A probability question on cards

Given a pack of $52$ cards of classes hearts, spades, diamonds and clubs. Each of this class has an Ace, numbers $2,\space 3, \space \cdots, \space 10$, a Knave, a Queen and a King.

$5$ cards are selected successively returning each card after selection before selecting the next one.

What is the probability of having at least an Ace and at least a Knave?

My approach:

We are trying to choose $i$ Ace(s) and $j$ Knave(s) such that $1\le i,j \le 4$ and $i+j \le 5$ and then choose $(5-i-j)$ card(s) from the remaining $44$ cards. And the probability will be $$\displaystyle \sum_{\substack{ 1\le i,j \le 4 \\ i+j\le 5 }}\cfrac{5!}{i!j!(5-i-j)!} \left(\cfrac 1{13}\right)^{i+j} \left(\cfrac {11}{13}\right)^{5-i-j}$$ $$= \sum_{\substack{ 1\le i \le 4 \\ }}\sum_{\substack{ 1\le j \le 5-i \\ }}\cfrac{5!}{i!j!(5-i-j)!} \left(\cfrac 1{13}\right)^{i+j} \left(\cfrac {11}{13}\right)^{5-i-j}$$

I am having problems computing this sum and I'm very sure that I'm missing something.

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Find the probability to have no ace. Find the probability to have no Knave. Find the probability to have no ace and no knave. Then, find the probability you need using these values –  barto Oct 29 '12 at 14:29

When working on the probability of having at least one Ace, it is easier to calculate the probability of having no aces and subtract from 1. In this case, each draw has $\frac {12}{13}$ chance of not being an ace, so the chance you have no aces is $1-\left(\frac{12}{13}\right)^5$
Now that you have to have at least one ace and at least one jack, the ways you can fail are to have no aces and at least one jack, no jacks and at least one ace, or none of either. If you just subtract $1-2\left(\frac{12}{13}\right)^5$ to account for not having aces or not having jacks, you have double subtracted the no aces and no jacks case. You need the inclusion-exclusion principle to deal with this. The chance of having no aces and no jacks is $\left(\frac{11}{13}\right)^5$, so the final answer is $1-2\left(\frac{12}{13}\right)^5+\left(\frac{11}{13}\right)^5$
@F'OlaYinka: Your approach is doable, but more work. You have missed the fact that the cards can come in any order. The case $i=j=1$ in your sum calculates the chance that you draw an ace first, then a jack, then three other cards, but you don't count the other orders, like other, ace, other, jack, other. For this assortment, there are 20 orders, but the count is different for different $i,j$. –  Ross Millikan Oct 29 '12 at 14:39
That's easy to fix, I can just add $\cfrac{5!}{i!j!(5-i-j)!}$ to my sum. Thanks. –  user31280 Oct 29 '12 at 14:48