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Chapter 3: Problem 12

A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the first actuarial exam in June. If she passes that exam, then she will take the second exam in July, and if she also passes that one, then she will take the third exam in September. If she fails an exam, then she is not allowed to take any others. The probability that she passes the first exam is \(.9 .\) If she passes the first exam, then the conditional probability that she passes the second one is \(.8,\) and if she passes both the first and the second exams, then the conditional probability that she passes the third exam is .7. (a) What is the probability that she passes all three exams? (b) Given that she did not pass all three exams, what is the conditional probability that she failed the second exam?

Short Answer

Expert verified
(a) The probability that she passes all three exams is 0.504. (b) Given that she did not pass all three exams, the conditional probability that she failed the second exam is approximately 0.3629.

Step by step solution

01

Identify relevant probabilities

We are given the following probabilities: - The probability of passing the first exam, P(A) = 0.9. - The conditional probability of passing the second exam given that she passed the first exam, P(B|A) = 0.8. - The conditional probability of passing the third exam given that she passed the first two exams, P(C|A鈭〣) = 0.7.
02

Calculate the probability of passing all three exams

We want to find the probability P(A鈭〣鈭〤), which represents the probability of passing all three exams. We can do this using the conditional probabilities: P(A鈭〣鈭〤) = P(C|A鈭〣) * P(A鈭〣). To find P(A鈭〣), we can use the conditional probability P(B|A): P(A鈭〣) = P(B|A) * P(A). Now we can calculate P(A鈭〣鈭〤): P(A鈭〣鈭〤) = P(C|A鈭〣) * P(A鈭〣) = 0.7 * (0.8 * 0.9) = 0.504. The probability that she passes all three exams is 0.504. b) Failing the second exam given not passing all three exams:
03

Identify relevant probabilities

We need to find the conditional probability P(B'|(A鈭〣鈭〤)'), where B' is the event of failing the second exam, and (A鈭〣鈭〤)' is the event of not passing all three exams.
04

Apply the conditional probability formula

Using the definition of conditional probability, we can write: P(B'|A鈭〣鈭〤') = P(B'鈭�(A鈭〣鈭〤')) / P(A鈭〣鈭〤').
05

Calculate the relevant probabilities

We need to find the probabilities in the numerator and the denominator. First, let's find P(A鈭〣鈭〤'). This represents the probability that she passes the first two exams but fails the third. We can find this by subtracting the probability of passing all three exams from the probability of passing the first two exams: P(A鈭〣鈭〤') = P(A鈭〣) - P(A鈭〣鈭〤) = (0.8 * 0.9) - 0.504 = 0.216. Now let's calculate P(B'鈭�(A鈭〣鈭〤')). This represents the probability that she fails the second exam and doesn't pass all three exams. Since failing the second exam implies not passing all three exams, this probability is equivalent to the probability of passing the first exam and failing the second exam: P(B'鈭�(A鈭〣鈭〤')) = P(A) * P(B'|A) = 0.9 * (1 - 0.8) = 0.18. Finally, we need to find P(A鈭〣鈭〤'). This is the probability of not passing all three exams, which is the complement of passing all three exams: P(A鈭〣鈭〤') = 1 - P(A鈭〣鈭〤) = 1 - 0.504 = 0.496.
06

Calculate the conditional probability

Now we can find the conditional probability P(B'|A鈭〣鈭〤'): P(B'|A鈭〣鈭〤') = P(B'鈭�(A鈭〣鈭〤')) / P(A鈭〣鈭〤') = 0.18 / 0.496 鈮� 0.3629. Given that she did not pass all three exams, the conditional probability that she failed the second exam is approximately 0.3629.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
Understanding conditional probability is crucial when dealing with sequential events where the outcome of one affects the outcome of another. In the case of our actuarial exam scenario, conditional probability is used to express the likelihood of passing subsequent exams given that previous exams have been passed.

For example, the conditional probability that the student passes the second exam, given she has passed the first, is denoted as P(B|A). This is not the same as the probability of passing the second exam without any prior conditions, as it only considers scenarios where the first exam has been passed. Similarly, P(C|A鈭〣) signifies the probability of passing the third exam, assuming the student has already succeeded in the first two exams.
Probability Theory
Probability theory is the mathematical framework for quantifying uncertainty and making predictions about random events. It's the backbone of actuarial science. In our textbook exercise, we apply probability theory to determine the likelihood of various outcomes for the actuarial exams. The multiplication rule, a fundamental concept in probability theory, allows us to calculate the probability of passing all three exams by multiplying individual conditional probabilities. To express this mathematically, the probability of passing all three exams, or P(A鈭〣鈭〤), is determined by P(C|A鈭〣) multiplied by P(A鈭〣) 鈥� clearly illustrating how each step relies on previous outcomes.
Exam Passing Probability
Exam passing probability involves calculating the chance that a student meets or exceeds the requirements to pass an examination or series of examinations. In the context of actuarial exams, which are taken in sequence, this involves a specific set of conditional probabilities. The probability for passing all three exams, denoted P(A鈭〣鈭〤), was found to be 0.504, or 50.4%. This relatively high probability of success reflects the student's strong preparation and understanding of the subject matter. When analyzing situations where all exams are not passed, the calculations become a bit more intricate, involving the complementary events and their associated probabilities, which in this example is a detailed use of conditional probability showcasing the student's likelihood of failing the second exam given the initial conditions.

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Most popular questions from this chapter

Independent flips of a coin that lands on heads with probability \(p\) are made. What is the probability that the first four outcomes are (a) \(H, H, H, H ?\) (b) \(T, H, H, H ?\) (c) What is the probability that the pattern \(T, H, H, H\) occurs before the pattern \(H, H, H, H ?\) Hint for part \((c):\) How can the pattern \(H, H, H, H\) occur first?

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