Draw a token from a bag of 9 blue tokens and 1 red token

Let X be the number of draws it takes to get the first red token.
P(X = 1) = 1/10
For X to be equal to 2, you must first draw a blue and then a red, so 9/10 * 2/10
If you write out a few more terms like that, a pattern emerges. In general: $$P(X = x) = \frac{9!}{(10-x)!}\frac{x}{10^{x}}$$$$ 1 \le x \le 10$$ If we plug in the numbers 1 through 10 into that formula, we can get the following p.m.f:

x	P(X = x)
1	.1
2	.18
3	.216
4	.2016
5	.1512
6	.09072
7	.042336
8	.0145152
9	.00326592
10	.00036288

If we take the cumulative sum of the second column, we’ll get the c.d.f:

x	P(X <= x)
1	.1
2	.28
3	.496
4	.6976
5	.8488
6	.93952
7	.981856
8	.9963712
9	.9996371
10	1

To find the expected value, we can use the p.m.f we found earlier and the formula: $$E(X) = \sum_{x = 1}^{10}{x*P(X = x)}$$ Summing all 10 terms together, we get E(X) = 3.660216.