best guess of this proportion in the first 98 tosses? What is your best guess for the proportion of heads in a random sampling of 80 tosses out of the 100?

In each case the answer is 0.5.

9. Aug 29, 2005

### CRGreathouse

This is why I assumed 19 instead of 20 stones, of which 4 instead of 5 have stones.

In any case, what numerical %age do you come up with? Let's see if we're actually disagreeing or just writing it differently.

10. Aug 30, 2005

### EnumaElish

Prob{Stones eaten > 1|stone in random cherry} = Prob{Stones eaten > 1 & stone in random cherry}/Prob{stone in random cherry}.

The denominator is 0.25 (because posterior prob of stone = prior prob of stone; see examples above in relation to red vs. blue balls and coin tosses).

The numerator is: Prob{Stones eaten > 1 & stone in random cherry} = Prob{Stones eaten = (1 or 2 or 3 or 4) & stone in random cherry} = Prob{Stones eaten = 1 & stone in random cherry} + Prob{Stones eaten = 2 & stone in random cherry} + Prob{Stones eaten = 3 & stone in random cherry} + Prob{Stones eaten = 4 & stone in random cherry} = $(5 \times 0.25 \times 0.75^4)(4/20)$ + $(10 \times 0.25^2 \times 0.75^3)(3/20)$ + $(10 \times 0.25^3 \times 0.75^2)(2/20)$ + $(5 \times 0.25^4 \times 0.75)(1/20)$ = 5 * 0.25 * 0.3164 * 0.2 + 10 * 0.0625 * 0.421875 * 0.15 + 10 * 0.015625 * 0.5625 * 0.1 + 5 * 0.003906 * 0.75 * 0.05 = 0.079102 + 0.039551 + 0.008789 + 0.000732 = 0.128174.

Numerator/Denominator = 0.128174/0.25 = 0.512695.

Your calculation gives 1 - (15/19)*(14/18)*(13/17)*(12/16)*(11/15) = 0.741744.

11. Aug 30, 2005

### balakrishnan_v

1)0.25
2)2875/3876

12. Aug 31, 2005

### EnumaElish

Do you mind writing out the conditional probability formula you have used for part (b)?

13. Aug 31, 2005

### balakrishnan_v

Pr(cherry contains stone)=1/4

Pr(atleast1/cherry has stone)=Pr(atleast1 and cherry has stone)/pr(cherry has stone)

={1/4-Pr(consumed 0 and cherry has stone)}/1/4=1-4.Pr(consumed 0 stones and cherry has stone)=1-4.C(15,5)/C(20,5).1/3=2875/3876

14. Sep 1, 2005

### EnumaElish

Hm. I must have made a calculation error in the numerator. On 2nd thought I am not sure of yours, either, bala: you are including the event that five stones were eaten, which has zero conditional probability.

"If there is any justice in the world," we should be getting the same result.

Last edited: Sep 1, 2005