Question

Amy is playing basketball. The probability that Amy makes a basket from the free throw line is p = 4/5. The probability of making any free throw is the same regardless of recent attempts made.

(a) (1 point) Is the event that Amy makes her second free throw independent of the event that she makes her first free throw?

(b) (3 points) A made free throw is worth one point. A miss scores zero points. If Amy takes three free throws, how many total points will she score on average?

(c) (3 points) What is the probability that Amy makes her first two free throws and misses her third?

(d) (3 points) What is the probability that Amy makes any two out of three free throws?

Answer 1

a) Yes because it is given that the  probability of making any free throw is the same regardless of recent attempts made.

b) Amy makes a basket from the free throw line is p = 4/5.

therefore average = n * p = 3*(4/5) = 12/5 = 2.4

c) P(Amy makes her first two free throws and misses her third) = p^2 (1 - p) = (16/25)*(1/5) = 16/125 = 0.128

d) there are ³C₂ = 3 combination of making 2 throws out of 3 throws.

So just multiply the answer of part c) by 3 so we get the required probability as 3*0.128 = 0.384