Question

Batting Average Problem Milt Famey has a baseball batting average of 300, which means that his probability of getting a hit at any one time at bat is 0.3. Suppose that Milt comes to bat 5 times during a game.
a. Calculate the probabilities that he gets exactly 0, 1, 2, 3, 4, and 5 hits.
b. What is Milt's mathematically expected number of hits in these 5 at-bats?

Answer 1

Answer:

a) $P(X=0)=(10C0)(0.3)^0 (1-0.3)^{5-0}=0.168$  

$P(X=1)=(10C1)(0.3)^1 (1-0.3)^{5-1}=0.360$  

$P(X=2)=(10C2)(0.3)^2 (1-0.3)^{5-2}=0.309$  

$P(X=3)=(10C3)(0.3)^3 (1-0.3)^{5-3}=0.132$  

$P(X=4)=(10C4)(0.3)^4 (1-0.3)^{5-4}=0.028$  

$P(X=5)=(10C5)(0.3)^5 (1-0.3)^{5-5}=0.00243$  

b) And we know that the expected value for a binomial distribution is given by:

$E(X) = np = 5*0.3= 1.5$

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem  

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=5, p=0.3)$  

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

Part a

$P(X=0)=(10C0)(0.3)^0 (1-0.3)^{5-0}=0.168$  

$P(X=1)=(10C1)(0.3)^1 (1-0.3)^{5-1}=0.360$  

$P(X=2)=(10C2)(0.3)^2 (1-0.3)^{5-2}=0.309$  

$P(X=3)=(10C3)(0.3)^3 (1-0.3)^{5-3}=0.132$  

$P(X=4)=(10C4)(0.3)^4 (1-0.3)^{5-4}=0.028$  

$P(X=5)=(10C5)(0.3)^5 (1-0.3)^{5-5}=0.00243$  

Part b

For this case We want to find the extecped number of hits in the 5 at-bat

And we know that the expected value for a binomial distribution is given by:

$E(X) = np = 5*0.3= 1.5$