An insurance company divides its customers into different risk groups. Suppose that 60% of the customers are in group A (low risk), 30% are in group B (medium risk) and 10% are in group C (high risk).

The probabilities of loss occurence for customers in the different groups are as follows:

  • Group A: 0.1
  • Group B: 0.25
  • Group C: 0.74

Calculate the probability that a loss occurs to a randomly chosen customer of the insurance company.

So first, lets start by noting what we already have (L - occurrence of loss):

P(A) = 0.6 given

P(B) = 0.3 given

P(C) = 0.1 given

P(L|A) = 0.1 given => P(L$^c$|A) = 1- 0.1 = 0.9

P(L|B) = 0.25 given => P(L$^c$|B) = 1 - 0.25 = 0.75

P(L|C) = 0.74 given => P(L$^c$|C) = 1 - 0.74 = 0.26

I'm not sure exactly whether the rule of total probability should be applied here to find the P(L) and if the rule of total probability is needed here, I'm rather confused as to how to calculate it with 3 different events? (A,B & C).

Hence, what I tried is:

P(L) = 1 - P(L$^c$) - event that no loss occurs for anyone

P(L$^c$) = P(L$^c$|A)*P(A)*P(L$^c$|B)*P(B)*P(L$^c$|C)*P(C) = 0.9*0.6*0.75*0.3*0.26*0.1 = 0.003159 the chance that no loss occurs to anyone

From here it follows that:

P(L) = 1 - 0.003159 = 0.996841 chance that loss occurs to a randomly chcosen customer of the company

So am I completely wrong or did I correctly solve this? I don't have a solution with which to compare it, hence, I'm asking you, so thank you in advance for your time! Any insights are much appreciated!


You can't multiply all the events like that to calculate $P(L')$. We start with first calculating the probability that he lies in the A loss group or the B loss group or the C loss group.

P(customer is in A loss group) = P(customer belongs to A) $\cdot$ P(an A customer is in loss)

$$P_{A} = 0.6\cdot 0.1 = 0.06$$

P(customer is in B loss group) = P(customer belongs to B) $\cdot$ P(a B customer is in loss)

$$P_{B} = 0.25\cdot 0.3 = 0.075$$

P(customer is in C loss group) = P(customer belongs to C) $\cdot$ P(a C customer is in loss)

$$P_{C} = 0.74\cdot 0.1 = 0.074$$

Since the customer can be in any loss group, we sum it all up to get

$$P = P_{A}+P_{B}+P_{C} = 0.06+0.075+0.074 = 0.209$$

  • $\begingroup$ Thank you for taking the time to help me out, it is much appreciated! :) $\endgroup$ – VRT Dec 11 '18 at 18:17
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    $\begingroup$ Then please feel free to accept the answer :) $\endgroup$ – Sauhard Sharma Dec 11 '18 at 18:22
  • $\begingroup$ Apologies, I'm new and wasn't sure about this and how to do it. But now I just did :) Thanks again! $\endgroup$ – VRT Dec 11 '18 at 18:24

You should suspect that $0.996841$ answer independently of how you arrived at it, because it makes no sense. It is almost $1$, but the probability that a random customer sustains a loss must surely be less than what it would be if they were all high risk.

The way to find the probability of a loss for a randomly chosen customer is as a weighted average over the kinds of customers, so $$ 0.6 \times 0.1 + 0.3 \times 0.25 + 0.1 \times 0.74 . $$

You can see that visually by drawing the probability tree with three branches for customer choice followed by two for loss/no loss.

If you were told that a customer had incurred a loss and asked the probabilities for which kind of customer it was that would be a different question.

Last observation: I am surprised to see the three loss probabilities add up exactly to $1$. That is a strange coincidence, or reflects some part of the question you haven't transcribed properly.

  • $\begingroup$ Thank you for the explanation and for taking the time to correct my mistake! It does make a lot of sense. I am actually self-studying for a statistics exam, hence, I don't have a teacher to guide me through it but just a suggested textbook that sometimes is failling to explain things (or could be just me, haha). On another note, no, I haven't written the question down incorrectly, it is word for word. But the 3 loss probabilities don't add up to exactly 1. They add up to 1.09 :) $\endgroup$ – VRT Dec 11 '18 at 18:16
  • $\begingroup$ @VRT You're welcome. Right. I misread $0.1$ as $0.01$. $\endgroup$ – Ethan Bolker Dec 11 '18 at 18:22

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