Answer to Question #269236 in Statistics and Probability for Sandy

Question

Over a long period of time, it has been observed that a professional basketball player can make a free throw on a given trial with probability equal to 0.8.

a) Suppose he shoots 4 free throws. What is the probability that making at least one free throw successful?

b) Suppose he shoots 12 free throws. What is the probability that he will make at least five free throws successfully?

c) Suppose he shoots 6 free throws. What is the probability that he will miss at least two free throws?

d) How many free throws he needs to throw to have a probability of 0.992 for at least one is success?

Answer 1

Let $X=$ the number of successful throws: $X\sim Bin(n,p).$

Given $p=0.8, q=1-p=0.2$

a) $n=4$

$$P(X\geq1)=1-P(X=0)=1-\dbinom{4}{0}(0.8)^0(0.2)^{4-0}$$

$$=0.9984$$

b) $n=12$

$$P(X\geq 5)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)-P(X=4)$$

$$=1-\dbinom{12}{0}(0.8)^0(0.2)^{12-0}-\dbinom{12}{1}(0.8)^1(0.2)^{12-1}$$ $$-\dbinom{12}{2}(0.8)^2(0.2)^{12-2}-\dbinom{12}{3}(0.8)^3(0.2)^{12-3}$$

$$-\dbinom{12}{4}(0.8)^4(0.2)^{12-4}=$$

$$=1-0.000000004096-0.000000196608$$

$$-0.000004325376-0.00005767168$$

$$-0.00051904512=0.99941875712$$

c) $n=6$

$$P(X\leq 4)=1-P(X=5)-P(X=6)$$

$$=1-\dbinom{6}{5}(0.8)^5(0.2)^{6-5}+\dbinom{6}{6}(0.8)^6(0.2)^{6-6}$$

$$=1-0.393216--0.262144=0.34464$$

d)

$$P(X\geq1)=1-P(X=0)$$

$$=1-\dbinom{n}{0}(0.8)^0(0.2)^{n-0}=0.992$$

$$(0.2)^n=0.008$$

$$n=3$$