Question

I was sick and when I came back to school I received this question as homework and I don't understand it...

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

draw a tree with roots :)

first
yes .6 no .4

second
yes,yes .6*.6 = .36
yes,no .6*.4 = .24
no, yes .4*.6 = .36 NOT allowed
no, no .4*.4 = .16 NOT allowed

so
.36 * 2 points + .24 * 1 point

in the second case
it is
.6 *.7 * 2 + .6 * .3* 1

Answer 2

I understand that I have to draw a tree, but the way you worded the response, I don't understand.