Zared plays basketball on his high school team. One of the things he needs to practice is his free throws. On his first shot, there is a probability of 0.6 that he will make the basket. If he makes a basket, his confidence grows and the probability he makes the next shot increases by 0.05. If he misses the shot, the probability he makes the next one decreases by 0.05.
He takes 5 shots. What is the probability he makes at least 3 shots? (Hint: a tree diagram might be a helpful strategy)
so, did you make the tree?
No, for the tree I start with a branch of 0.6 and 0.4 and from there branch off to .35,.45,.55, and .65?
Send detailed explanation ASAP
answer? please
To find the probability that Zared makes at least 3 shots out of 5, we can use a tree diagram. Let's break down the possible outcomes for each shot and calculate the probabilities step by step.
First, let's define the probabilities for each shot: P(make) = 0.6 and P(miss) = 0.4. Since these probabilities change based on Zared's performance, we need to calculate them for each shot.
For the first shot:
- The probability Zared makes the shot is P(make) = 0.6.
- The probability Zared misses the shot is P(miss) = 0.4.
For the second shot:
- If Zared makes the first shot, his confidence grows. Therefore, the probability Zared makes the second shot is increased by 0.05 to become 0.65 (P(make) = 0.6 + 0.05).
- If Zared misses the first shot, the probability Zared makes the second shot is decreased by 0.05 to become 0.55 (P(make) = 0.6 - 0.05).
For the third shot:
- If Zared made the first two shots, his confidence continues to grow. Therefore, the probability Zared makes the third shot is increased by another 0.05 to become 0.7 (P(make) = 0.65 + 0.05).
- If Zared missed one of the first two shots, the probability Zared makes the third shot is decreased by 0.05 to become 0.5 (P(make) = 0.55 - 0.05).
For the fourth shot:
- If Zared made the first three shots, the probability Zared makes the fourth shot is increased by another 0.05 to become 0.75 (P(make) = 0.7 + 0.05).
- If Zared missed two of the first three shots, the probability Zared makes the fourth shot is decreased by 0.05 to become 0.45 (P(make) = 0.5 - 0.05).
For the fifth shot:
- If Zared made the first four shots, the probability Zared makes the fifth shot is increased by another 0.05 to become 0.8 (P(make) = 0.75 + 0.05).
- If Zared missed three or more of the first four shots, the probability Zared makes the fifth shot is decreased by 0.05 to become 0.4 (P(make) = 0.45 - 0.05).
Now, let's calculate the probability of each outcome:
1) Zared makes at least 3 shots.
- This could happen in three ways: (make, make, make, make, make), (make, make, make, make, miss), (make, make, make, miss, make). Let's calculate the probabilities for each case and add them together.
- P(make, make, make, make, make) = 0.6 * 0.65 * 0.7 * 0.75 * 0.8
- P(make, make, make, make, miss) = 0.6 * 0.65 * 0.7 * 0.75 * 0.4
- P(make, make, make, miss, make) = 0.6 * 0.65 * 0.7 * 0.25 * 0.8
- Add the probabilities together: P(make at least 3 shots) = P(make, make, make, make, make) + P(make, make, make, make, miss) + P(make, make, make, miss, make)
2) Zared misses more than 2 shots.
- This could happen in three ways: (make, make, make, miss, miss), (make, make, miss, miss, miss), (make, miss, miss, miss, miss). Let's calculate the probabilities for each case and add them together.
- P(make, make, make, miss, miss) = 0.6 * 0.65 * 0.7 * 0.25 * 0.6
- P(make, make, miss, miss, miss) = 0.6 * 0.65 * 0.3 * 0.2 * 0.4
- P(make, miss, miss, miss, miss) = 0.6 * 0.35 * 0.2 * 0.15 * 0.4
- Add the probabilities together: P(miss more than 2 shots) = P(make, make, make, miss, miss) + P(make, make, miss, miss, miss) + P(make, miss, miss, miss, miss)
Finally, calculate the required probability:
P(make at least 3 shots) / (P(make at least 3 shots) + P(miss more than 2 shots))
Note: The tree diagram is a helpful visual tool to understand the problem and calculate the probabilities step by step.