Question

STATISTICS HELP. PLEASE I WILL MARK BRAINLIEST!!

Jenny is a 60 percent free-throw shooter She gets to shoot a second free throw if and only if she makes her first shot. She can score 0 points( if she misses her first shot), 1 point (she makes her first but misses her second), or 2 points ( she makes both shots.)

Assuming that the probability for each shot is independent, what number of points is Jenny most likely to make?

How many points on average, can Jenny expect to make when she shoots a one-and-one (when she only gets a second shot if she makes the first)?

Suppose Jenny gains confidence after making her first free throw and the probability of her making the second free throw goes up to 70%. Now, what number of points is Jenny most likely to make when she shoots a one-and-one?

Thanks!! :)

Answer 1

There is a 40% possibility of making 0 points.

There is a (.6)(.4) = 24% possibility of making 1 point.

There is a (.6)(.6) = 36% possibility of making 2 points.

Expected value = 0 * .40 + 1 * .24 + 2 * .36 = 0.96  

a) She is most likely to make 0.96 points.

b)  

Expected value = 0 * ( .4) + 1 * (.6)(.3) + 2 * (.6)(.7)  

she is most likely to make 1.8 points.

Hope I helped.

Could you please mark me brainliest.

Answer 2

The expected a number of points will be 1 if Jenny is a 60 percent free-throw shooter.

What is probability?

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

We have:

Jenny is a 60 percent free-throw shooter She gets to shoot a second free throw if and only if she makes her first shot.

P(0 point) = 0.4

P(1 point) = 0.6×0.3 = 0.18

P(2 point) = 0.6×0.7 = 0.42

Expected number of points = (0×0.4) + (1×0.18) + (2×0.42)

= 1.02 ≈ 1

Thus, the expected number of points will be 1 if Jenny is a 60 percent free-throw shooter.

Learn more about the probability here:

brainly.com/question/11234923

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