Probability Notes and Worksheets - Lindsay Bowden - Free Printable

Let's solve each problem step by step, using the principles of conditional probability and interpreting the Venn diagram.

---

1. What is the probability of two dice landing on 4 if you already know the first die has landed on 4?

We are given that the first die is a 4, and we want the probability that both dice land on 4 under this condition.

- Since the first die is already 4, we only care about the second die.
- The second die must also be 4.
- A fair die has 6 sides, so the probability of rolling a 4 on the second die is:

$$
P(\text{Second die = 4}) = \frac{1}{6}
$$

✔ Answer: $ \frac{1}{6} $

---

2. What is the probability of three quarters landing on tails if you know the first one has landed on tails?

We are told the first quarter is tails, and we want the probability that all three are tails.

- We're looking for: $ P(\text{All 3 tails} \mid \text{First is tails}) $
- Given the first is tails, we need the remaining two to be tails.
- Each coin flip is independent.
- So, probability that the second and third are both tails:

$$
P(\text{Second = T}) \times P(\text{Third = T}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
$$

✔ Answer: $ \frac{1}{4} $

---

Now let’s analyze the Venn diagram:

The diagram shows:
- Only A: 21
- A ∩ B: 3
- Only B: 33
- Outside both: 12

So total number of students:
$$
21 + 3 + 33 + 12 = 69
$$

Let’s define:
- Set A: 9th graders → Total in A = 21 + 3 = 24
- Set B: Students who have pets → Total in B = 33 + 3 = 36
- Intersection (A ∩ B): 3 students who are 9th graders and have pets
- Outside both: 12 students who are not 9th graders and don’t have pets

---

3. Find $ P(A|B) $

Conditional probability formula:
$$
P(A|B) = \frac{P(A \cap B)}{P(B)}
$$

From the diagram:
- $ P(A \cap B) = \frac{3}{69} $
- $ P(B) = \frac{36}{69} $

So:
$$
P(A|B) = \frac{3/69}{36/69} = \frac{3}{36} = \frac{1}{12}
$$

✔ Answer: $ \frac{1}{12} $

---

4. Find $ P(B|A) $

$$
P(B|A) = \frac{P(A \cap B)}{P(A)}
$$

- $ P(A \cap B) = \frac{3}{69} $
- $ P(A) = \frac{24}{69} $

$$
P(B|A) = \frac{3/69}{24/69} = \frac{3}{24} = \frac{1}{8}
$$

✔ Answer: $ \frac{1}{8} $

---

5. If A represents 9th graders and B represents students that have pets, what does $ P(A|B) $ represent in the context of this problem?

This is the probability that a student is a 9th grader, given that they have a pet.

In other words:
> "What is the chance that a student who has a pet is a 9th grader?"

✔ Answer: The probability that a student is a 9th grader given that they have a pet.

---

6. What does $ P(B|A) $ represent in the context of this problem?

This is the probability that a student has a pet, given that they are a 9th grader.

> "What is the chance that a 9th grader has a pet?"

✔ Answer: The probability that a student has a pet given that they are a 9th grader.

---

7. What do the 12 students who are not in sets A or B represent in the context of this problem?

These students are:
- Not in A → Not 9th graders
- Not in B → Do not have pets

So they are students who are not 9th graders and do not have pets.

✔ Answer: Students who are not 9th graders and do not have pets.

---

8. How many students have pets?

Students in set B = those who have pets.

From the diagram:
- Only B: 33
- A ∩ B: 3
- So total with pets = 33 + 3 = 36

✔ Answer: 36 students have pets.

---

✔ Final Answers Summary:

1. $ \frac{1}{6} $
2. $ \frac{1}{4} $
3. $ \frac{1}{12} $
4. $ \frac{1}{8} $
5. Probability a student is a 9th grader given they have a pet.
6. Probability a student has a pet given they are a 9th grader.
7. Students who are not 9th graders and do not have pets.
8. 36 students have pets.

Let me know if you'd like this formatted as a printable answer sheet!