There's a group of 10 diseases such that each one has a 1/10 chance of affecting an animal during its life. What is the probability of that animal to get infected by at least one of those diseases?
I know that it may be a little basic for this site standards, but i'm really struggling with this. I thought that i could calculate it by adding the probability of getting affected by one, two,..., and ten of the diseases. But then i realized that if did it that way i should have use the inclusion-exclusion principle. The problem is that i didn't even learned to do that and i don't have a very clear idea of what to take account. I also considered find the probability of that animal to not get affected by any of those diseases but i really don't know how to find it. Please, help.
$p_i = \frac{1}{10}$ is probability for animal to have disease $i$. What is probability animal doesn't have disease $i$? The answer is $q = q_i=1-p_i=\frac{9}{10}$. The probability for animal to be healthy is product $q_1q_2...q_{10}$. Animal is not healthy if it has a least one disease. The complement of "animal is healthy" is animal is not healthy hence $P(\text{"Animal has a least one disease"}) = 1 - q^{10}$.(See below about what is complement event).
When you think about why $P(\text{"Animal is healthy"})=q^{10}$. You can think about population of $10^{10}$ animals. How many animals out of $10^{10}$ don't have disease 1? Answer if $10^{10} \cdot \frac{9}{10} = 9 \cdot 10^{10}$. Out of these how many don't have disease 2? Again multiply by $\frac{9}{10}$. In the end you will get $9^{10}$ animals are healthy. Percentage wise it is $(\frac{9}{10})^{10} \cdot 100\%$.
Complement events are mutually exclusive. If one happens other can't happen. If $B$ is complement of $A$ then $P(A) = 1 - P(B)$.