Question

A basketball player makes 40% of his shots from the free throw line. suppose that each of his shots can be considered independent and that he throws 3 shots. let x = the number of shots that he makes. what is the probability that he makes 2 shots or less?

Answer 1

The probability of making 2 or fewer shots is 0.936.

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P(≤2) = 1-P(3) = 1-(0.4)^3 = 1 -0.064 = 0.936

Answer 2

Using the binomial distribution, it is found that there is a 0.936 = 93.6% probability that he makes 2 shots or less.

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For each throw, there are only two possible outcomes. Either the player makes it, or he misses it. The probability of making a throw is independent of any other throw, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

• x is the number of successes.
• n is the number of trials.
• p is the probability of a success on a single trial.

In this question:

• The player makes 40% of the shots, thus $p = 0.4$
• 3 shots, thus $n = 3$

The probability of making 2 or less is the probability that he does not make all of them, that is:

$P(X < 3) = 1 - P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.4)^{3}.(0.6)^{0} = 0.064$

$P(X < 3) = 1 - P(X = 3) = 1 - 0.064 = 0.936$

0.936 = 93.6% probability that he makes 2 shots or less.

A similar problem is given at https://brainly.com/question/15557838