# Probability of an event, discrete uniform distribution

There are 100 slips of paper in a hat, each of which has one of the numbers 1,2,...100 written on it, with no number appearing more than once. Five of the slips are drawn, one at a time. By sampling with replacement:

What is the probability that the number 100 is drawn at least once?

My line of reasons goes: the draws are independent and each draw has the same probability of having 100 on it, that is $1 \over 100$. Because I want to know the probability of at least one of them I can think of this as an operation of unions and since they are independent I can add the results, so probably in non-standard notation I'm thinking of this as:

$$P(\cup_{i=1}^{5}A_{i}=100)$$ $$= \sum_{i=1}^{5} P(A_{i} = 100)$$ $$={5\over 100} = 0.05$$

The solution to this problem, as stated in the book is:

$$1- (99/100)^{5} \approx 0.049$$

These numbers are oddly close but nonetheless not the same. So my question is, is my line of reasoning OK? Where am I going wrong?

## 2 Answers

Your reasoning is wrong because you are using $P(A \cup B)=P(A)+P(B)$, and that is true only if $A$ and $B$ are disjoint. In general $P(A \cup B)=P(A)+P(B)-P(A \cap B)$. Forgetting the intersection, as you did, would be like supposing that the number 100 can't be drawn more than once, since $A_i \cap A_j$ is the event of the number 100 being drawn in the ith and the jth round. However, since probability of drawing 100 more than once is small, your result is close (but not equal) to the correct result.

Using union and intersection to solve this problem would be rather complicated, since you need to take in account a lot of intersections. It's easier to compute the probability of not drawing the number 100 in any round and subtract that probability from 1.

• i see, i was pooling independence and disjoint together – atomsmasher Mar 17 '18 at 23:31

The reason for the answer in the book is that the probability of 100 at least once is the opposite of the probability that 100 never occurs in 5 tries which because the trials are independent and the numbers are replaced is (99/100)$^5$ making the answer 1 minus that. Your answer assumes the events are disjoint which means for example that getting 100 on the first or second trial is the same as the probability of getting it on the first plus the probability of getting it on the second. But that double counts the event the it occurs on both.