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!! :)

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

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

To answer your questions, let's break down the problem step by step:

Question 1: What number of points is Jenny most likely to make assuming a 60% free-throw shooting ability?

To find the number of points Jenny is most likely to make, we need to calculate the probability of making 0, 1, or 2 points.

1. Probability of making 0 points: If Jenny misses her first shot, she won't get a chance to shoot the second shot. The probability of this happening is 40% (1 - 60%). Therefore, the probability of making 0 points is 40%.

2. Probability of making 1 point: Jenny can only make 1 point if she makes her first shot but misses the second one. The probability of making her first shot is 60%, and the probability of missing the second shot is 40% (1 - 60%). Therefore, the probability of making 1 point is 60% * 40% = 24%.

3. Probability of making 2 points: Jenny can make 2 points if she makes both her first and second shots. The probability of making both shots is 60% * 60% = 36%.

Now, let's compare the probabilities to find the number of points Jenny is most likely to make. Since the probability of making 2 points (36%) is higher than both the probabilities of making 0 points (40%) and making 1 point (24%), Jenny is most likely to make 2 points.

Therefore, the number of points Jenny is most likely to make is 2.

Question 2: How many points on average can Jenny expect to make when shooting a one-and-one?

To calculate the average number of points, we need to multiply the probability of each outcome by the number of points it corresponds to and sum them.

1. Probability of making 0 points: 40% probability * 0 points = 0 points.
2. Probability of making 1 point: 60% probability * 1 point = 0.6 points.
3. Probability of making 2 points: 36% probability * 2 points = 0.72 points.

Now, let's sum up the expected points: 0 points + 0.6 points + 0.72 points = 1.32 points.

Therefore, Jenny can expect to make an average of 1.32 points when shooting a one-and-one.

Question 3: What number of points is Jenny most likely to make when shooting a one-and-one with a 70% shooting probability for the second shot?

Similar to the first question, let's calculate the probabilities of making 0, 1, or 2 points.

1. Probability of making 0 points: Same as before, if Jenny misses her first shot, she won't get a chance to shoot the second shot. The probability of this happening is 40% (1 - 60%).

2. Probability of making 1 point: Jenny can only make 1 point if she makes her first shot but misses the second one. The probability of making her first shot is 60%, and the probability of missing the second shot is 30% (1 - 70%).

3. Probability of making 2 points: Jenny can make 2 points if she makes both her first and second shots. The probability of making both shots is 60% * 70% = 42%.

Now, comparing the probabilities, we can see that the probability of making 2 points (42%) is higher than both the probabilities of making 0 points (40%) and making 1 point (18%).

Therefore, the number of points Jenny is most likely to make when shooting a one-and-one with a 70% shooting probability for the second shot is 2.

I hope this helps! Let me know if you have any further questions.

To solve these questions, we need to calculate the probabilities of each outcome and determine the most likely number of points Jenny will make.

Let's go step by step:

1. Assuming Jenny is a 60% free-throw shooter, we can calculate the probabilities for each outcome. The probability of making the first shot is 0.6, while the probability of missing is 0.4. Since the second shot is only allowed if the first shot is made, the probability of making the second shot is also 0.6.

To determine the most likely number of points, we need to compare the probabilities of each outcome.

- The probability of scoring 0 points is 0.4 (missing the first shot).
- The probability of scoring 1 point is 0.6 * 0.4 = 0.24 (making the first shot and missing the second).
- The probability of scoring 2 points is 0.6 * 0.6 = 0.36 (making both shots).

Since the probability of scoring 2 points (0.36) is the highest among the three outcomes, we can conclude that Jenny is most likely to make 2 points.

2. To determine the average number of points Jenny can expect to make when shooting a one-and-one, we can calculate the weighted average of the possible outcomes.

- The probability of scoring 0 points is 0.4 (missing the first shot).
- The probability of scoring 1 point is 0.6 * 0.4 = 0.24 (making the first shot and missing the second).
- The probability of scoring 2 points is 0.6 * 0.6 = 0.36 (making both shots).

Now we multiply each outcome by its corresponding probability and sum them up:

Expected points = (0 points * 0.4) + (1 point * 0.24) + (2 points * 0.36)
Expected points = 0 + 0.24 + 0.72
Expected points = 0.96 points

Jenny can expect to make an average of 0.96 points when shooting a one-and-one.

3. If Jenny's probability of making the second free throw increases to 70%, we need to adjust the calculations.

- The probability of making the first shot is still 0.6, while the probability of missing is 0.4.
- With an increased probability of making the second shot (0.7), the probability of missing the second shot becomes 0.3.

- The probability of scoring 0 points remains 0.4 (missing the first shot).
- The probability of scoring 1 point becomes 0.6 * 0.3 = 0.18 (making the first shot and missing the second).
- The probability of scoring 2 points becomes 0.6 * 0.7 = 0.42 (making both shots).

Now we compare the probabilities of each outcome.

Since the probability of scoring 2 points (0.42) is the highest among the three outcomes, we can conclude that Jenny is most likely to make 2 points when shooting a one-and-one with her increased free-throw percentage.

I hope this helps! Let me know if you have any further questions.