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A basketball player makes 60% of his shots from the free throw line. Suppose that each of his shots can be considered independent and that he throws 4 shots. Let x = the number of shots that he makes. What is the probability that he makes 3 shots?.

Sagot :

facundo3141592

The probability of making 3 out of 4 shots is:

P = 0.3456

How to find the probability?

We know that the player makes 60% of his shots, then we have:

  • Probability of making the shot = 0.6
  • Probability of missing the shot = 0.4.

Because the shots are independent, making 3 shots means that, for example, he makes the first 3, and then misses the fourth, so the probability of this event is:

P = (0.6)*(0.6)*(0.6)*(0.4)

But take in mind that this is only one permutation, here the missed shot is the last one, but it also could be the first one, second, or third, so we have 4 possible permutations.

So the probability of making 3 shots is:

P(x = 3) = 4*(0.6)*(0.6)*(0.6)*(0.4) = 0.3456

If you want to learn more about probability, you can read:

https://brainly.com/question/251701

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