I've got a 2 advanced probability questions that I'm having trouble with so just asking to confirm an answer/method

  1. A computer company provides an insurance policy for one of its systems. If the system fails during the first year the policy pays £3000. The benefit decreases by £1000 each year until it reaches zero. If the system has not failed at the beginning of a year, the probability that it fails during the year is 0.1. How much should the company charge for the insurance policy so that, on average, its net gain per policy be £100?

  2. A man aiming at a target receives 10 points if his shot is within 1cm of the center of the target, five points if it is between 1cm and 3cm, and three points if it is between 3cm and 5cm. What is the expected number of points scored if the man's shot is uniformly distributed in a circle of radius 8cm centers on the target?

  • 2
    $\begingroup$ Where are you having "trouble"? What have you tried? Both questions require you to know a definition of expected value. For the first question, you might find drawing a probability tree to be helpful. The second is really a geometry question in disguise; the asker likely is expecting incorrect answers of the form "there must be a 1 in 8 chance of hitting the center" whereas the chance is really only 1 in 64. $\endgroup$
    – whuber
    Sep 2, 2013 at 17:53
  • $\begingroup$ I've got no clue as to how to solve them. "definition of expected value"? what do you mean? Even drawing a tree has left me nothing for the first one. I got 18 for no.2 but that seems wrong $\endgroup$
    – Kei Cheung
    Sep 2, 2013 at 18:00
  • $\begingroup$ OK, that's the cause of your difficulties: you need to learn what an expected value (or expectation is) before you tackle these problems. You could start at stats.stackexchange.com/questions/30365 and if you need more information, search our site. $\endgroup$
    – whuber
    Sep 2, 2013 at 18:05
  • $\begingroup$ ok, I understand the method but a little trouble applying it. Should I get 10x1/8 + 5x2/8 + 3x2/8 + 0x3/8? You said 1/64 but could you expand on that? Also, any tips for question 1? $\endgroup$
    – Kei Cheung
    Sep 2, 2013 at 18:20
  • 1
    $\begingroup$ You wrote down precisely the incorrect answer I described, so it shows you're applying the expectation calculation correctly but you're not computing the probabilities correctly. 1/64 comes from formulas for the area of a circle. My tip for question 1 is to draw a probability tree. In your tree the values on the leaves will be changing from one year to the next. An example of a tree where numbers are changing (both the values and the probabilities change over time) appears at stats.stackexchange.com/a/68638. $\endgroup$
    – whuber
    Sep 2, 2013 at 18:23

1 Answer 1


Studying for an FIA actuarial exam? 8-)

As @whuber said, knowing the expected value is key. What is an expected value? Simply (and probably non-rigorously), it can be thought of as the long-term average. If the company wrote this policy for one million blocks of three years, then the total amount they paid over those blocks divided by a million would be very close to the "expected value".

Insurance works on the principle that even though any one person's risk is too volatile to measure with any certainty, a pool of risks is much more tractable and measurable. So while said insurer would not want to write just one policy, if they write 100,000 such policies, they will have a much better chance of having the actual average loss they experience be close to the expected value. Because there is volatility, one must add a contingency/volatility/profit load which in this case is the $£100$.

Now, to the case in question - the expected value in this case would be the sum of the expected payments over the three years. There is a $10\%$ chance the system fails in year $1$. If it does, the payout is $£3000$ so the expected payment in year 1 is $£300$. For the system to fail in year $2$ it must have survived year $1$, so the expected loss is $0.9\cdot0.1\cdot£2000$ or $£180$. Similarly for year $3$ the system has to survive for two years and then fail, so the expected loss is $0.9^2\cdot0.1\cdot£1000$ or $£81$. So the total expected loss for the insurance policy is $£561$. What this means is if the company writes one million of these policies, they can be rather sure that their sum total loss will be very close to $£561,000,000$. Since the company wants to make an average profit of $£100$ one each policy, they should charge $£661$.

As for the second question, it is a matter of understanding the proportion that each ring and center bullseye is of the the total area of the circle, as having a uniform distribution means that the shooter is rather wild and can land the shot on any point within that circle equally. The area of the total circle is $64\pi$. The area of the bullseye is $1^2\pi$, the area of the five-point ring is the area of its circle minus the area of the bullseye, which is is $3^2\pi - 1^2\pi = 8\pi$. Similarly, the area of the three-point ring is $5^2\pi - 3^2\pi = 16\pi$ and there are no points in the remaining $8^2\pi - 5^2\pi = 39\pi$. The $\pi$ all cancel in the ratios, and we have the following probability: $$ \frac{0\cdot 39 + 3 \cdot 16 + 5 \cdot 8 + 10 \cdot 1}{64} = \frac{98}{64} \approx 1.53 $$

So this persons expected "points-per-shot" is $\frac{98}{64}$ or about $1.53$.

  • $\begingroup$ No, it's some practice questions I found on the net for Msc Statistics. 1. Oh right, I got my tree incorrect. but calculations correct. made the fail rate + 10% each year instead (ugh) 2. ok, I failed on basic circle theory ... I did 2^2π for the second calculation. <br/>Thanks a lot for clearing my mistakes! $\endgroup$
    – Kei Cheung
    Sep 2, 2013 at 19:56

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