Question

If a baseball player has a batting average of 0.375, what is the probability that the player will get the following number of hits in the next four times at bat?
A. Exactly 2 hits(Round to 3 decimal places as needed)
B. At least 2 hits (Round to 3 decimal places as needed)

Answer 1

Answer:

a) $P(X=2)=(4C2)(0.375)^2 (1-0.375)^{4-2}=0.330$  

b) $P(X\geq 2)=1-P(X< 2)=1-[P(X=0)+P(X=1)]$

$P(X=0)=(4C0)(0.375)^0 (1-0.375)^{4-0}=0.153$  

$P(X=1)=(4C1)(0.375)^1 (1-0.375)^{4-1}=0.366$  

And replacing we got:

$P(X\geq 2)=1-P(X< 2)=1-[0.153+0.366]=0.481$

Step-by-step explanation:

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

$X \sim Binom(n=4, p=0.375)$  

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

$P(X=2)=(4C2)(0.375)^2 (1-0.375)^{4-2}=0.330$  

Part b

$P(X\geq 2)=1-P(X< 2)=1-[P(X=0)+P(X=1)]$

$P(X=0)=(4C0)(0.375)^0 (1-0.375)^{4-0}=0.153$  

$P(X=1)=(4C1)(0.375)^1 (1-0.375)^{4-1}=0.366$  

And replacing we got:

$P(X\geq 2)=1-P(X< 2)=1-[0.153+0.366]=0.481$