Compute the following conditional probabilities

Question

but assume that Pr[E] = 0.3, Pr[F] = 0.35, and Pr[E′∩ F] = 0.15.
Compute the following conditional probabilities:

(1) Pr[E | F] =

(2) Pr[F | E] =

Answer 1

Given:

1. Pr⁡[E]=0.3
2. Pr⁡[F]=0.35
3. Pr⁡[E′∩F]=0.15.

Let us first find the the total probability, Pr[F]

Pr[F]=Pr[E∩F]+Pr[E′∩F]

Plugging in the values , we get

0.35=Pr[E∩F] + 0.15

Pr[E∩F]=.0.35-0.15 = 0.20

Let us now compute Pr⁡[E∣F] ( which is the conditional probability of event E given event F, so we use the formula for conditional probability here))

Pr[E∣F]=Pr[E∩F]​ / Pr[F] =0.20/0.35 = 0.571

Next we compute Pr(FIE) ( which probability of event F given Event E, so we will again use the formula for conditional probability).

Pr[F∣E] = Pr[E∩F]​ I Pr[E] = 0.20/..30 = 0.667

Hope this is helpful.

Answer 2

I find these problems easier to answer if I write them in plain English.

So the probability of choosing E is 0.3. This is the probability of only E (and not F) plus the probability of both E and F. We can rewrite it like this:

P(E) = 0.3 = P(E and not F) + P(E and F)

The probability of F is 0.35. Let's rewrite it using the same logic.

P(F) = 0.35 = P(F and not E) + P(E and F)

They also give us P[E' and F] = 0.15. The prime symbol just means not. We've actually encountered P(F and not E) somewhere. In the second equation!

P(F) = 0.35 = 0.15 + P(E and F)

P(E and F) = 0.2

Now we have enough information to solve the problems.

1) P(E | F). This means the probability of getting E if you've already gotten F. It is equal to P(E and F)/P(F)

which equals 0.2/0.35 = 0.57

P(E | F) = 0.57

2)P(F | E). this might look similar to P(E | F), but they are not the same. What we're looking for is the probability of getting F when we already have E. It is equal to P(E and F)/P(E)

which equals 0.2/0.3 = 0.67

P(F | E) = 0.67