Sure, let's tackle the problem step by step, making calculations directly.
### Part (a): Probability of Drawing an Ace and a King
First, we need to consider the cases separately:
1. Drawing an Ace first and then a King:
- Probability of drawing an Ace first:
There are 4 Aces in a deck of 52 cards.
[tex]\[
\text{Probability of drawing an Ace first} = \frac{4}{52}
\][/tex]
- After drawing an Ace, there are 51 cards left, with 4 of them being Kings.
[tex]\[
\text{Probability of drawing a King second} = \frac{4}{51}
\][/tex]
- Therefore, the combined probability of drawing an Ace first and a King second is:
[tex]\[
\left(\frac{4}{52}\right) \times \left(\frac{4}{51}\right) = \frac{4}{52} \times \frac{4}{51} \approx 0.006033182503770739
\][/tex]
2. Drawing a King first and then an Ace:
- Probability of drawing a King first:
There are 4 Kings in a deck of 52 cards.
[tex]\[
\text{Probability of drawing a King first} = \frac{4}{52}
\][/tex]
- After drawing a King, there are 51 cards left, with 4 of them being Aces.
[tex]\[
\text{Probability of drawing an Ace second} = \frac{4}{51}
\][/tex]
- Therefore, the combined probability of drawing a King first and an Ace second is:
[tex]\[
\left(\frac{4}{52}\right) \times \left(\frac{4}{51}\right) = \frac{4}{52} \times \frac{4}{51} \approx 0.006033182503770739
\][/tex]
To find the total probability of drawing one Ace and one King (in any order):
[tex]\[
\text{Total Probability} = \text{Probability of Ace first, King second} + \text{Probability of King first, Ace second}
= 0.006033182503770739 + 0.006033182503770739 = 0.012066365007541479
\][/tex]
### Part (b): Probability of Drawing Two Aces
- Probability of drawing an Ace first:
There are 4 Aces in a deck of 52 cards.
[tex]\[
\text{Probability of drawing an Ace first} = \frac{4}{52}
\][/tex]
- After drawing the first Ace, there are 3 Aces left in a deck of 51 cards.
[tex]\[
\text{Probability of drawing a second Ace} = \frac{3}{51}
\][/tex]
- Therefore, the combined probability of drawing two Aces is:
[tex]\[
\left(\frac{4}{52}\right) \times \left(\frac{3}{51}\right) = \frac{4}{52} \times \frac{3}{51} \approx 0.004524886877828055
\][/tex]
### Summary:
(a) Probability of getting an Ace and a King: [tex]\(\approx 0.012066365007541479\)[/tex]
(b) Probability of getting two Aces: [tex]\(\approx 0.004524886877828055\)[/tex]
Thus, the probabilities have been calculated successfully.
