Question

Michael Beasley is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next one, and he makes his free throws 75%, percent of the time. What is the probability of Michael Beasley making all of his next 4 free throw attempts?

A. .75^8
B. .375^4
C. .75^4
D. 1.50^2

Answer 1

C. 0.75^4

Explanation:

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Answer 2

Answer: C.
`0.75^4`

Explanation:

Let x be the binomial variable that denotes the number of makes.

Since each throw is independent from the other throw , so we can say it follows Binomial distribution .

So
`X\sim Bin(n=4 , p=0.75)`

Binomial distribution formula: The probability of getting x success in n trials :

`P(X=x)=^nC_xp^n(1-p)^(n-x)` , where p = probability of getting success in each trial.

Then, the probability of Michael Beasley making all of his next 4 free throw attempts will be :

`P(X=4)=^4C_4(0.75)^4(1-0.75)^(0)`

`=(1)(0.75)^4(1)\ \ [\because\ ^nC_n=1]\\\\=(0.75)^4`

Thus, the probability of Michael Beasley making all of his next 4 free throw attempts is
`=0.75^4`

Hence, the correct answer is C.
`0.75^4`.