The Problem:
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)
3 marks for showing possible outcomes
5 marks for showing work
1 mark for a correct strategy to find final probability
1 mark for correct final probability
To solve this problem, we can use a tree diagram to represent all the possible outcomes of Zared's free throw shots. Each branch of the tree will represent whether Zared makes or misses the shot, and we will calculate the probability associated with each branch.
Let's start by drawing the tree diagram:
---- M (0.6) ---- M (0.65) ---- M (0.7) ---- M (0.75) ---- M (0.8)
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------ M (0.6) ---- M (0.65) ---- M (0.7) ---- M (0.75) ---- H (0.8)
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----- M (0.6) ---- M (0.65) ---- M (0.7) ---- H (0.75)
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S (0.6) -- H (0.65)
In the tree diagram:
- 'M' represents making the shot,
- 'H' represents missing the shot,
- The numbers in parentheses represent the probability of that outcome.
Now, let's calculate the probability for each branch starting from the initial shot.
1. If Zared makes the first shot (M - 0.6), there are four possible outcomes (M - 0.65, M - 0.7, M - 0.75, and M - 0.8). We multiply the probabilities of each outcome to get the probability of this branch: 0.6 * 0.65 * 0.7 * 0.75 * 0.8.
2. If Zared misses the first shot (H - 0.4), there are two possible outcomes (M - 0.65 and M - 0.7). We multiply the probabilities of each outcome to get the probability of this branch: 0.4 * 0.65 * 0.7.
3. If Zared misses the first two shots (H - 0.4 - H - 0.35), there is only one possible outcome (M - 0.7). We multiply the probability of this outcome to get the probability of this branch: 0.4 * 0.35 * 0.7.
4. If Zared misses the first three shots (H - 0.4 - H - 0.35 - H - 0.3), there are no more shots, so the probability of this branch is simply the product of the probabilities of each outcome: 0.4 * 0.35 * 0.3.
Adding up the probabilities of all the branches where Zared makes at least three shots, we get:
Probability = (0.6 * 0.65 * 0.7 * 0.75 * 0.8) + (0.4 * 0.65 * 0.7) + (0.4 * 0.35 * 0.7) + (0.4 * 0.35 * 0.3)
Now, we can calculate this value to find the final probability.