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There was actually another question like this but the final answer a person mentioned was incorrect and I was confused as to how he got it. Can any answers explain how they got there? I'd like to understand what's going on so these types of problems are easier for me in the future.

Q: 10 cards are drawn with replacement at random from a 52 card deck. What is the probability that at least 2 queens are drawn?

Here's the link that asked the same question: probability of selecting cards

Thanks for any help in advance.

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Probability Space: Drawing 10 cards with replacement from a standard deck.

Favoured Event: Drawing at least 2 queens.

Complementary Event: Drawing less than 2 queens

Let $Q$ be the count of queens drawn.

Assuming the deck is well shuffled after each replacement, then the probability of getting a queen on any draw is independent of every other draw, and equal to $1/13$.   This gives $Q$ a binomial distribution.

$$\begin{align} Q \sim \mathcal{Bin}(10, 1/13) \\[2ex] \mathsf P(Q=k) & = {10\choose k}\frac{12^{10-k}}{13^{10}} \\[2ex] \mathsf P(Q \geq 2) & = 1 - \mathsf P(Q<2) \\[1ex] & = 1 - {10\choose 0}\frac{12^{10}}{13^{10}} - {10\choose 1}\frac{12^{9}}{13^{10}} \\[1ex] & = \frac{13^{10}-12^{10}-10\cdot 12^{9}}{13^{10}} \end{align}$$


From your comment.

What is a binomial distribution? Is it similar to the binomial theorem? EDIT: This looks like the binomial theorem, actually, Kind of.

Indeed it does. If a trial has probability of success $p$ and failure $q=1-p$, then the probability of $k$ successes in $n$ trials is measured by the $p^k$ term of the binomial expansion of:

$$(q+p)^n = q^n + n p q^{n-1} + \frac{n(n-1)}{2}p^2 q^{n-2}p^2 + \cdots + {n\choose k}p^k q^{n-k} + \cdots + p^n$$

That is: $$\begin{align} \mathsf P(X=k) &= {n\choose k}p^k q^{n-k} \\ & = {n\choose k}p^k (1-p)^{n-k} \end{align}$$

This is why we call such a probability distribution a binomial distribution.

Alternatively: the probability of $k$ successes and $n-k$ failures in a certain order is $p^k q^{n-k}$, and there are ${n\choose k}$ ways to order such events. So the probability of $k$ successes and $n-k$ failures in any order is:${n\choose k}q^{n-k}p^k$

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  • $\begingroup$ What is a binomial distribution? Is it similar to the binomial theorem? EDIT: This looks like the binomial theorem, actually. Kind of. $\endgroup$ – John Oct 31 '14 at 1:36
  • $\begingroup$ I really don't understand what's going on. $\endgroup$ – John Oct 31 '14 at 1:42
  • $\begingroup$ See the addendum to my answer $\endgroup$ – Graham Kemp Oct 31 '14 at 2:23
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So, we need to calculate the probability of drawing out at least two queens in 10 draws with replacement.

A note on the line of thought - If we start counting all the cases, viz two queens, three queens, four queens, etc, we will have to handle a lot of cases. So, it is a good idea to calculate the total number of ways in which we can draw the cards and remove the uninteresting cases, namely, zero queens and one queen.

Total no. of ways in which we can draw 10 cards with replacement = $52^{10}$ (Since each card can be selected in 52 ways)

Total no. of ways in which we can draw 10 cards with replacement without any queen = $48^{10}$ (We leave out the 4 queens, so we have 48 ways of selecting each card)

Total no. of ways in which we can select 10 cards with replacement with exactly one being queen - (Selecting a location for the queen card = 10 places) $\times$ (Ways of drawing a queen card = 4 colors) $\times$ (Selecting non-queen cards for the remaining 9 places = $48^{9}$) = $10 \times 4 \times 48^9$

So, no. of ways of drawing out at least two queens in 10 draws with replacement = $52^{10} - 10 \times 4 \times 48^9 - 48^{10}$

And the probability of drawing out at least two queens in 10 draws with replacement = $\dfrac{52^{10} - 10 \times 4 \times 48^9 - 48^{10}}{52^{10}}$ = $\dfrac{13^{10} - 10 \times 12^9 - 12^{10}}{13^{10}}$

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