Question

During the 2015 baseball season, the Detroit Tigers' Miguel Cabrera hit for the elusive Triple Crown by having the highest batting average, most home runs, and most runs batted in. Most baseball players average four at-bats per game. Using Cabrera's batting average of 0.338, the probability that he would get no hits in one game would be 0.192, one hit would be 0.392, two hits would be 0.300, and three hits would be 0.102. Using the probability distribution table given, what is the probability that Miguel Cabrera gets a hit in all four at-bats?
x 0 1 2 3 4
p(x) 0.192 0.392 0.300 0.102

Answer 1

Answer:

The probability that Miguel Cabrera gets a hit in all four at-bats is 0.014

Step-by-step explanation:

Given

$\begin{array}{cccccc}x & {0} & {1} & {2} & {3} & {4} \ \\ {P(x)} & {0.192} & {0.392} & {0.300} & {0.102} & { } \ \end{array}$

Required

P(x = 4)

To do this, we use the following formula:

$\sum\limit^n_{x=i} P(x) = 1$

In this case, the above formula can be interpreted as:

$P(x = 0) + P(x = 1) + P(x = 2) + P(x =3) + P(x = 4) = 1$

This gives:

$0.192 + 0.392 + 0.300 +0.102 + P(x = 4) = 1$

$0.986 + P(x = 4) = 1$

Collect like terms

$P(x = 4) = -0.986 +1$

$P(x = 4) = 0.014$