i'm confused by the following problem could someone walk me through it, so i can understand

(a) 10 cards are drawn at random one at a time with replacement from an ordinary deck of cards.

  1. What is the sample space?
  2. What is the probability that no Ace appears on any of the draws?
  3. What is the probability that at least one King appears in 10 draws?
  4. What is the probability that at least 2 Queens appear in the 10 draws?

(b) What are the corresponding probabilities in (a) if the drawing is done without replacement?

Solution: (a) The sample space is a sequence of ten cards, and its size is 5210

No Ace has probability (48/52)10

By looking at the complement, at least one King has probability 1-(48/52)10

There's a (10 choose 1)(4/52)(48/52)9 probability of getting exactly one queen, so by compliment at least 2 Queens has probability 1 - (48/52)10 - (10 choose 1)(4/52)(48/52)9

(b) Since we are drawing without replacement, it is easier to think of drawing (un-ordered) subsets, and the sample space is all hands of ten cards;

Thus, no Ace has probability (48 choose 10)/(52 choose 10)

At least one King has probability 1 - [(48 choose 10)/(52 choose 10)]

At least 2 Queens has probability 1 - [((48 choose 10)+4*(48 choose 9))/(52 choose 10)]

  • 1
    $\begingroup$ To be picky, the sample space is the set of all sequences of $10$ cards. The expressions you obtained are all correct. If they are what you wrote, there is no point in writing a formal answer that says you are in fact not confused. If you are transcribing someone else's solution, it would help to know what you have trouble with. $\endgroup$ Feb 25 '13 at 7:37

I will attempt to answer with the assumption that you do not have the answers, so you can see the thinking process.

(a) 1

The sample space should be defined as a hand of $10$ cards. I like to use the term "a vector of $10$ cards". As you mentioned, the sample space has $52^{10}$ possibilities. The final answer is $$52^{10}$$

(a) 2

The sample space is defined as a hand of $10$ cards with no Aces, $A\clubsuit \space A\spadesuit \space A\heartsuit \space A\diamondsuit $. There are $48$ different draws per card of a hand. Hence, the sample space has $48^{10}$ possibilities. The final answer is $$\frac{48^{10}}{52^{10}} = \begin{pmatrix}\frac{48}{52}\end{pmatrix}^{10}$$

(a) 3

For this question, instead of considering the sample space with $1K, 2K, 3K, 4K$...only consider the space where there are $0K$. What you can do is use the answer from (a) 2...simply swap out $A$ to $K$. The final answer is $$1-\frac{48^{10}}{52^{10}} = 1-\begin{pmatrix}\frac{48}{52}\end{pmatrix}^{10}$$

(a) 4

To obtain the number of possibilities with $1Q$, consider this:

There are 4 ways to draw a $Q$ and 48 ways to draw any other card. The number of possibilities to draw will be $$\begin{bmatrix}\frac{4}{52} \times \begin{pmatrix}\frac{48}{52}\end{pmatrix}^9\end{bmatrix}$$

However, there are $10$ ways to arrange these draws. Divide this by the whole sample space (with no restrictions), you get $$\frac{\begin{bmatrix}\frac{4}{52} \times \begin{pmatrix}\frac{48}{52}\end{pmatrix}^9\end{bmatrix}}{52^{10}}$$

  • $\begingroup$ For $(4)$, Please see this $\endgroup$
    – Nick
    Oct 31 '14 at 2:15

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