Question

Suppose when a baseball player gets a hit, a single is twice as likely as a double which is twice as likely as a triple which is twice as likely as a home run. Also, the player’s batting average, i.e., the probability the player gets a hit, is 0.300. Let B denote the number of bases touched safely during an at-bat. For example, B = 0 when the player makes an out, B = 1 on a single, and so on. What is the PMF of B?

Answer 1

Answer:

The PMF of B is given by

P(B=0) = 0.7

P(B=1) = 0.16

P(B=2) = 0.08

P(B=3) = 0.04

P(B=4) = 0.02

Step-by-step explanation:

Let x denote P(B=1), we know that

P(B=0) = 1-0.3 = 0.7

P(B=1) = x

P(B=2) = x/2

P(B=3) = x/4

P(B=4) = x/8

Also, the probabilities should sum 1, thus

0.7+x+x/2+x/4+x/8 = 1

15x/8 = 0.3

x = 0.16

As a result, the PMF of B is given by

P(B=0) = 0.7

P(B=1) = 0.16

P(B=2) = 0.08

P(B=3) = 0.04

P(B=4) = 0.02