Krdytkyu

I've come across the following problem in a probability text that I haven't seen before and am generally unsure about my approach to solving:

"The annual numbers of thefts a homeowners insurance policyholder experiences are analyzed over three years.

Define the following events:

i) A = the event that the policyholder experiences no thefts in the three
years.

ii) B = the event that the policyholder experiences at least one theft in the
second year.

iii) C = the event that the policyholder experiences exactly one theft in the
first year.

iv) D = the event that the policyholder experiences no thefts in the third
year.

v) E = the event that the policyholder experiences no thefts in the second
year, and at least one theft in the third year.

Determine which three events satisfy the condition that the probability of
their union equals the sum of their probabilities."

At first, because it asks which events satisfy condition that the probability that their union equals the sum of their probabilities, that I could just try to reason it out which ones are mutually exclusive. That didn't get me far.

Then I decided to draw a grid with the columns being years 1-3 and the rows being $0, =1$, and $geq 1$ thefts.

I ended up with this grid where the columns go from year 1 to 3 from left to right and an x or xx denotes no information for that cell:

=0 | A | AE | AD|

=1 | C | xx | xx |

$geq 1$ | x | B | E |

The answer is the union of A,B,E

By my grid, the union is 5/9 cells but the sum of the probabilities is P(A) = $dfrac{3}{9}$, P(B) = $dfrac{1}{9}$, P(E) = $dfrac{2}{9}$ and the sum of these is $dfrac{6}{9}$.

Can anyone point me in the direction of how to solve a question like this in general, or hi-light a flaw in my reasoning?

Thanks

The probability of a union of $3$ events $A, B, C$ is
$$P(Acup B cup C)$$

Usually you will need to compute

$$P(Acup B cup C)= P(A) + P(B) + P(C) - P(Acap B)- P(Acap C)- P(Bcap C) + P(Acap B cap C)$$

However if $A, B, C$ are all mutually exclusive with each other you can simply solve

$$P(Acup B cup C)= P(A) + P(B) + P(C) - 0- 0- 0 + 0$$

$$P(Acup B cup C)= P(A) + P(B) + P(C)$$

Hence, as the SOA solution states "The probability a union of three events equals the sum of their probabilities if and only if they
are mutually exclusive, that is, no two of them can both occur. "

So you must find the events that are mutually exclusive. To find that, one simply counts them:

$$A=P(T=0)P(T=0)P(T=0)$$
$$B=P(T=t)P(Tge 1)P(T=t)$$
$$C=P(T=1)P(T=t)P(T=t)$$
$$D=P(T=t)P(T=t)P(T=0)$$
$$E=P(T=0)P(T=0)P(Tge 1)$$

As you can see, the events that are mutually exclusive are $A,B,E$. We do not care at all about what the actual probabilities are, only which ones are mutually exclusive.