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

Answer:

$P(Homeroom|Casey)= \frac{1}{21}$

$P(Rob|Hit)= \frac{23}{50}$

Step-by-step 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:

$P(A|B) = \frac{P(AnB)}{P(B)}$

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:

$P(Homeroom|Casey)= \frac{P(HomeroomnCasey)}{P(Casey)}$

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

$P(Homeroom|Casey)= \frac{P(HomeroomnCasey)}{P(Casey)}= \frac{0.025}{0.525} = \frac{1}{21}= 0.0476$

2)

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

$P(Rob|Hit)= \frac{P(RobnHit)}{P(Hit)}$

The probability of the intersection is

P(Rob∩Hit)= 0.2875

P(Hit)= 0.2875+0.3375= 0.625

$P(Rob|Hit)= \frac{P(RobnHit)}{P(Hit)}= \frac{0.2875}{0.625} = \frac{23}{50} = 0.46$

I hope this helps!