Probabilities using a deck of cards

Question

So I have a deck with $36$ cards containing $4$ aces. After extracting $3$ random cards:

1. What is the probability of being EXACTLY $1$ ace in those $3$ cards?
2. What is the probability of being AT LEAST an ace in those $3$ cards?

I have trouble differentiating the $2$ problems. So for the first problem, the first card drawn has a chance of $\frac{32}{36}$. Assuming it wasn't an ace the second one is $\frac{31}{35}$, and the third one $\frac{30}{34}$. The probability is $P = 1 - \frac{32}{36} * \frac{31}{35} * \frac{30}{34}$. Am I approaching this correctly? And as far as the second one goes, I have no idea. What is the correct approach for this type of problem?

Answer 1

The first question asks you to find the probability that there is exactly one ace, no more, no less. Your sample space consists of one out of four aces and two non aces in this case.
The second part asks you to find the probability that there is at least one ace. In this case, your sample space can contain

1. One out of four aces, and two non aces.
2. Two out of four aces, and one non ace.
3. Three out of four aces.

Easier way to approach the second problem would be to consider the complement of the sample space, i.e., no card is an ace. Can you take it further?

As for your solution, you counted the probability for first three cards being a non ace and subtracted from $1$. That means you are counting the probability of complement of an event $E$ where $E$ is no card is an ace. Hence $E^C$ means at least one card is an ace. So indirectly, you solved the second problem.

Hint: For the first part, pick any one ace in $4\choose 1$ ways and select remaining $2$ cards out of $32$ in $32 \choose 2$ ways.

Answer 2

I believe for part b) you need to consider all the cases where you have at least 2 aces. You could get all 3 aces, what's the probability of that? You could get one Ace, what's the probability of that? You want to know what the probability of ANY of the success cases occurring is. Part a) seems like a part of b). I think you might find this website helpful: Random Services

Answer 3

For $i=1,2,3$ let $A_i$ denote the event that the $i$-th draw results in an ace.

In 1) you are asked to find: $$P(A_1\cap A_2^{\complement}\cap A_3^{\complement})+P(A_1^{\complement}\cap A_2\cap A_3^{\complement})+P(A_1^{\complement}\cap A_2^{\complement}\cap A_3)$$

The first term can be calculated as:$$P(A_1\cap A_2^{\complement}\cap A_3^{\complement})=P(A_1)P(A_2^{\complement}\mid A_1)P(A_3^{\complement}\mid A_1\cap A_2^{\complement})=\frac4{36}\frac{32}{35}\frac{31}{34}$$

Symmetry tells us that for the other terms we get the same outcome so the final answer is: $$3\times\frac4{36}\frac{32}{35}\frac{31}{34}$$

(Also you can go for finding every term separately in order to convince yourself that symmetry works here)

In 2) you are asked to find:$$P(A_1\cup A_2\cup A_3)$$

In cases like this (characterized by "at least" or "at most") very often it is handsome to for the complement:$$P(A_1\cup A_2\cup A_3)=1-P(A_1^{\complement}\cap A_2^{\complement}\cap A_3^{\complement})=1-P(A_1^{\complement})P(A_2^{\complement}\mid A_1^{\complement})P(A_3^{\complement}\mid A_1^{\complement}\cap A_2^{\complement})=$$$$1-\frac{32}{36}\frac{31}{35}\frac{30}{34}$$