A basketball player makes 60% of the free throws that he attempts, except that if he has just tried and missed a free throw then his chances of making a second one go down to only 30%. Suppose he has just been awarded two free throws.
Find the probability that he makes both.
Find the probability that he makes at least one. (A tree diagram could help.)
Cannot draw tree diagram here.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
"at least" = one or both
P(both) = .6 * .6
P(only second) = .4 * .3
P(only first) = .6 * .7
Either-or probabilities are found by adding the individual probabilities. Add the three probabilities.
Well, let's break this down step by step.
First, let's look at the probability of the basketball player making both free throws. We have two scenarios: either he makes the first one and then makes the second one, or he misses the first one and then makes the second one.
The probability of making the first one is 60%, so that's a 0.6 chance. If he makes the first one, then the probability of making the second one is also 60%, so that's another 0.6 chance.
The probability of missing the first one is 40%, so that's a 0.4 chance. If he misses the first one, then the probability of making the second one drops to 30%, which is a 0.3 chance.
To find the overall probability of making both free throws, we add up the probabilities of the two scenarios:
(0.6 * 0.6) + (0.4 * 0.3) = 0.36 + 0.12 = 0.48
So, the probability that he makes both free throws is 48%.
Now let's move on to the probability of making at least one free throw. To find this, we need to look at the complementary event of missing both free throws. In other words, we'll find the probability of missing both throws (which we already know is 52%) and then subtract that from 100%.
100% - 52% = 48%
So, the probability that he makes at least one free throw is 48%.
I hope that answers your question! When it comes to basketball, it's all about percentages, right?
To find the probability that the basketball player makes both free throws, we can consider two cases: the first free throw is made, and the first free throw is missed.
Case 1: The first free throw is made:
- The probability of making the first free throw is 60% or 0.60.
- If the first free throw is made, the probability of making the second free throw is also 60% or 0.60.
- Therefore, the probability of making both free throws in this case is 0.60 * 0.60 = 0.36.
Case 2: The first free throw is missed:
- The probability of missing the first free throw is 40% or 0.40.
- With a missed first free throw, the chances of making the second free throw decrease to 30% or 0.30.
- Therefore, the probability of making both free throws in this case is 0.40 * 0.30 = 0.12.
Now, we can calculate the overall probability of making both free throws by adding the probabilities from each case:
0.36 + 0.12 = 0.48
Therefore, the probability that the basketball player makes both free throws is 0.48.
To find the probability that he makes at least one free throw, we can use a tree diagram:
M (Missed)
/ \
M (30%) S (70%) S (Success)
/ \
M (30%) S (Success)
The probability of making at least one free throw is the sum of the probabilities of each path that leads to success:
- Path 1: Missed, Missed = 0.40 * 0.30 = 0.12
- Path 2: Missed, Success = 0.40 * 0.70 = 0.28
- Path 3: Success = 0.60
Adding all probabilities:
0.12 + 0.28 + 0.60 = 1.00
Therefore, the probability that the basketball player makes at least one free throw is 1.00.
To find the probability that the basketball player makes both free throws, we need to consider a few scenarios:
Scenario 1: The player makes the first free throw (60% chance) and also makes the second free throw (60% chance):
Probability = (60% * 60%) = 0.6 * 0.6 = 0.36
Scenario 2: The player misses the first free throw (40% chance) and then makes the second free throw (30% chance since he just missed one):
Probability = (40% * 30%) = 0.4 * 0.3 = 0.12
Now, let's use the concept of the tree diagram to find the probability that the player makes at least one free throw.
In this case, we have two branches - one where the player makes the first free throw and another where the player misses the first free throw.
For the branch where the player makes the first free throw:
Probability = (60% * 100%) = 0.6
For the branch where the player misses the first free throw:
Probability = (40% * 30%) = 0.4 * 0.3 = 0.12
The probabilities for each branch indicate the likelihood of each event occurring. We can now calculate the probability of making at least one free throw by adding the probabilities of both branches.
Probability of making at least one free throw = 0.6 + 0.12 = 0.72
So, the probability that the basketball player makes both free throws is 0.36 (36%) and the probability that he makes at least one free throw is 0.72 (72%).