How to solve this statistics problem using binomial coefficient?

Question

I thought of a problem today. There are 100 names in a bucket. Your name is in the bucket. If I randomly draw 5 names, without replacement, what is the probability that your name will be drawn?

I can solve the problem without binomial coefficient (n choose k).

Let P(A) be the probability that your name will be drawn and let P(B) be the probability that your name will not drawn.

$$ P(B) = \frac{99}{100}*\frac{98}{99}*\frac{97}{98}*\frac{96}{97}*\frac{95}{96} = 0.95$$

$$ P(A) = 1 - P(B) = 0.05 $$

I then tried to figure out the answer using binomial coefficient.

$ \binom{n}{k} = \frac{100!}{5!(95!)} = 75287520 $

Since the answer I got was .05, that must mean there are 75287520 * .05 = 3764376 groups of 5 names that include your name. The problem is, I do not know how to get that number besides writing down all the combinations. How to get the number 3764376 is what I'd like to know.

Answer 1

In order to find a group that includes your name, choose the 4 other members of that group among the 99 people who are not you.

$$ \binom{99}{4} = 3,764,376 $$

Answer 2

another way to find how many groups of $5$ elements contain $a$ is to do:

$\binom{100}{5}-\binom{99}{5}$.

Answer 3

Let $A_{1},A_{2},A_{3},A_{4},A_{5}$ be the events that you name is on the first, second, third, fourth or fifth slips.

$\begin{align} P(A_{1}\cup{A_{2}}\ldots{A_{5}})&=P(A_{1})+\ldots+P(A_{5})\text{ ; since these are disjoint}\\ &=\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+\frac{1}{100}\\ &=\frac{5}{100} \end{align}$

We can just as well think of first choosing five random slips out of $100$ blank slips and then randomly write names on the slips, which gives a $5/100$ chance that your name is on one of those five chosen slips.