Solved – How to calculate the probability of this claim

probability

An insurance company insures a good driver and a bad driver on the same policy. The table below gives the probability of a given number of claims occurring for each of these drivers in the next ten years.

The number of claims occurring for the two drivers are independent.
Calculate the mode of the distribution of the total number of claims occurring on this policy in the next ten years.

The solution shows this:
The probability needs to be calculated for each total number of claims.
0: 0.5(0.2) = 0.10, 1: 0.5(0.3) + 0.3(0.2) = 0.21
, 2: 0.5(0.4) + 0.3(0.3) + 0.2(0.2) = 0.33
, 3: 0.5(0.1) + 0.3(0.4) + 0.2(0.3) + 0.0(0.2) = 0.23
At this point there is only 0.13 probability remaining, so the mode must be at 2.

I don't understand how they got 0.5(0.3) + 0.3(0.2) for total claim for 1, can someone explain how they got these numbers and did this calculation?

Thanks in advance!

Best Answer

You might find it useful to write out a table of the bivariate distribution:

                Good driver number of claims
                 0       1       2       3   
             0  0.1    0.06 ...
 Bad driver  1
 number of   2
 claims      3

and then add the probabilities for the individual cells that have the same number of total claims. (e.g. so P(T=1) will be P(G=0,B=1)+P(G=1,B=0); you'll be adding along "diagonals" running this way $\it{/}$ )

This is not the fastest possible way to do it, but when you're first encountering these kinds of problems it's perhaps the way which makes it clearest what's going on.

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Solved – Probability for selecting centroids – K-means++

The paper (see section 2.2) suggests that you use the squared distance when computing probabilities. In fact, you can try distance$^{\ell}$ using any exponent $\ell$ greater than 1. It stands to reason that as $\ell$ increases, the likelihood of initial centroids being close together will go to zero.

Solved – Has this expectation been computed correctly

Sometimes it's clearest and simplest just to enumerate the (equiprobable) possibilities. After all, there are only $\binom{10}{2}=45$ subsets of two interviewees. I'll be thorough rather than efficient in order to illustrate some of the techniques and to serve as a check on the arithmetic.

How can the difference be zero? When either both interviewees are Independents ($\binom{5}{2}$ possibilities) or one is Republican and the other Democrat ($3\times 2$ possibilities).
How can the difference be one? One interviewee is Independent and the other is not: $5\times 5$ possibilities.
How can the difference be two? Both interviewees must be in the same party ($\binom{3}{2} + \binom{2}{2}$ possibilities).

As a check, the total is

$$\left(\binom{5}{2} + 3\times 2\right) + 5\times 5 + \left(\binom{3}{2}+\binom{2}{2}\right) = (10+6)+25+(3+1)=45=\binom{10}{2}.$$

The expectation is the probability-weighted average. We needn't include the value of zero because it contributes nothing. That leaves

$$\frac{1}{\binom{10}{2}}\left((5\times 5) \times 1 + \left(\binom{3}{2}+\binom{2}{2}\right)\times 2\right) = \frac{33}{45} = \frac{11}{15}.$$

In retrospect, I suspect you might have shared only the beginning of a solution. The full solution might go like this:

There's a 1/2 chance the first interviewee is Independent. Conditional on that, there's a 4/9 chance one of the remaining nine voters is one of the remaining 4 Independents and a 5/9 chance they belong to a party. The conditional expectation of the absolute difference is $$(4/9)\times 0 + (5/9)\times 1 = 5/9.$$
There's a 3/10 chance the first interviewee is Republican. Conditional on that, there's a 5/9 chance the second is Independent (absolute difference is 1), a 2/9 chance they are Republican (absolute difference is 2), and a 2/9 chance they are Democrat (absolute difference is 0). The conditional expectation is $$(5/9) \times 1 + (2/9)\times 2 + (2/9)\times 0 = 1.$$
A comparable analysis of the last case, a 2/10 chance of a Democrat, gives a conditional expectation of $$(5/9)\times 1 + (3/9)\times 0 + (1/9)\times 2 = 7/9.$$

The expectation itself is weighted by the chances of each scenario, giving

$$(1/2) \times (5/9) + (3/10)\times (1) + (2/10) \times (7/9) = 11/15.$$

We can ask the computer to do the enumeration, too. Here's one way (in R). It is coded to be a little different than either of the previous solutions--otherwise it wouldn't be a very good check. I hope it's clear enough. Its output is

Expectation is 11 / 15

voters <- c(rep("Republican", 3), 
            rep("Democrat", 2), 
            rep("Independent", 5))          # The voters
polls <- combn(voters, 2)                   # All possible subsets of two voters
republicans <- colSums(polls=="Republican")
democrats <- colSums(polls=="Democrat")
X <- abs(republicans - democrats)
expectation <- mean(X)
cat("Expectation is", 15*expectation, "/ 15")

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