Conditional Probability: Acturary Exam P
Conditional Probability measures the likelihood (probability) of one event given that another event has occurred. It’s useful in many fields, including finance, healthcare, and politics. Here, we look at some examples from the Acturary Probability Exam Sample Questions.
The first step is always to read the question and figure out “what is given” and “what is the goal”. Once we have a clear starting point and goal, the problem becomes easier. When doing so, we might feel it takes so long to solve a problem. Over time, you will master the process and become faster and faster. Remember,
Be Clear, Not Clever.
Question 1
A public health researcher examines the medical records of a group of 937 men who died in 1999 and discovers that 210 of the men died from causes related to heart disease. Moreover, 312 of the 937 men had at least one parent who suffered from heart disease, and, of these 312 men, 102 died from causes related to heart disease.
Calculate the probability that a man randomly selected from this group died of causes related to heart disease, given that neither of his parents suffered heart disease.
Solution
Given:
Let \(H\) = event that a death is due to heart disease, \(F\) = event that at least one parent suffered from heart disease.
Then based on the medical records,
\(N(H) = 210\) = the number of those who died from causes related to heart disease,
\(N(H \cap F) = 102\) = the number of those who died from causes related to heart disease and had at least one parent who suffered from heart disease, and
\(N(F) = 317\) = the number of those who died and had at least one parent who suffered from heart disease.
Lastly, \(N(F \cup F’) = 937 \) = the total number of men in the medical group who died in 1999.
Goal:
The conditional probability \(P[H | F’]\) means the probability that a man randomly selected from this group died of causes related to heart disease, given that neither of his parents suffered heart disease.
Let us try to achieve the goal:
Note that $$P[H \cap F’] = \frac{N(H \cap F’)}{937} = \frac{N(H) – N(H \cap F)}{937}= \frac{210-102}{937} = \frac{108}{937}$$ and
$$ P[F’] = \frac{N(F’)}{937} = \frac{937 – N(F)}{937}= \frac{937 – 312}{937} = \frac{625}{937}$$
Therefore, the probability that a man randomly selected from this group died of causes related to heart disease, given that neither of his parents suffered heart disease, is the following:
$$P[H | F’] = \frac{P[H \cap F’]}{P[F’]} = \frac{108/937}{625/937} = \frac{108}{625} = 0.173$$
Question 2
The number of injury claims per month is modeled by a random variable \(N\) with $$P[N=n] = \frac{1}{(n+1)(n+2)}, \mbox{for nonnegative integers, } n$$
Calculate the probability of at least one claim during a particular month, given that there have been at most four claims during that month.
Solution
This question is fairly straightforward. The crucial step is to express the desired probability using the conditional probability expression.
Let us try to achieve the goal:
The probability of at least one claim during a particular month, given that there have been at most four claims during that month, can be expressed as \( P[N\geq 1 | N \leq 4]\). Therefore,
\(\begin{aligned}
P[N\geq 1 | N \leq 4] &= \frac{P[1 \leq N \leq 4 }{P[N \leq 4]} \\
&= \frac{\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}}{\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}}\\
&= \frac 25
\end{aligned}\)
Question 3
An actuary is studying the prevalence of three health risk factors, denoted by \(A\), \(B\), and \(C\), within a population of women. For each of the three factors, the probability is 0.1 that a woman in the population has only this risk factor (and no others). For any two of the three factors, the probability is 0.12 that she has exactly these two risk factors (but not the other). The probability that a woman has all three risk factors, given that she has \(A\) and \(B\), is 1/3.
Calculate the probability that a woman has none of the three risk factors, given that she does not have risk factor \(A\).
Keynotes
- Don’t forget that the Venn diagram is a useful tool for easy and clear illustration.
Solution
Given:
For each of the three factors, the probability is 0.1 that a woman in the population has only this risk factor (and no others). For any two of the three factors, the probability is 0.12 that she has exactly these two risk factors (but not the other).
Although we can express the given information using probability notation and symbols, we instead use the Venn diagram for quick and easy illustration.
Let \(x\) denote the unknown probability \(P[A \cap B \cap C] = x\). The information that the probability that a woman has all three risk factors, given that she has A and B, is 1/3, can be written as $$ P[A \cap B \cap C | A \cap B] = \frac{x}{x + 0.12} = \frac 13$$
Goal:
The probability that a woman has none of the three risk factors, given that she does not have risk factor \(A\), is \( P[ (A \cup B \cup C)’ | A’]\)
Let us try to achieve the goal:
From the Venn diagram, we see that once we have the probability \(P[A \cap B \cap C]\), we can easily find the probabilities \(P[A \cup B \cup C]\) and \(P[A]\) and hence \( P[ (A \cup B \cup C)’ | A’]\). Let us go ahead and solve for \(x\). $$ P[A \cap B \cap C | A \cap B] = \frac{x}{x + 0.12} = \frac 13$$ shows that \( 3x = x+ 0.12\) and thus, $$x = 0.06$$.
The probability that a woman has none of the three risk factors, given that she does not have risk factor \(A\), is $$\begin{aligned}
P[ (A \cup B \cup C)’ | A’] &= \frac{P[ (A \cup B \cup C)’ ]}{P[A’]} \\
& = \frac{1 – P[ A \cup B \cup C ]}{1 – P[A]} \\
&= \frac{1 – 3(0.1) – 3(0.12)-0.06}{1-0.1 – 2(0.12)-0.06}\\
&= \frac{0.28}{0.60} \\
&=0.467
\end{aligned}$$
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