**Probability and Binomial Distribution: Calculating Success Rates**A professional basketball player has an 81% success rate when shooting free throws. Let the random variable \(X\) represent the number of free throws he makes in a random sample of 10 free throws (assume this experiment meets all binomial requirements). What is the probability that he makes exactly 7 of the 10 free throws? Note: You do not need to solve for the exact probability; you just need to choose the equation below that correctly represents how you would solve for this probability.### Options for the Correct Equation:**a.**\[ P(X = 7) = \frac{10!}{3!7!} (0.81)^7(0.19)^{10} \]**b.**\[ P(X = 7) = \frac{10!}{7!} (0.81)^7(0.19)^3 \]**c.**\[ P(X = 7) = \frac{10!}{3!} (0.81)^7(0.19)^3 \]**d.**\[ P(X = 7) = \frac{10!}{3!7!} (0.81)^3(0.19)^7 \]**e.**\[ P(X = 7) = \frac{7!}{3!} (0.81)^3(0.19)^7 \]**f.**\[ P(X = 7) = \frac{10!}{7!3!} (0.81)^7(0.19)^3 \]To solve such a probability problem using the binomial distribution, the correct formula involves calculating the binomial coefficient and multiplying it by the success and failure probabilities raised to the appropriate powers. The correct choice for this scenario is **f.**### Explanation:For the binomial distribution, the probability \( P(X = k) \) of getting exactly \( k \) successes in \( n \) trials is given by:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Where:- \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient.- \(p\) is the probability of success on an individual trial.- \(

A professional basketball player has an 81% success rate when shooting free throws. Let the random variable X represent the number of free throws he makes in a random sample of 10 free throws (assume this experiment meets all binomial requirements). What is the probability that he makes exactly 7 of the 10 free throws? Note: You do not need to solve for the exact probability; you just need to choose the equation below that correctly represents how you would solve for this probability. 10! -(0.81)7(0.19)10 3!7! P(X=7) = а. 10! P(X=7) = Ob. -(0.81)7(0.19)3 7! P(X=7) = Oc. 10! -(0.81)7(0.19)3 3! P(X=7) = d. 10! -(0.81)³(0.19)7 3!7! P(X =7) = (0.81)P(0.19)7 P(X=7) = 3!

A professional basketball player has an 81% success rate when shooting free throws. Let the random variable X represent the number of free throws he makes in a random sample of 10 free throws (assume this experiment meets all binomial requirements). What is the probability that he makes exactly 7 of the 10 free throws? Note: You do not need to solve for the exact probability; you just need to choose the equation below that correctly represents how you would solve for this probability. 10! -(0.81)7(0.19)10 3!7! P(X=7) = а. 10! P(X=7) = Ob. -(0.81)7(0.19)3 7! P(X=7) = Oc. 10! -(0.81)7(0.19)3 3! P(X=7) = d. 10! -(0.81)³(0.19)7 3!7! P(X =7) = (0.81)P(0.19)7 P(X=7) = 3!

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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