$5$-Letter Word with $d$ occurring at least once

Question

Find the probability that a $5$-letter word (in English) contains the letter $d$ at least once.

I know that we can take the complement so the answer is $$1-\left(\frac{25}{26}\right)^{5}$$

However, I am interested in solving it without taking the compliment, in other words by counting.

So far my reasoning is this: $\quad$ There are $5$ positions $d$ can be in, so the answer should be $$\frac{5\cdot 26^4}{26^5}$$

But it isn't. I am double counting. How else can I proceed? Or how can I fix the double counting?

Any help is greatly appreciated :)

Answer 1

As an alternative to inclusion-exclusion, consider the probability of a d in each position if the previous positions are d-free, to get:

$$\frac{1}{26}+\frac{25}{26}\frac{1}{26} +\left(\frac{25}{26}\right)^2\frac{1}{26}+\left(\frac{25}{26}\right)^3\frac{1}{26}+\left(\frac{25}{26}\right)^4\frac{1}{26}$$

Answer 2

I have a way of doing it but it makes use of the fact binomial expansion $(a+b)^n = \sum_{k=0}^n {n\choose k}a^kb^{n-k} = b^n + \sum_{k=1}^n {n\choose k}a^kb^{n-k}$. Note that the summation starts from $k=0$ which is key here.

Note that $d$ is in the 5-digit word follows a binomial distribution with a random variable $D\sim B(5,p)$ where $p = \frac{1}{26}$. Then probability that letter $d$ comes atleast once is simply:

$P(D \geq 1) = \sum_{k=1}^{5}{5\choose k}p^k(1-p)^{5-k} = (p+(1-p))^5 - \underbrace{(1-p)^5}_{k=0 \ \text{case}} = 1 - \left(\frac{25}{26}\right)^5$.

Hope this answers your question.