/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 A recent college graduate is pla... [FREE SOLUTION] | 91影视

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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
The probability of the student passing all three exams is 0.504. 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

Define the events and probabilities

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

Find the probability of passing all three exams (P(A鈭〣鈭〤))

To find P(A鈭〣鈭〤), we can use the conditional probabilities. We know that P(A鈭〣鈭〤) = P(C|A鈭〣) * P(A鈭〣). Since P(A鈭〣) = P(B|A) * P(A), we can substitute and get: P(A鈭〣鈭〤) = P(C|A鈭〣) * P(B|A) * P(A) Now, plug in the given probabilities: P(A鈭〣鈭〤) = 0.7 * 0.8 * 0.9 = 0.504 So, the probability of the student passing all three exams is 0.504.
03

Find P(B'|A鈭〣'鈭狝'鈭〣')

To find this conditional probability, we will use the formula: P(B'|A鈭〣'鈭狝'鈭〣') = P(B'鈭�(A鈭〣'鈭狝'鈭〣')) / P(A鈭〣'鈭狝'鈭〣') First, we need to find P(A鈭〣'鈭狝'鈭〣'). This is equal to 1 - P(A鈭〣鈭〤), which is the complement of passing all three exams. We have already calculated P(A鈭〣鈭〤) = 0.504 in Step 2. So, P(A鈭〣'鈭狝'鈭〣') = 1 - 0.504 = 0.496 Next, we need to find P(B'鈭�(A鈭〣'鈭狝'鈭〣')). We know that A'鈭〣' is the event of not passing the first exam, and in this case, the student cannot take the second exam. Therefore, P(A'鈭〣') = 0. So, we only need to find P(B'鈭〢鈭〣'). Since P(B'|A) = 1 - P(B|A) = 1 - 0.8 = 0.2, we can say that: P(B'鈭〢鈭〣') = P(B'|A) * P(A) Now, plug in the given probabilities: P(B'鈭〢鈭〣') = 0.2 * 0.9 = 0.18 Finally, we can calculate the conditional probability: P(B'|A鈭〣'鈭狝'鈭〣') = P(B'鈭�(A鈭〣'鈭狝'鈭〣')) / P(A鈭〣'鈭狝'鈭〣') = 0.18 / 0.496 = 0.3629 So, given that the student 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.

Understanding Conditional Probability
Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. Think of it as updating your beliefs about the chances of something happening in light of new information. To calculate the conditional probability of an event B occurring given that event A has happened, we use the formula:

\[\begin{equation} P(B|A) = \frac{P(A \cap B)}{P(A)}\end{equation}\]
Where:
  • P(B|A) is the conditional probability of B given A
  • P(A \cap B) is the joint probability of A and B occurring together
  • P(A) refers to the probability of A occurring
In our actuarial exam problem, we calculated the conditional probability that the student passes the second exam given that she has passed the first by using the provided joint and individual probabilities. It's essential to correctly apply this concept, as it forms the basis of understanding complex probabilistic scenarios often encountered in actuarial science as well as in other fields that involve risk and uncertainty.
Joint Probability and its Significance
Joint probability comes into play when we are concerned with the likelihood of two events occurring simultaneously. This is expressed as:

\[\begin{equation} P(A \cap B)\end{equation}\]
The actuarial exam problem provides an excellent illustration of joint probability. To find the probability that our graduate passes all three exams, we calculate the joint probability of passing each consecutive exam, based on the conditional probabilities. As we are dealing with dependent events 鈥� where the outcome of one influences another 鈥� understanding joint probability enables actuaries to predict the likelihood of an individual facing various risks all at once. Always keep in mind that for dependent events, the joint probability is found by multiplying the probability of the first event by the conditional probability of the second event, given the first.
The Complement Rule Explained
The complement rule is a straightforward yet powerful concept in probability theory. It states that the probability of an event not occurring (the complement of the event) is 1 minus the probability of the event occurring.

\[\begin{equation} P(A') = 1 - P(A)\end{equation}\]
In terms of our actuarial exam scenario, we applied the complement rule to determine the probability of the student not passing all three exams 鈥� the complement of our previous calculation for passing all three exams. By subtracting the joint probability of passing all exams from 1, we arrived at the probability of at least one failure. This illustrates how the complement rule allows us to switch perspectives from success to failure (or vice-versa), which can be particularly useful in the field of insurance and risk management where understanding the full spectrum of outcomes is crucial.

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Most popular questions from this chapter

A total of 500 married working couples were polled about their annual salaries, with the following information resulting: $$\begin{array}{lcc} \hline & \multicolumn{2}{c} {\text { Husband }} \\ \\)\cline { 2 - 3 }\\( \text { Wife } & \begin{array}{c} \text { Less than } \\ \$ 25,000 \end{array} & \begin{array}{c} \text { More than } \\ \$ 25,000 \end{array} \\ \hline \text { Less than \$25,000 } & 212 & 198 \\ \text { More than \$25,000 } & 36 & 54 \\ \hline \end{array}$$ For instance, in 36 of the couples, the wife earned more and the husband earned less than \(\$ 25,000 .\) If one of the couples is randomly chosen, what is (a) the probability that the husband earns less than \(\$ 25,000 ?\) (b) the conditional probability that the wife earns more than \(\$ 25,000\) given that the husband earns more than this amount? (c) the conditional probability that the wife earns more than \(\$ 25,000\) given that the husband earns less than this amount?

An engineering system consisting of \(n\) components is said to be a \(k\) -out- of- \(n\) system \((k \leq n)\) if the system functions if and only if at least \(k\) of the \(n\) components function. Suppose that all components function independently of one another. (a) If the \(i\) th component functions with probability \(P_{i}, i=\) \(1,2,3,4,\) compute the probability that a 2 -out-of- 4 system functions. (b) Repeat part (a) for a 3 -out-of- 5 system. (c) Repeat for a \(k\) -out-of- \(n\) system when all the \(P_{i}\) equal \(p\) (that is, \(\left.P_{i}=p, i=1,2, \ldots, n\right)\)

With probability \(.6,\) the present was hidden by mom; with probability \(4,\) it was hidden by dad. When mom hides the present, she hides it upstairs 70 percent of the time and downstairs 30 percent of the time. Dad is equally likely to hide it upstairs or downstairs. (a) What is the probability that the present is upstairs? (b) Given that it is downstairs, what is the probability it was hidden by dad?

\(A\) and \(B\) play a series of games. Each game is independently won by \(A\) with probability \(p\) and by \(B\) with probability \(1-p .\) They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner of the series. (a) Find the probability that a total of 4 games are played. (b) Find the probability that \(A\) is the winner of the series.

If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is \(i ?\) Compute for all values of \(i\) between 2 and 12.
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