Exam P sample questions, with full solutions

1. Question 1General Probability

   Of the policyholders at an insurer, 64% hold an auto policy, 38% hold a homeowners policy and 22% hold both. One policyholder is chosen at random.

   Calculate the probability that the policyholder holds exactly one of the two policies.

   A0.20B0.42C0.58D0.64E0.80

   Show the solution

   1. At least one policy: $P(A \cup H)$ $= 0.64 + 0.38 - 0.22$ $= 0.80$.
   2. Exactly one is at least one, minus both: $0.80 - 0.22$ $= 0.58$.
   3. Check: auto only is $0.64 - 0.22$ $= 0.42$ and homeowners only is $0.38 - 0.22$ $= 0.16$, and $0.42 + 0.16$ $= 0.58$.
2. Question 2General Probability

   An insurer sorts its drivers into three risk classes: 50% are low risk, 35% are medium risk and 15% are high risk. In a year, the probability of at least one claim is 0.05 for a low-risk driver, 0.12 for a medium-risk driver and 0.30 for a high-risk driver.

   A driver chosen at random makes at least one claim this year. Calculate the probability that the driver is high risk.

   A0.150B0.300C0.375D0.402E0.638

   Show the solution

   1. Law of total probability: $P(C)$ $= 0.50(0.05)$${}+ 0.35(0.12)$${}+ 0.15(0.30)$ $= 0.025 + 0.042 + 0.045$ $= 0.112$.
   2. Bayes' theorem: $P(\text{high} \mid C)$ $= \dfrac{0.045}{0.112}$ $= 0.402$.
3. Question 3General Probability

   A working group of 4 people is chosen at random from 7 actuaries and 5 underwriters.

   Calculate the probability that the group contains at least 2 underwriters.

   A0.010B0.141C0.354D0.424E0.576

   Show the solution

   1. Equally likely groups: $\binom{12}{4}$ $= 495$.
   2. Two underwriters: $\binom{5}{2}\binom{7}{2}$ $= 10 \times 21$ $= 210$. Three: $\binom{5}{3}\binom{7}{1}$ $= 70$. Four: $\binom{5}{4}$ $= 5$.
   3. $P$ $= \dfrac{210 + 70 + 5}{495}$ $= \dfrac{285}{495}$ $= 0.576$.
4. Question 4Univariate Random Variables

   The number of claims reported to an insurer in a week has a Poisson distribution with mean 2.5.

   Given that at least one claim is reported in a week, calculate the probability that at least three claims are reported that week.

   A0.456B0.497C0.544D0.713E0.918

   Show the solution

   1. $P(N \le 2) = e^{-2.5}\left(1 + 2.5 + \dfrac{2.5^2}{2}\right) = 0.08208 \times 6.625 = 0.54381$, so $P(N$ $\ge 3)$ $= 0.45619$.
   2. $P(N$ $\ge 1)$ $= 1 - e^{-2.5}$ $= 0.91792$.
   3. Since $N$ $\ge 3$ already means $N$ $\ge 1$: $P(N$ $\ge 3 \mid N$ $\ge 1)$ $= \dfrac{0.45619}{0.91792}$ $= 0.497$.
5. Question 5Univariate Random Variables

   A loss $X$ has an exponential distribution with mean 2,000. An insurance policy pays the part of the loss above a deductible of 500, up to a maximum payment of 3,000.

   Calculate the expected payment per loss.

   A1,054B1,111C1,210D1,558E1,652

   Show the solution

   1. The payment is 0 for $X$ $\le 500$, $X - 500$ for $500$ $< X$ $\le 3500$, and 3,000 for $X$ $> 3500$.
   2. So the payment is $\min(X, 3500)$${}- \min(X, 500)$, and for an exponential $E[\min(X, m)] = \theta\left(1 - e^{-m/\theta}\right)$.
   3. $E[\text{payment}] = 2000\left(e^{-500/2000} - e^{-3500/2000}\right) = 2000(0.77880 - 0.17377) = 1{,}210$.
6. Question 6Univariate Random Variables

   Total annual claims for a block of policies are normally distributed with mean 50,000 and standard deviation 8,000. The insurer sets a reserve so that the probability total claims exceed the reserve is 0.05.

   Calculate the reserve.

   A50,000B58,000C60,252D63,159E65,680

   Show the solution

   1. We need $P(Z$ $> z)$ $= 0.05$, so $P(Z$ $< z)$ $= 0.95$: the table's list of common values gives $z$ $= 1.6449$.
   2. Reserve $= 50{,}000$${}+ 1.6449 \times 8{,}000$ $= 63{,}159$. With z rounded to 1.645 it is 63,160, and the closest choice is still 63,159.
7. Question 7Univariate Random Variables

   A random variable $X$ has density $f(x)$ $= \dfrac{3x^2}{64}$ for $0$ $< x$ $< 4$, and 0 otherwise.

   Calculate the median of $X$.

   A2.00B2.52C3.00D3.17E3.36

   Show the solution

   1. $F(x) = \displaystyle\int_0^x \frac{3t^2}{64}\,dt = \frac{x^3}{64}$ for $0$ $< x$ $< 4$.
   2. The median $m$ solves $F(m)$ $= 0.5$: $m^3$ $= 32$.
   3. $m$ $= 32^{1/3}$ $= 3.17$.
8. Question 8Multivariate Random Variables

   The joint probability function of $X$ and $Y$ is given in the table.

   | | Y = 0 | Y = 1 | Y = 2 | |---|---|---|---| | X = 0 | 0.20 | 0.15 | 0.05 | | X = 1 | 0.10 | 0.25 | 0.25 |

   Calculate $\operatorname{Cov}(X, Y)$.

   A0.00B0.15C0.24D0.60E0.75

   Show the solution

   1. Marginals: $P(X$ $= 1)$ $= 0.60$, so $E[X]$ $= 0.60$. $P(Y$ $= 1)$ $= 0.40$ and $P(Y$ $= 2)$ $= 0.30$, so $E[Y]$ $= 0.40 + 0.60$ $= 1.00$.
   2. $E[XY]$: only $X$ $= 1$ contributes, $1(0.25) + 2(0.25)$ $= 0.75$.
   3. $\operatorname{Cov}(X, Y)$ $= 0.75$${}- 0.60 \times 1.00$ $= 0.15$.
9. Question 9Multivariate Random Variables

   An insurer has 100 independent policies. Each policy's annual claim amount has mean 400 and standard deviation 300.

   Using the normal approximation, calculate the probability that total annual claims exceed 43,000.

   A0.0000B0.1587C0.3413D0.4602E0.8413

   Show the solution

   1. The total $S$ has mean $100 \times 400$ $= 40{,}000$ and variance $100 \times 300^2$ $= 9{,}000{,}000$, so its standard deviation is 3,000.
   2. $z$ $= \dfrac{43{,}000 - 40{,}000}{3{,}000}$ $= 1.00$.
   3. $P(S$ $> 43{,}000)$ $\approx 1 - \Phi(1.00)$ $= 1 - 0.8413$ $= 0.1587$.
10. Question 10Multivariate Random Variables

    A device has two components that work independently. Component 1 lasts an exponential time with mean 6 years, and component 2 an exponential time with mean 3 years. The device fails as soon as either component fails.

    Calculate the probability that the device works for more than 3 years.

    A0.2231B0.3679C0.5134D0.6065E0.7513

    Show the solution

    1. The device lasts past 3 years only if both components do: $P(T$ $> 3)$ $= P(T_1$ $> 3)P(T_2$ $> 3)$.
    2. $e^{-3/6} \times e^{-3/3}$ $= e^{-1.5}$ $= 0.2231$.
    3. Equivalently, the first failure is exponential with rate $\tfrac16 + \tfrac13$ $= \tfrac12$, a mean of 2 years.

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