Sec 2 2 Probability Independent And Dependent Probabilities - Fill ... - Free Printable

It looks like you've uploaded an image containing a worksheet titled "Section 5.2 Probability: Independent & Dependent Probability" with several problems related to probability, including independent and dependent events, conditional probability, and tree diagrams.

Let me go through each problem and solve it step by step, explaining the reasoning clearly.

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Problem 1: Determine the following probabilities.

We are given a table of outcomes for rolling two dice:

| Outcome | P(x) |
|--------|------|
| 2 | 1/36 |
| 3 | 2/36 |
| 4 | 3/36 |
| 5 | 4/36 |
| 6 | 5/36 |
| 7 | 6/36 |
| 8 | 5/36 |
| 9 | 4/36 |
| 10 | 3/36 |
| 11 | 2/36 |
| 12 | 1/36 |

#### a) \( P(6 \text{ and } 7) \)

This is asking for the probability that both a 6 and a 7 occur. But in a single roll of two dice, only one sum can occur. So this is impossible unless we're rolling twice.

But since the context is rolling two dice once, you cannot get both a 6 and a 7 in a single roll.

So:
\[
P(6 \text{ and } 7) = 0
\]

> ✔ Answer: \( 0 \)

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#### b) \( P(6 \text{ or } 7) \)

This is the probability of getting either a 6 or a 7.

From the table:
- \( P(6) = \frac{5}{36} \)
- \( P(7) = \frac{6}{36} \)

Since these are mutually exclusive (can't roll both at once), we add:
\[
P(6 \text{ or } 7) = \frac{5}{36} + \frac{6}{36} = \frac{11}{36}
\]

> ✔ Answer: \( \frac{11}{36} \)

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#### c) \( P(\text{not } 6) \)

This is the complement of rolling a 6.

\[
P(\text{not } 6) = 1 - P(6) = 1 - \frac{5}{36} = \frac{31}{36}
\]

> ✔ Answer: \( \frac{31}{36} \)

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#### d) Find the P(5) assuming 3 and 7 are independent events, and explain if this assumption is valid.

Wait — this part seems confusing.

If we're rolling two dice, the outcome of a single roll (like a sum of 5) is not "independent" of other outcomes in the sense that they are mutually exclusive. But perhaps the question is asking whether rolling a 5 is independent of rolling a 3 or 7?

But let's read carefully: *"Find the P(5) assuming 3 and 7 are independent events..."*

That doesn’t make sense — 3 and 7 are outcomes, not events that can be independent in the usual way. Also, the probability of rolling a 5 is already given as \( \frac{4}{36} \), so why assume anything?

Possibly a misstatement.

Alternatively, maybe the intent was to ask: *Are the events “rolling a 3” and “rolling a 7” independent?* But they are mutually exclusive, so they cannot be independent unless both have probability zero.

But since they are mutually exclusive and both have positive probability, they are not independent.

So the assumption that 3 and 7 are independent is invalid because:

- Two events A and B are independent if \( P(A \cap B) = P(A)P(B) \)
- But \( P(3 \cap 7) = 0 \), while \( P(3)P(7) = \frac{2}{36} \cdot \frac{6}{36} = \frac{12}{1296} > 0 \)

So they are not independent.

Therefore, the assumption is invalid.

> ✔ Answer:
> \( P(5) = \frac{4}{36} = \frac{1}{9} \)
> The assumption that 3 and 7 are independent is invalid because they are mutually exclusive and both have non-zero probability.

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Problem 2: Pulling a 2 on a standard die and then a 6

This is a sequence of two rolls (assuming with replacement, or independent rolls).

- \( P(\text{2}) = \frac{1}{6} \)
- \( P(\text{6}) = \frac{1}{6} \)

Since the rolls are independent:
\[
P(\text{2 then 6}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}
\]

> ✔ Answer: \( \frac{1}{36} \)

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Problem 3: Rolling a 2 on a standard die and pulling a red card from a standard deck of cards

These are independent events.

- \( P(\text{rolling a 2}) = \frac{1}{6} \)
- \( P(\text{red card}) = \frac{26}{52} = \frac{1}{2} \)

So:
\[
P(\text{2 and red card}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}
\]

> ✔ Answer: \( \frac{1}{12} \)

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Problem 4: Drawing a 3 from a deck of cards, replacing it, and drawing another 3

This is with replacement → independent events.

- \( P(\text{first 3}) = \frac{4}{52} = \frac{1}{13} \)
- \( P(\text{second 3}) = \frac{4}{52} = \frac{1}{13} \)

So:
\[
P(\text{both 3s}) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}
\]

> ✔ Answer: \( \frac{1}{169} \)

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Problem 5: If you were to flip a coin and roll a die, what is the probability that you would get heads and a 4?

Independent events.

- \( P(\text{heads}) = \frac{1}{2} \)
- \( P(\text{roll a 4}) = \frac{1}{6} \)

So:
\[
P(\text{heads and 4}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}
\]

> ✔ Answer: \( \frac{1}{12} \)

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Problem 6: You are chosen at random from the staff group. If you selected rank 1, 2, or 3, what is the chance that you were male?

We are given a table:

| Rank | Male | Female |
|------|------|--------|
| 1 | 4 | 6 |
| 2 | 10 | 15 |
| 3 | 12 | 18 |

Total males in ranks 1–3: \( 4 + 10 + 12 = 26 \)

Total females in ranks 1–3: \( 6 + 15 + 18 = 39 \)

Total people in ranks 1–3: \( 26 + 39 = 65 \)

We want: \( P(\text{male} \mid \text{rank 1, 2, or 3}) \)

Since all people in the group are in these ranks, this is just:
\[
P(\text{male}) = \frac{26}{65} = \frac{2}{5} = 0.4
\]

> ✔ Answer: \( \frac{2}{5} \) or \( 0.4 \)

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Problem 7: Smoking and Lung Cancer

Given:
- 45% of people smoke → \( P(S) = 0.45 \)
- 75% of smokers develop lung cancer → \( P(C|S) = 0.75 \)
- 25% of non-smokers develop lung cancer → \( P(C|\neg S) = 0.25 \)

We are asked: What is the probability that a person has lung cancer?

Use the Law of Total Probability:

\[
P(C) = P(C|S)P(S) + P(C|\neg S)P(\neg S)
\]
\[
= (0.75)(0.45) + (0.25)(0.55)
\]
\[
= 0.3375 + 0.1375 = 0.475
\]

> ✔ Answer: \( 0.475 \) or \( 47.5\% \)

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Problem 8: Conditional Probability – Tree Diagram

Given:
- 70% of people who take the test are sick → \( P(S) = 0.7 \)
- 80% of sick people test positive → \( P(T^+|S) = 0.8 \)
- 90% of healthy people test negative → \( P(T^-|\neg S) = 0.9 \) → so \( P(T^+|\neg S) = 0.1 \)

We want: \( P(\text{sick} \mid \text{positive test}) \)

Use Bayes’ Theorem:

\[
P(S|T^+) = \frac{P(T^+|S)P(S)}{P(T^+)}
\]

First compute \( P(T^+) \):

\[
P(T^+) = P(T^+|S)P(S) + P(T^+|\neg S)P(\neg S)
= (0.8)(0.7) + (0.1)(0.3) = 0.56 + 0.03 = 0.59
\]

Now:
\[
P(S|T^+) = \frac{0.56}{0.59} \approx 0.9492
\]

> ✔ Answer: Approximately \( 0.949 \) or \( 94.9\% \)

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Problem 9: Conditional Probability – Coin Toss

A fair coin is tossed until a head appears. We are told that the first toss was a tail.

We want: What is the probability that the second toss is a head, given that the first was a tail?

Since tosses are independent:

\[
P(\text{H on second} \mid \text{T on first}) = P(\text{H}) = \frac{1}{2}
\]

> ✔ Answer: \( \frac{1}{2} \)

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Problem 10: Probability of choosing a student who is a girl AND likes math

We are given:
- 30 students total
- 12 girls, 18 boys
- 15 like math, 15 don’t
- 9 girls like math

We want: \( P(\text{girl and likes math}) \)

Number of students who are girls and like math = 9

So:
\[
P(\text{girl and likes math}) = \frac{9}{30} = \frac{3}{10} = 0.3
\]

> ✔ Answer: \( 0.3 \)

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Problem 11: Are the events "being a girl" and "liking math" independent?

Check if:
\[
P(\text{girl and likes math}) = P(\text{girl}) \cdot P(\text{likes math})
\]

- \( P(\text{girl}) = \frac{12}{30} = 0.4 \)
- \( P(\text{likes math}) = \frac{15}{30} = 0.5 \)
- \( P(\text{girl and likes math}) = 0.3 \)

Now check:
\[
0.4 \times 0.5 = 0.2 \ne 0.3
\]

So they are not independent.

> ✔ Answer: No, they are not independent.

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Final Answers Summary

| Problem | Answer |
|--------|--------|
| 1a | 0 |
| 1b | \( \frac{11}{36} \) |
| 1c | \( \frac{31}{36} \) |
| 1d | \( P(5) = \frac{1}{9} \); assumption invalid |
| 2 | \( \frac{1}{36} \) |
| 3 | \( \frac{1}{12} \) |
| 4 | \( \frac{1}{169} \) |
| 5 | \( \frac{1}{12} \) |
| 6 | \( \frac{2}{5} \) |
| 7 | \( 0.475 \) |
| 8 | \( \approx 0.949 \) |
| 9 | \( \frac{1}{2} \) |
| 10 | \( 0.3 \) |
| 11 | Not independent |

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