The best free throw shooter on a basketball team has a 60% chance of making the 1st shot when shooting 2 free throws. If he makes the 1st shot, he has an 80% chance of making the 2nd. If he misses the 1st shot, he only has a 40% chance of making the 2nd shot. What are the possible outcomes and the probability of each outcome?
H -- hit
M -- miss
possible outcomes
HH -- (.6)(.8) = .48
HM -- (.6)(.2) = .12
MH -- (.4)(.4) = .16
MM -- (.4)(.6) = .24
notice the sum = 1 , as expected
List equally likely outcomes in the sample space for the experiment of rolling a pair of dice. One student list 5,6 and 6,5. Another student staye
List equally likely outcomes in the sample space for the experiment of rolling a pair of dice. One student list 5,6 and 6,5. Another student states only one of these need to be listed. Which student is correct and why and what activity could be described to help further explain why.
Thanks Reiny!
To find the possible outcomes and their probabilities, we can use a probability tree.
Let's start by representing the possible outcomes of the first shot:
1st Shot: Made (60%)
-> 2nd Shot: Made (80%)
-> 2nd Shot: Missed (20%)
1st Shot: Missed (40%)
-> 2nd Shot: Made (40%)
-> 2nd Shot: Missed (60%)
Now we can calculate the probabilities of each outcome by multiplying the probabilities along the corresponding branches of the tree.
1. Making both shots: (60%) * (80%) = 0.48 (48%)
2. Making the first shot and missing the second: (60%) * (20%) = 0.12 (12%)
3. Missing the first shot and making the second: (40%) * (40%) = 0.16 (16%)
4. Missing both shots: (40%) * (60%) = 0.24 (24%)
Therefore, the possible outcomes and their corresponding probabilities are:
1. Making both shots: 48%
2. Making the first shot and missing the second: 12%
3. Missing the first shot and making the second: 16%
4. Missing both shots: 24%