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.
- What is the sample space?
- What is the probability that no Ace appears on any of the draws?
- What is the probability that at least one King appears in 10 draws?
- 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)]
Best Answer
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 $$\frac{\begin{bmatrix}\frac{4}{52} \times \begin{pmatrix}\frac{48}{52}\end{pmatrix}^9\end{bmatrix}}{52^{10}}$$