7. What is the probability of drawing an ace OR a jack from a standard 52-card deck?
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![### Probability Question
**Question 7**: What is the probability of drawing an ace OR a jack from a standard 52-card deck?
#### Explanation
In a standard deck of 52 playing cards, there are 4 aces and 4 jacks. To find the probability of drawing an ace or a jack, you can use the formula for the probability of either of two mutually exclusive events happening:
\[ P(A \text{ or } J) = P(A) + P(J) \]
Here, \( P(A) \) represents the probability of drawing an ace, and \( P(J) \) represents the probability of drawing a jack.
- The probability of drawing an ace (\( P(A) \)):
\[ P(A) = \frac{\text{Number of Aces}}{\text{Total Number of Cards}} = \frac{4}{52} \]
- The probability of drawing a jack (\( P(J) \)):
\[ P(J) = \frac{\text{Number of Jacks}}{\text{Total Number of Cards}} = \frac{4}{52} \]
Since drawing an ace and drawing a jack are mutually exclusive events (they cannot happen at the same time), you can simply add the probabilities:
\[ P(A \text{ or } J) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} \]
Simplifying the fraction:
\[ P(A \text{ or } J) = \frac{8}{52} = \frac{2}{13} \]
Therefore, the probability of drawing an ace or a jack from a standard deck of 52 cards is \(\frac{2}{13}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e662ae8-5544-47c6-b13f-443ca92f6b6f%2F827b3177-f5d0-4e2e-970f-01e90f04a21a%2Fnkr2v5l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability Question
**Question 7**: What is the probability of drawing an ace OR a jack from a standard 52-card deck?
#### Explanation
In a standard deck of 52 playing cards, there are 4 aces and 4 jacks. To find the probability of drawing an ace or a jack, you can use the formula for the probability of either of two mutually exclusive events happening:
\[ P(A \text{ or } J) = P(A) + P(J) \]
Here, \( P(A) \) represents the probability of drawing an ace, and \( P(J) \) represents the probability of drawing a jack.
- The probability of drawing an ace (\( P(A) \)):
\[ P(A) = \frac{\text{Number of Aces}}{\text{Total Number of Cards}} = \frac{4}{52} \]
- The probability of drawing a jack (\( P(J) \)):
\[ P(J) = \frac{\text{Number of Jacks}}{\text{Total Number of Cards}} = \frac{4}{52} \]
Since drawing an ace and drawing a jack are mutually exclusive events (they cannot happen at the same time), you can simply add the probabilities:
\[ P(A \text{ or } J) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} \]
Simplifying the fraction:
\[ P(A \text{ or } J) = \frac{8}{52} = \frac{2}{13} \]
Therefore, the probability of drawing an ace or a jack from a standard deck of 52 cards is \(\frac{2}{13}\).
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