Question

3. Jay is taking a statistics course with two instructors: Tom and Jerry. If Tom designs the final exam problems, Jay will get A with probability .8. If Jerry designs the final exam problems, Jay will get A with probability .6. The proportion of times that Tom designs the final is .4. (1) What's the probability that Jay will get A? (10 points) (2) If Jay doesn't get A, what is the probability that Jerry designs the final? (10 points)

Answer 1

Let, us consider the following events:

A = Jay gets A grade

Ac = Jay doesn’t get A grade.

B = Jerry designs Final Question

Bc = Tom designs Final Question (or Jerry doesn’t design Final Question)

We are provided the following information:

If Tom designs the final question Jay gets A with probability 0.8 i.e.

P[A|Bc] =0.8

If Jerry designs the final question Jay will get A with probability 0.6 i.e.

P[A|B] = 0.6

The proportion of times Tom Designs Final questions is 0.4

P[Bc] = 0.4

So, P[B] =1- P[Bc] = 1- 0.4 =0.6

The proportion of times Jerry Designs Final questions is 0.6

(1) We need to find the probability that Jay will get A.

From, Total probability theorem,

So, P[A] = P[A|Bc] × P[Bc] + P[A|B] × P[B]

                   = 0.8 × 0.4 + 0.6 × 0.6

                   = 0.68

(2) We need to find the probability of Jerry Designing the paper if Jay doesn’t get A.

So, we need to calculate P[B|Ac]

P[B|A] = P[BRA] P[A]

                            

P[AB] P[A]

                            

_P[A°|B] ~ P[B] P[A]

[The above is the proof for Bayes theorem used here)

P[Ac] = P[Ac|Bc] × P[Bc] + P[Ac|B] × P[B]

          = (1- P[A|Bc]) × P[Bc] + (1- P[A|B]) × P[B]

          = (1- 0.8) × 0.4 + (1- 0.6) × 0.6

          = 0.32

So, P[BA] = (1 – 0.6) 0.6 F = 0.75 0.32