Addition Rule Of Probability Independent Practice Worksheet - Fill ... - Free Printable

Problem Analysis and Solution

The worksheet involves problems related to the Addition Rule of Probability for independent events. The Addition Rule states:

- For two mutually exclusive events \( A \) and \( B \):
\[
P(A \cup B) = P(A) + P(B)
\]
- For two non-mutually exclusive events \( A \) and \( B \):
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]

Let's solve each problem step by step.

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#### Problem 1:
In a group of 52 students, 40 have brown eyes and 35 have hazel eyes. Find the probability that a student picked from this group at random either has brown or hazel eyes.

##### Solution:
1. Define the events:
- Let \( A \): Event that a student has brown eyes.
- Let \( B \): Event that a student has hazel eyes.
- We are asked to find \( P(A \cup B) \).

2. Given data:
- Total students = 52
- \( P(A) = \frac{\text{Number of students with brown eyes}}{\text{Total students}} = \frac{40}{52} \)
- \( P(B) = \frac{\text{Number of students with hazel eyes}}{\text{Total students}} = \frac{35}{52} \)

3. Check for overlap (non-mutually exclusive events):
- Since some students may have both brown and hazel eyes, we need to account for the overlap \( P(A \cap B) \).
- However, the problem does not provide the number of students who have both brown and hazel eyes. Assuming no overlap (for simplicity unless stated otherwise):
\[
P(A \cap B) = 0
\]

4. Apply the Addition Rule:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
\[
P(A \cup B) = \frac{40}{52} + \frac{35}{52} - 0
\]
\[
P(A \cup B) = \frac{40 + 35}{52} = \frac{75}{52}
\]

5. Final Answer:
\[
\boxed{\frac{75}{52}}
\]

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#### Problem 2:
John is playing cards. What is the probability of pulling an ace or queen from the deck?

##### Solution:
1. Define the events:
- Let \( A \): Event of drawing an ace.
- Let \( B \): Event of drawing a queen.
- We are asked to find \( P(A \cup B) \).

2. Given data:
- Total cards in a deck = 52
- Number of aces = 4
- Number of queens = 4

3. Calculate probabilities:
- \( P(A) = \frac{\text{Number of aces}}{\text{Total cards}} = \frac{4}{52} \)
- \( P(B) = \frac{\text{Number of queens}}{\text{Total cards}} = \frac{4}{52} \)

4. Check for overlap (mutually exclusive events):
- Aces and queens are distinct cards, so there is no overlap:
\[
P(A \cap B) = 0
\]

5. Apply the Addition Rule:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
\[
P(A \cup B) = \frac{4}{52} + \frac{4}{52} - 0
\]
\[
P(A \cup B) = \frac{4 + 4}{52} = \frac{8}{52} = \frac{2}{13}
\]

6. Final Answer:
\[
\boxed{\frac{2}{13}}
\]

---

#### Problem 3:
In a group of 62 students, 27 are normal, 13 are abnormal, and 22 are normal-abnormal. Find the probability that a student picked from this group at random is either normal or abnormal.

##### Solution:
1. Define the events:
- Let \( N \): Event that a student is normal.
- Let \( A \): Event that a student is abnormal.
- We are asked to find \( P(N \cup A) \).

2. Given data:
- Total students = 62
- Normal students (\( N \)) = 27
- Abnormal students (\( A \)) = 13
- Normal-abnormal students (\( N \cap A \)) = 22

3. Calculate probabilities:
- \( P(N) = \frac{\text{Number of normal students}}{\text{Total students}} = \frac{27}{62} \)
- \( P(A) = \frac{\text{Number of abnormal students}}{\text{Total students}} = \frac{13}{62} \)
- \( P(N \cap A) = \frac{\text{Number of normal-abnormal students}}{\text{Total students}} = \frac{22}{62} \)

4. Apply the Addition Rule:
\[
P(N \cup A) = P(N) + P(A) - P(N \cap A)
\]
\[
P(N \cup A) = \frac{27}{62} + \frac{13}{62} - \frac{22}{62}
\]
\[
P(N \cup A) = \frac{27 + 13 - 22}{62} = \frac{18}{62} = \frac{9}{31}
\]

5. Final Answer:
\[
\boxed{\frac{9}{31}}
\]

---

#### Problem 4:
George is playing cards. What is the probability of pulling an ace or jack from the deck?

##### Solution:
1. Define the events:
- Let \( A \): Event of drawing an ace.
- Let \( J \): Event of drawing a jack.
- We are asked to find \( P(A \cup J) \).

2. Given data:
- Total cards in a deck = 52
- Number of aces = 4
- Number of jacks = 4

3. Calculate probabilities:
- \( P(A) = \frac{\text{Number of aces}}{\text{Total cards}} = \frac{4}{52} \)
- \( P(J) = \frac{\text{Number of jacks}}{\text{Total cards}} = \frac{4}{52} \)

4. Check for overlap (mutually exclusive events):
- Aces and jacks are distinct cards, so there is no overlap:
\[
P(A \cap J) = 0
\]

5. Apply the Addition Rule:
\[
P(A \cup J) = P(A) + P(J) - P(A \cap J)
\]
\[
P(A \cup J) = \frac{4}{52} + \frac{4}{52} - 0
\]
\[
P(A \cup J) = \frac{4 + 4}{52} = \frac{8}{52} = \frac{2}{13}
\]

6. Final Answer:
\[
\boxed{\frac{2}{13}}
\]

---

#### Problem 5:
In a group of 100 people, 19 were drinking punch and 23 were eating fish. Find the probability that a person picked from the group at random is either drinking punch or eating fish.

##### Solution:
1. Define the events:
- Let \( P \): Event that a person is drinking punch.
- Let \( F \): Event that a person is eating fish.
- We are asked to find \( P(P \cup F) \).

2. Given data:
- Total people = 100
- People drinking punch (\( P \)) = 19
- People eating fish (\( F \)) = 23

3. Calculate probabilities:
- \( P(P) = \frac{\text{Number of people drinking punch}}{\text{Total people}} = \frac{19}{100} \)
- \( P(F) = \frac{\text{Number of people eating fish}}{\text{Total people}} = \frac{23}{100} \)

4. Check for overlap (non-mutually exclusive events):
- The problem does not specify how many people are both drinking punch and eating fish. Assuming no overlap (for simplicity unless stated otherwise):
\[
P(P \cap F) = 0
\]

5. Apply the Addition Rule:
\[
P(P \cup F) = P(P) + P(F) - P(P \cap F)
\]
\[
P(P \cup F) = \frac{19}{100} + \frac{23}{100} - 0
\]
\[
P(P \cup F) = \frac{19 + 23}{100} = \frac{42}{100} = \frac{21}{50}
\]

6. Final Answer:
\[
\boxed{\frac{21}{50}}
\]

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Final Answers:

1. \(\boxed{\frac{75}{52}}\)
2. \(\boxed{\frac{2}{13}}\)
3. \(\boxed{\frac{9}{31}}\)
4. \(\boxed{\frac{2}{13}}\)
5. \(\boxed{\frac{21}{50}}\)