A basketball player is fouled while attempting to make a basket and receives two

Question

A basketball player is fouled while attempting to make a basket and receives two free throws. The opposing coach believes there is a 27% chance that the player will miss both shots, a 25% chance that he will make one of the shots, and a 48% chance that he will make both shots.

Construct the appropriate probability distribution. (Round your answers to 2 decimal places.)

What is the probability that he makes no more than one of the shots? (Round your answer to 2 decimal places.)

What is the probability that he makes at least one of the shots? (Round your answer to 2 decimal places.)

A basketball player is fouled while attempting to make a basket and receives two free throws. The opposing coach believes there is a 27% chance that the player will miss both shots, a 25% chance that he will make one of the shots, and a 48% chance that he will make both shots.

Explanation / Answer

Solution :

We are given that : A basketball player is fouled while attempting to make a basket and receives two free throws.

The opposing coach believes there is a 27% chance that the player will miss both shots,

That is P( Missing both) = P( X = 0) = 0.27

a 25% chance that he will make one of the shots,

That is :P( Making one of the shots) = P(X = 1) = 0.25

and a 48% chance that he will make both shots.

That is : P( Making both Shots) = P( X = 2) = 0.48

Part a) Construct the appropriate probability distribution

Let X= number of shots make.

Then probability distribution of X is :

Part b) What is the probability that he makes no more than one of the shots?

That is we have to find : P( X 1 ) =..........?

Thus P( X 1 ) = P( X = 0) + P( X = 1)

       P( X 1 ) = 0.27 + 0.25

        P( X 1 ) = 0.52

Part c) What is the probability that he makes at least one of the shots?

That is we have to find : P( X 1) = ...........?

P( X 1) = P( X= 1) + P( X = 2 )

P( X 1) = 0.25 + 0.48

P( X 1) = 0.73

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