Question

Rob and casey are batting practice . The table below shows the joint relative frequencies for their misses , fouls ,hits and home runs. Determine the totals then find find each probability

Answer 1

`P(Homeroom|Casey)= (1)/(21)`

`P(Rob|Hit)= (23)/(50)`

Explanation:

Hello!

1)

Given the table of probabilities you have to calculate the conditional probability that the result was a homeroom given that Casey was batting.

Following the general model you can calculate this probability as:

`P(A|B) = (P(AnB))/(P(B))`

Given two events A and B that are not independent, the probability of A given that B has occurred is equal to the division between the intersection between A and B by the probability of B.

In terms of this excersise:

`P(Homeroom|Casey)= (P(HomeroomnCasey))/(P(Casey))`

Reading the table, the "P(Casey)" is found on the marginals of the table, is calculated as the summation of all battings done by Casey:

P(Casey)= 0.1125+0.05+0.3375+0.025= 0.525

P(Homeroom∩Casey)= 0.025

`P(Homeroom|Casey)= (P(HomeroomnCasey))/(P(Casey))= (0.025)/(0.525) = (1)/(21)= 0.0476`

2)

You have to calculate the probability of the batting being made by Rob given that it was a hit.

`P(Rob|Hit)= (P(RobnHit))/(P(Hit))`

The probability of the intersection is

P(Rob∩Hit)= 0.2875

P(Hit)= 0.2875+0.3375= 0.625

`P(Rob|Hit)= (P(RobnHit))/(P(Hit))= (0.2875)/(0.625) = (23)/(50) = 0.46`

I hope this helps!