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

To find the probability of getting exactly 2 hits, use the binomial probability formula. To find the probability of getting at least 2 hits, calculate the probabilities of getting 2, 3, and 4 hits and sum them up. The probabilities are 0.259 and 0.697, respectively.

Step-by-step explanation:

To find the probability that the player will get exactly 2 hits in the next four times at bat, we can use the binomial probability formula. The probability of getting exactly 2 hits is given by:

P(X = 2) = C(4, 2) * (0.375)^2 * (1 - 0.375)^(4 - 2)

Where C(4, 2) is the number of ways to choose 2 hits from 4 times at bat, 0.375 is the batting average, and (1 - 0.375) is the probability of not getting a hit. Plugging in the values, we get:

P(X = 2) = 6 * (0.375)^2 * (0.625)^2 = 0.259.

To find the probability that the player will get at least 2 hits in the next four times at bat, we need to calculate the probabilities of getting 2, 3, and 4 hits and sum them up:

P(X >= 2) = P(X = 2) + P(X = 3) + P(X = 4)

Using the same formula as before, we can calculate:

P(X = 3) = C(4, 3) * (0.375)^3 * (1 - 0.375)^(4 - 3) = 0.352

P(X = 4) = C(4, 4) * (0.375)^4 * (1 - 0.375)^(4 - 4) = 0.086

Adding up the probabilities:

P(X >= 2) = 0.259 + 0.352 + 0.086 = 0.697.

Answer 2

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=(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`