A child sits at a computer keyboard and types each of the 26 letters of the alphabet exactly once, in a random order. A child sits at a computer keyboard and types each of the 26 letters of the alphabet
exactly once, in a random order.
How many independent children typists would you need such that the probability
that the word ‘exam’ appears is at least 0.9?
Probability of getting EXAM is 23!/26!. So I decided to do (23!/26!)^n ≥ 0.9. Is that correct ?
 A: The total number of combinations is $26!$
The number of combinations containing the word EXAM is $23!$

The probability of a combination containing the word EXAM is $\frac{23!}{26!}$
The probability of a combination not containing the word EXAM is $1-\frac{23!}{26!}$
The probability of $n$ combinations not containing the word EXAM is $\Big(1-\frac{23!}{26!}\Big)^n$

You need to calculate the smallest integer value of $n$ such that $\Big(1-\frac{23!}{26!}\Big)^n\leq1-0.9$:
$\Big(1-\frac{23!}{26!}\Big)^n\leq1-0.9 \implies \Big(\frac{15599}{15600}\Big)^n\leq0.1 \implies n\geq\frac{\ln(0.1)}{\ln\Big(\frac{15599}{15600}\Big)} \implies n=35920$
A: As you've mentioned the probability of getting 'exam' is $\frac{23!}{26!}$.
Then the probability of the mistake is $\left(1 - \frac{23!}{26!}\right)$. Then the probability of the mistake for all of $n$ children is $\left(1 - \frac{23!}{26!}\right)^n$. It remains to evaluate such $n$ that 
$$\left(1 - \frac{23!}{26!}\right)^n \le 0.1$$
A: You're right. The probability of getting "EXAM" is $p=\frac{23!}{26!}$. Now, suppose, at least $n$ children are needed. If the total number of occurings we get is $X$, then $P(X=r)={n \choose r} \cdot p^r \cdot (1-p)^{(n-r)}$, according to the theory of Barnoulli Trials. Now, $P(X \geq 1) \geq 0.9 \implies P(X=0) \leq 0.1 \implies {n \choose 0} \cdot p^0 \cdot (1-p)^{(n-0)} \leq 0.1 \implies n \log { \left(1-\frac{23!}{26!}\right)} \leq \log 0.1 \implies n \geq 35920$
So, you'll need at least $35920$ children.
