Math Problem Statement

Peter wants to pass his driving test. The probability that he passes at his first attempt is 0.7. When Peter passes his driving test, he does not take it again. If he fails, the probability that he passes at the next attempt is 0.8. Calculate the probability that Peter needs exactly two attempts to pass his driving test and the probability that Peter passes at his third or fourth attempt.

Solution

The problem revolves around calculating probabilities based on a scenario where Peter attempts a driving test multiple times, with given success and failure probabilities. Here's a breakdown of the tasks:

(a) Completing the Probability Tree Diagram

The diagram shows:

  • First attempt:
    • Success probability P(Pass)=0.7P(\text{Pass}) = 0.7,
    • Failure probability P(Fail)=0.3P(\text{Fail}) = 0.3.
  • Second attempt (if he fails the first):
    • Success probability P(Pass|Fail)=0.8P(\text{Pass|Fail}) = 0.8,
    • Failure probability P(Fail|Fail)=0.2P(\text{Fail|Fail}) = 0.2.

This matches the given probabilities in the tree.

(b) Probability Peter Needs Exactly Two Attempts

For Peter to pass on his second attempt, the sequence of events is:

  1. Fail the first attempt: P(Fail first)=0.3P(\text{Fail first}) = 0.3,
  2. Pass the second attempt: P(Pass second | Fail first)=0.8P(\text{Pass second | Fail first}) = 0.8.

The combined probability is:

= 0.3 \cdot 0.8 = 0.24.$$ --- ### **(c) Probability Peter Passes on the Third or Fourth Attempt** To calculate this, we need to analyze the sequence of events: 1. **Pass on third attempt:** - Fail the first and second attempts: $$P(\text{Fail first}) \cdot P(\text{Fail second | Fail first}) = 0.3 \cdot 0.2 = 0.06.$$ - Pass the third attempt: $$P(\text{Pass third | Fail twice}) = 0.8.$$ - Combined probability: $$P(\text{Pass on third}) = 0.06 \cdot 0.8 = 0.048.$$ 2. **Pass on fourth attempt:** - Fail the first, second, and third attempts: $$P(\text{Fail first}) \cdot P(\text{Fail second | Fail first}) \cdot P(\text{Fail third | Fail twice}) = 0.3 \cdot 0.2 \cdot 0.2 = 0.012.$$ - Pass the fourth attempt: $$P(\text{Pass fourth | Fail three times}) = 0.8.$$ - Combined probability: $$P(\text{Pass on fourth}) = 0.012 \cdot 0.8 = 0.0096.$$ Adding the probabilities for passing on the third or fourth attempt: $$P(\text{Pass on third or fourth}) = 0.048 + 0.0096 = 0.0576.$$ --- ### Final Answer: - **(b)** Probability Peter needs exactly two attempts: $$0.24$$. - **(c)** Probability Peter passes on the third or fourth attempt: $$0.0576$$. --- Let me know if you'd like further clarification or a detailed explanation of these calculations! --- ### **5 Related Questions:** 1. What is the total probability that Peter passes the driving test within the first four attempts? 2. How would the probabilities change if the success rate on any subsequent attempt was reduced to 0.5? 3. What is the probability that Peter fails all his first four attempts? 4. How does altering the initial pass probability to 0.8 affect the outcomes? 5. Can this type of problem be modeled using a geometric probability distribution? ### **Tip:** For probability trees, always ensure that the probabilities for each branch sum to 1, as this checks for calculation errors.

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Math Problem Analysis

Mathematical Concepts

Probability
Conditional Probability
Tree Diagrams

Formulas

P(Event A and Event B) = P(A) * P(B|A)
P(Passing on nth attempt) = P(Failing all previous attempts) * P(Passing on nth)

Theorems

Multiplication Rule for Probability
Conditional Probability Theorem

Suitable Grade Level

Grades 10-12

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