What is the chance the jam is toxic

Question

A woman picks 7 apples in the forest, but unfortunately two of them are poisonous. Three apples are used in a pie, which turns out to be toxic. Another two are used in a jam. What is the chance that the jam is also toxic?

Answer 1

Say we give each of the seven apples labels a₁, a₂, a₃, a₄, a₅, a₆, a₇. Suppose apples a₁, a₂, a₃ are used to make the pie and a₄, a₅ are used to make the jam and a₆, a₇ are our left over apples. We can denote the toxic apples as being any two apples a_i, a_j. Our sample space is defined as the collection of every possible choice of those two toxic apples:

S = {{a_i, a_j} : a_i, a_j ∈ {a₁, a₂, a₃, a₄, a₅, a₆, a₇}}.

|S| = C(7,2) = 21 is the total number of outcomes for choosing toxic apples.

We want to know Pr(Jam is toxic | Pie is toxic). Define the events in our conditional probability:

P^c : At least one apple in the pie is toxic.

J^c : At least one apple in the jam is toxic.

And define the complements of those events:

P: All the apples in the pie are non-toxic.

J: All the apples in the jam are non-toxic.

To arrive Pr(J^c | P^c ), we start by finding Pr(P), Pr(J), and Pr(P∩J).

P entails that a_i, a_j can't be a₁, a₂, or a₃. In set notation, we can define P by all the ways a_i, a_j can be excluded the pie:

P = {{a_i, a_j} : a_i, a_j ∈ {a₄, a₅, a₆, a₇}}.

Pr(P) = |P| / |S| = C(4,2) / |S| = 6/21 = 2/7

For J we have that a_i, a_j can't be a₄, or a₅. Set J can be defined:

J = {{a_i, a_j} : a_i, a_j ∈ {a₁, a₂, a₃, a₆, a₇}}.

Pr(J) = |J| / |S| = C(5,2) / |S| = 10/21

P∩J occurs if all the apples in the pie and the jam are non-toxic. This means that a_i, a_j are excluded from being a₁, a₂, a₃, a₄, or a₅ and the logical entailment is that a_i, a_j must be the left over apples a₆, a₇.

P∩J = {{a₆, a₇}}

Pr(P∩J) = |P∩J| / |S| = 1/21

Now using Bayes and negation rule:

Pr(P|J) = Pr(P∩J)/Pr(J) = (1/21)/(10/21) = 1/10

Pr(P^c|J) = 1 - Pr(P|J) = 1 - 1/10 = 9/10

Pr(J|P^c) = (Pr(P^c|J) * Pr(J)) / Pr(P^c) = ((9/10) * (10/21)) / (5/7) = 3/5

Pr(J^c|P^c) = 1 - Pr(J|P^c) = 1 - (3/5) = 2/5

So Pr(Jam is toxic | Pie is toxic) = 2/5

Answer 2

Define:

P_t : the pie is toxic

J_t : the jam is toxic

Pr ( J_t | P_t): the probability that the jam is toxic given that the pie is toxic.

Pr ( J_t | P_t) = Pr( J_t ∩ P_t) ⁄ Pr (P_t)

J_t ∩ P_t only occurs when each has one of the toxic apples.

So, Pr( J_t ∩ P_t) = Pr(the jam has one toxic apple) * Pr(the pie has one toxic apple)

Pr(the jam has one toxic apple) = 2 * 2/7 * 5/6 (2 ways to choose a toxic apple and a non-toxic apple)

Pr(the pie has one toxic apple) = 3 * 2/7 * 5/6 * 4/5 (3 ways to choose a toxic apple and two non-toxic apples)

Pr (P_t) = (3 * 2/7 * 5/6 * 4/5) + (3 * 2/7 * 1/6) = 5/7 (either one or two toxic apples in the pie)

So, Pr ( J_t | P_t) = [(2 * 2/7 * 5/6) * (3 * 2/7 * 5/6 * 4/5)] / 5/7 = 8/21

Answer 3

The only condition that is possible for both the pie and the jam to be toxic (N=nonT, T=toxic) is

2N,T in the pie and N,T in the jam (If the pie has 2T, the jam can't be toxic)

The probability of both of these happening is the product of each individual probability:

₅C_{2 2}C₁ * ₃C₁ * 1

_______ ______ (There are 5N take 2, and 2T take 1 out of 7 apples take 3)

₇C_{3 4}C₂ (After 3 are used for the pie, there are 4 left: 3N, 1T)

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