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

Answer: C.  $0.75^4$

Step-by-step 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.  $0.75^4$.