1. ## Lost about a probability problem

Hey there, this is such a great forum, so happy I found it! I honestly have no idea whatsoever how to solve the two problems (except for the theorem prrof, that's straight forward).

3.
A bowl contains twenty cherries, exactly fifteen of which have had their stones removed. A greedy pig eats five whole cherries, picked at random, 1
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 cherry contains a stone?
(b) Given that this cherry contains a stone, what is the probability that
the pig consumed at least one stone?

4.
(a) State and prove Bayes theorem for calculating posterior probabilities from prior probabilities and some observed events.
(b) A committee of 3 people has been formed by random selection from five left wingers and four right wingers. The committee members then vote for or against a strike whenever there is a dispute. Each left winger votes for a strike three our of four times in strike votes, whereas each right winger votes for a strike only once out of three times in strike votes. If it is known that the committee has decided against one particular strike, what is the probability that there is a majority of right wingers on the committee?

Does anyone know how to approach these?

Hey there, this is such a great forum, so happy I found it! I honestly have no idea whatsoever how to solve the two problems (except for the theorem prrof, that's straight forward).

3. A bowl contains twenty cherries, exactly fifteen of which have had their stones removed. A greedy pig eats five whole cherries, picked at random, 1
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 cherry contains a stone?
(b) Given that this cherry contains a stone, what is the probability that
the pig consumed at least one stone?

[snip]
Let X be the random variable number of cherries with stones eaten by pig.

(a) Pr(picked cherry contains stone) = Pr(picked cherry contains stone | X = 0) Pr(X = 0) + Pr(picked cherry contains stone | X = 1) Pr(X = 1) + ...... + Pr(picked cherry contains stone | X = 5) Pr(X = 5).

I'll calculate Pr(picked cherry contains stone | X = 1) Pr(X = 1):

Pr(picked cherry contains stone | X = 1) = 4/15.

$\Pr(X = 1) = \frac{ {5 \choose 1} {15 \choose 4}}{ {20 \choose 5}} = \frac{2275}{5168}$.

Therefore Pr(cherry contains stone | X = 1) Pr(X = 1) = 455/3876.

--------------------------------------------------------------------------

(b) Pr(X > 0 | picked cherry contains stone) = Pr(X > 0 and picked cherry contains stone)/Pr(picked cherry contains stone).

The denominator has been calculated in part (a).

Pr(X > 0 and picked cherry contains stone)

= Pr(X = 1 and picked cherry contains stone) + Pr(X = 2 and picked cherry contains stone) + ...... + Pr(X = 5 and picked cherry contains stone)

= Pr(picked cherry contains stone | X = 1) Pr(X = 1) + ...... + Pr(picked cherry contains stone | X = 5) Pr(X = 5).

These have been calculated in part (a).

[snip]

4.
(a) State and prove Bayes theorem for calculating posterior probabilities from prior probabilities and some observed events.

(b) A committee of 3 people has been formed by random selection from five left wingers and four right wingers. The committee members then vote for or against a strike whenever there is a dispute. Each left winger votes for a strike three our of four times in strike votes, whereas each right winger votes for a strike only once out of three times in strike votes. If it is known that the committee has decided against one particular strike, what is the probability that there is a majority of right wingers on the committee?

Does anyone know how to approach these?
(b) Let X be the random variable number of right wingers on committee.

Pr(X > 1 | committee decides against strike) = Pr( X > 1 and committee decides against strike)/Pr(committee decides against strike).

Pr(committee decides against strike) = Pr(committee decides against strike | X = 0) Pr(X = 0) + Pr(committee decides against strike | X = 1) Pr(X = 1) + Pr(committee decides against strike | X = 2) Pr(X = 2) + Pr(committee decides against strike | X = 3) Pr(X = 3).

I'll calculate Pr(committee decides against strike | X = 2) Pr(X = 2):

$\Pr(X = 2) = \frac{{4 \choose 2} {5 \choose 1} }{{9 \choose 3}} = \frac{5}{14}$.

Committe decides against a strike => either 2 or 3 votes against strike:

Pr(3 votes against) = $\left( \frac{1}{4}\right)^2 \left( \frac{2}{3}\right) = \frac{1}{24}$.

Pr(2 votes against) = $\left( \frac{1}{4}\right)^2 \left( \frac{1}{3}\right) + 2 \left( \frac{1}{4}\right) \left( \frac{3}{4}\right) \left( \frac{2}{3}\right) = \frac{13}{48}$.

Therefore Pr(committee decides against strike | X = 2) = $\frac{15}{48}$.

Pr( X > 1 and committee decides against strike) = Pr( X = 2 and committee decides against strike) + Pr( X = 3 and committee decides against strike)

= Pr(committee decides against strike | X = 2) Pr(X = 2) + Pr(committee decides against strike | X = 3) Pr(X = 3).

These will already have been calculated - above.

Note: I reserve the right for careless mistakes in this reply.