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A password is either 4, 5, 6 characters long, and can only contain 2, 3, 5, 7 and must contain at least 1 two. Find the probability that your friend guesses the password within 10 attempts.

There are a total of: $5364 - 1044 = 4320$ possible passcodes that work.

But how do I fit in the 10 attempts?

First try + second try + ... + tenth try?

$=1/4320 + 1/4319 + .. + ?$

Im not sure, would like help!

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    A style comment on writing out numbers: it's confusing to say that it must contain "at least 1 two". Much clearer and more standard would be, "at least one 2". Compare to the situation where the password entries are not integers but symbols: $\square, \bigcirc, \triangle$. Which is better, "at least 1 square", or "at least one $\square$"? – 2017-07-28

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The total number of passwords should be the number of 4-6 digit passwords using the characters 2, 3, 5, and 7 minus the number of passwords using just 3, 5, and 7. For this you compute

$$(4^4 + 4^5 + 4^6) - (3^4 + 3^5 + 3^6) = 5376 - 1053 = 4323.$$

The probability that your friend guesses on the first attempt is $\frac{1}{4323}$, as you said.

The probability that your friend guesses your password on the second attempt is the probability that she guesses wrong on the first attempt and right on the second attempt. That's $\frac{4322}{4323}\cdot\frac{1}{4322} = \frac{1}{4323}$.

Similarly, let's say you look at the third attempt: you want the probability that your friend guesses wrong on the first and second tries, and then guesses your password on the third try. That happens with probability $\frac{4322}{4323}\cdot\frac{4321}{4322}\cdot\frac{1}{4321} = \frac{1}{4323}$.

The same pattern continues: assuming your friend guesses a different password every time, the probability that she guesses on the $n^\text{th}$ attempt is $\frac{1}{4323}$.

Now, as you suggested, you can add the probabilities corresponding to the first try + second try + ... + tenth try, since you want to know the odds that she guesses right on the first try or the second try or the third try, and so on.

This gives you a probability of

$$10 \cdot \frac{1}{4323} = \frac{10}{4323}.$$

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    Rather than counting each attempt individually with fractions, one can simply skip to the end and say that there are $10$ distinct attempts out of a possible $4323$ combinations, for a final probability of $\frac{10}{4323}$. – 2017-07-28
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assuming you counted correctly:

The probability that he misses the first attempt is $\dfrac{4319}{4320}$, the probability he misses the second is $ \dfrac{4318}{4319}$ and so on.

The probability he doesn't guess is the product of all these fractions, almost everything cancels and you get $\dfrac{4310}{4320}$. Therefore the probability he guesses is $\dfrac{10}{4320}=\dfrac{1}{432}$