(a)
To determine: The probability model for the number of points the player may score on a one-and-one opportunity.
The probability model for the number of points the player may score on a one-and-one opportunity is:
Y | 0 | 1 | 2 |
P(Y) | 0.2 | 0.16 | 0.64 |
Given information:
Overall, a certain player makes 80% of his foul shots.
Formula used:
P(miss) = 100%-P(made)
Calculation:
Let Y = number of points scored by the player on a one and one opportunity.
The player has an 80% chance of making a foul shot When the first shot is missed, then 0 points are scored.
P(0) =P(miss) = 100%-P(made) = 100%-80% = 20% = 0.20
1 point will be scored when the first shot is made and the second shot is missed.
P(1) = P(made)×P(miss) = 0.80×0.20 = 0.16
2 points will be scored when the first two shots are made.
P(2) = P(made)×P(made) = 0.80×0.80 = 0.64
Let us combine the found probabilities in a table:
Y | 0 | 1 | 2 |
P(Y) | 0.2 | 0.16 | 0.64 |
(b)
To determine: The expected number of points.
The expected number of points.is 1.44.
Given information:
Overall, a certain player makes 80% of his foul shots.
Formula used:
P(miss) = 100%-P(made)
Calculation:
Let Y = number of points scored by the player on a one and one opportunity.
The player has an 80% chance of making a foul shot When the first shot is missed, then 0 points are scored.
P(0) =P(miss) = 100%-P(made) = 100%-80% = 20% = 0.20
1 point will be scored when the first shot is made and the second shot is missed.
P(1) = P(made)×P(miss) = 0.80×0.20 = 0.16
2 points will be scored when the first two shots are made.
P(2) = P(made)×P(made) = 0.80×0.80 = 0.64
Let us combine the found probabilities in a table:
Y | 0 | 1 | 2 |
P(Y) | 0.2 | 0.16 | 0.64 |
The expected value (or mean) is the sum of the product of the possibilities and its probability P(x):
Substitute
Hence, the expected number of points.is 1.44.
Chapter 10 Solutions
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