The number of injury claims per month is modeled by a random variable [math]N[/math] with

[[math]] \operatorname{P}[N=n] = \frac{1}{(n+1)(n+2)} [[/math]]

, for nonnegative integers, [math]n[/math]. Calculate the probability of at least one claim during a particular month, given that there have been at most four claims during that month.

• 1/3
• 2/5
• 1/2
• 3/5
• 5/6

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Step 1: Understand the Problem and Conditional Probability Formula

The problem asks for the probability of at least one claim during a particular month, given that there have been at most four claims during that month. This can be expressed using conditional probability notation as [math]\operatorname{P}[ N \ge 1 | N \le 4 ][/math]. The formula for conditional probability is:

[[math]]\operatorname{P}[A|B] = \frac{\operatorname{P}[A \cap B]}{\operatorname{P}[B]}[[/math]]

In this specific case:

• Event [math]A[/math] is "at least one claim," so [math]N \ge 1[/math].
• Event [math]B[/math] is "at most four claims," so [math]N \le 4[/math].
• The intersection [math]A \cap B[/math] is "at least one claim AND at most four claims," which means [math]1 \le N \le 4[/math].

Therefore, the required probability is:

[[math]]\operatorname{P}[ N \ge 1 | N \le 4 ] = \frac{\operatorname{P}[1 \le N \le 4]}{\operatorname{P}[N \le 4]}[[/math]]

The probability mass function (PMF) for [math]N[/math] is given by:

[[math]]\operatorname{P}[N=n] = \frac{1}{(n+1)(n+2)}[[/math]]

for non-negative integers [math]n[/math].

Step 2: Calculate Individual Probabilities [math]\operatorname{P}[N=n][/math] for Relevant Values

We need to calculate [math]\operatorname{P}[N=n][/math] for [math]n = 0, 1, 2, 3, 4[/math].

Individual Probabilities

[math]n[/math]	[math]\operatorname{P}[N=n] = \frac{1}{(n+1)(n+2)}[/math]	Value (Fraction)	Value (Decimal)
0	[math]\frac{1}{(0+1)(0+2)} = \frac{1}{1 \cdot 2}[/math]	[math]\frac{1}{2}[/math]	0.5
1	[math]\frac{1}{(1+1)(1+2)} = \frac{1}{2 \cdot 3}[/math]	[math]\frac{1}{6}[/math]	0.1667
2	[math]\frac{1}{(2+1)(2+3)} = \frac{1}{3 \cdot 4}[/math]	[math]\frac{1}{12}[/math]	0.0833
3	[math]\frac{1}{(3+1)(3+4)} = \frac{1}{4 \cdot 5}[/math]	[math]\frac{1}{20}[/math]	0.05
4	[math]\frac{1}{(4+1)(4+5)} = \frac{1}{5 \cdot 6}[/math]	[math]\frac{1}{30}[/math]	0.0333

Step 3: Calculate the Probability of At Most Four Claims, [math]\operatorname{P}[N \le 4][/math]

This is the sum of probabilities for [math]N=0, 1, 2, 3, 4[/math]:

[[math]]\operatorname{P}[N \le 4] = \operatorname{P}[N=0] + \operatorname{P}[N=1] + \operatorname{P}[N=2] + \operatorname{P}[N=3] + \operatorname{P}[N=4][[/math]]

[[math]]\operatorname{P}[N \le 4] = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30}[[/math]]

To sum these fractions, we find a common denominator, which is 60:

[[math]]\operatorname{P}[N \le 4] = \frac{30}{60} + \frac{10}{60} + \frac{5}{60} + \frac{3}{60} + \frac{2}{60}[[/math]]

[[math]]\operatorname{P}[N \le 4] = \frac{30 + 10 + 5 + 3 + 2}{60} = \frac{50}{60} = \frac{5}{6}[[/math]]

Step 4: Calculate the Probability of At Least One Claim and At Most Four Claims, [math]\operatorname{P}[1 \le N \le 4][/math]

This is the sum of probabilities for [math]N=1, 2, 3, 4[/math]:

[[math]]\operatorname{P}[1 \le N \le 4] = \operatorname{P}[N=1] + \operatorname{P}[N=2] + \operatorname{P}[N=3] + \operatorname{P}[N=4][[/math]]

[[math]]\operatorname{P}[1 \le N \le 4] = \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30}[[/math]]

Using the common denominator of 60:

[[math]]\operatorname{P}[1 \le N \le 4] = \frac{10}{60} + \frac{5}{60} + \frac{3}{60} + \frac{2}{60}[[/math]]

[[math]]\operatorname{P}[1 \le N \le 4] = \frac{10 + 5 + 3 + 2}{60} = \frac{20}{60} = \frac{1}{3}[[/math]]

Step 5: Calculate the Conditional Probability

Now we can compute the conditional probability using the results from Step 3 and Step 4:

[[math]]\operatorname{P}[ N \ge 1 | N \le 4 ] = \frac{\operatorname{P}[1 \le N \le 4]}{\operatorname{P}[N \le 4]}[[/math]]

[[math]]\operatorname{P}[ N \ge 1 | N \le 4 ] = \frac{\frac{1}{3}}{\frac{5}{6}}[[/math]]

To divide fractions, we multiply by the reciprocal of the denominator:

[[math]]\operatorname{P}[ N \ge 1 | N \le 4 ] = \frac{1}{3} \times \frac{6}{5}[[/math]]

[[math]]\operatorname{P}[ N \ge 1 | N \le 4 ] = \frac{6}{15} = \frac{2}{5}[[/math]]

The final answer is [math]\frac{2}{5}[/math].

Key Insights

• Conditional Probability Definition: Always start by correctly setting up the conditional probability formula [math]P[A|B] = P[A \cap B] / P[B][/math] and clearly identifying events [math]A[/math], [math]B[/math], and their intersection.
• PMF Analysis: The given PMF [math]P[N=n] = \frac{1}{(n+1)(n+2)}[/math] can be simplified using partial fraction decomposition: [math]\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}[/math]. This property is very useful for summing probabilities, as it forms a telescoping series, simplifying calculations for sums like [math]\operatorname{P}[N \le k][/math] or [math]\operatorname{P}[N \ge k][/math].
• Summation of Probabilities: When summing fractions, finding the least common denominator is crucial for efficient and accurate calculation.
• Careful Calculation: Ensure accuracy in summing fractions and performing fraction division. Double-check each step to avoid arithmetic errors.

This article was generated by AI and may contain errors. If permitted, please edit the article to improve it.

Solution: B

Observe

[[math]] \begin{align*} \operatorname{P}[ N ≥ 1 | N ≤ 4 ] = \frac{\operatorname{P}[1 ≤ N ≤ 4]}{\operatorname{P}[ N ≤ 4]} &= \frac{1/6 + 1/12 + 1/20 + 1/30}{1/2 + 1/6 +1/12 + 1/20 + 1/30} \\ &= \frac{10 + 5 + 3 + 2}{30 + 10 + 5 + 3 + 2} \\ &= \frac{20}{50} \\ &= \frac{2}{5}. \end{align*} [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.