Compound Probability Error Analysis worksheet with six problems requiring students to identify and correct math errors in probability calculations.

It looks like you've uploaded a worksheet titled "Compound Probability Error Analysis", which contains several probability problems, each with a mistake in the solution. Your task is to:

1. Identify the mistake in each problem.
2. Explain why it's wrong.
3. Provide the correct answer.

Let’s go through each problem one by one and analyze the errors.

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Problem 1:

> A bag contains 7 blue, 5 red, and 3 yellow marbles. Charlotte draws two marbles, records the color, and then replaces them. What is the probability of drawing a blue marble and then a yellow marble?

Student's Solution:
$$
\frac{7}{15} \times \frac{3}{14} = \frac{21}{210} = \frac{1}{10}
$$

Error:
The student used without replacement logic (since they used $ \frac{3}{14} $ instead of $ \frac{3}{15} $), but the problem states that the marbles are replaced after each draw.

✔ Correct Approach: Since the draws are independent due to replacement, the probability remains the same for both draws:
$$
P(\text{blue then yellow}) = \frac{7}{15} \times \frac{3}{15} = \frac{21}{225} = \frac{7}{75}
$$

✘ Mistake: Used conditional probability (no replacement) when replacement was specified.

✔ Correct Answer: $ \boxed{\frac{7}{75}} $

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Problem 2:

> You spin the spinner below twice. What is the probability of spinning a 3 both times?

Spinner has numbers: 1, 2, 3, 4, 5

Student's Solution:
$$
\frac{1}{5} + \frac{1}{5} = \frac{2}{5}
$$

Error:
The student added the probabilities instead of multiplying them. This is a common mistake when dealing with independent events.

✔ Correct Approach: Since the spins are independent:
$$
P(\text{3 and 3}) = \frac{1}{5} \times \frac{1}{5} = \frac{1}{25}
$$

✘ Mistake: Added probabilities instead of multiplying for compound events.

✔ Correct Answer: $ \boxed{\frac{1}{25}} $

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Problem 3:

> You have two choosing partners from a class by drawing names out of a bag. There are 12 boys and 13 girls. The first person drawn is a girl. What is the probability that the teacher will draw a boy first and then a girl?

Wait — this wording is confusing. Let’s read carefully:

> "The first person drawn is a girl." → But then asks: "What is the probability that the teacher will draw a boy first and then a girl?"

This is contradictory. If the first person drawn is a girl, then “boy first” cannot happen.

But looking at the student's work:
$$
\frac{12}{25} \times \frac{13}{24} = \frac{156}{600} = \frac{13}{50}
$$

That’s actually the correct way to compute P(boy first, then girl), assuming no replacement.

So what’s the issue?

Ah! The problem says: "The first person drawn is a girl." So we're given that the first person is a girl.

Then it asks: "What is the probability that the teacher will draw a boy first and then a girl?" — this is impossible if the first is already a girl.

So either:
- The question is misworded, or
- The student ignored the condition.

But the student computed P(boy then girl), not conditioning on the first being a girl.

Mistake: The problem gives a condition: "The first person drawn is a girl", so the event "boy first" cannot occur. Therefore, the probability of "boy first and then girl" is zero.

But the student didn't use the condition and just computed the unconditional probability.

✔ Correct Interpretation: Given that the first person drawn is a girl, the probability that the first is a boy is 0.

So the probability of "boy first and then girl" is 0.

✘ Mistake: Ignored the given condition that the first person drawn is a girl.

✔ Correct Answer: $ \boxed{0} $

---

Problem 4:

> Gin is playing a game at a carnival. She gets to spin a wheel with 8 equal sections. 3 are labeled "win" and 5 are labeled "lose". She gets to spin the wheel twice. What is the probability that she wins both times?

Student's Solution:
$$
\frac{3}{8} \times \frac{3}{8} = \frac{9}{64}
$$

Wait — this looks correct? But let's check.

Assuming spins are independent (no mention of removal), yes, the probability of winning both times is:
$$
\frac{3}{8} \times \frac{3}{8} = \frac{9}{64}
$$

But the worksheet says there's an error.

Looking closely: the student wrote:
$$
\frac{3}{8} \times \frac{3}{8} = \frac{9}{64}
$$

But the teacher's note says: *"The student correctly found the probability of winning both times."* — Wait, no, the student's answer is correct, but perhaps the problem is misstated?

Wait — look at the actual image — the student wrote:
$$
\frac{3}{8} \times \frac{3}{8} = \frac{9}{64}
$$

But the correct answer box says: $ \frac{3}{8} \times \frac{3}{8} = \frac{9}{64} $ — so it’s marked as correct?

But the worksheet says “Find the mistake”, so something must be wrong.

Wait — perhaps the wheel is not independent? Or maybe it’s with replacement?

No — spinning a wheel twice is independent.

Unless... the student did not reduce or something?

Wait — the student did reduce correctly.

But the correct answer is shown as $ \frac{3}{8} \times \frac{3}{8} = \frac{9}{64} $ — so the student got it right.

But the worksheet says "find the mistake" — so maybe there's a typo?

Wait — looking at the original problem: It says "She gets to spin the wheel twice."

But the student’s work is correct.

Hmm.

Wait — maybe the student’s answer is correct, but the teacher’s note says: “The student correctly found the probability of winning both times.” — so perhaps the error is in the explanation?

No — the student’s math is correct.

Wait — perhaps the mistake is in the assumption that spins are independent?

But that’s standard.

Alternatively — could the wheel be spun only once, and she gets two chances?

No — it says "spin the wheel twice".

I think this might be a trick — the student’s answer is actually correct, so maybe the mistake is not in the math, but in the interpretation?

Wait — look again at the student's written work:

$$
\frac{3}{8} \times \frac{3}{8} = \frac{9}{64}
$$

Yes — correct.

But the worksheet says: “Find the mistake.”

Perhaps the student should have used conditional probability? No — independent spins.

Wait — unless the wheel changes? But no.

Wait — maybe the student forgot to multiply? No — they did.

Wait — perhaps the problem says "she wins at least once"? But no — it says "wins both times".

So I think the student is correct, and the worksheet may have an error — or perhaps I'm missing something.

But the teacher’s note says: “The student correctly found the probability of winning both times.” — so maybe the error is elsewhere?

Wait — no — the student’s answer is correct, so perhaps this problem is not intended to have an error?

But the instructions say each problem has a mistake.

Wait — perhaps the student wrote the answer as 9/64, but the correct answer is different?

No — 3/8 × 3/8 = 9/64.

Unless the wheel has more than 8 sections?

Wait — the problem says: "a wheel with 8 equal sections. 3 are labeled 'win' and 5 are labeled 'lose'." So yes, 3/8.

So I think this one is correct, but maybe the mistake is in the explanation?

Wait — perhaps the student assumed independence, but the problem doesn’t specify whether the wheel is fair or not?

But it says "equal sections", so it is fair.

So I think this one is correct, and the worksheet may have a typo.

But since the instructions say every problem has a mistake, maybe I’m missing something.

Wait — look at the student’s work — they wrote:
$$
\frac{3}{8} \times \frac{3}{8} = \frac{9}{64}
$$

But the correct answer is $ \frac{9}{64} $, so no mistake.

But the teacher’s note says: “The student correctly found the probability…” — so perhaps this is not an error?

But the worksheet says each problem has a mistake.

Wait — maybe the problem is asking for something else?

No — it clearly says: “What is the probability that she wins both times?”

So I think this one is correct, and the mistake might be in the problem statement.

But let’s move on.

---

Problem 5:

> A bag contains eight white and six black marbles. A marble is randomly selected and not replaced. Then a second marble is randomly selected. What is the probability that both marbles are white?

Student’s Solution:
$$
\frac{8}{14} \times \frac{8}{13} = \frac{64}{182} = \frac{32}{91}
$$

Error:
The student used $ \frac{8}{13} $ for the second draw, but after removing one white marble, there are now 7 white marbles left out of 13 total.

So the second probability should be $ \frac{7}{13} $, not $ \frac{8}{13} $.

✔ Correct Calculation:
$$
P(\text{white then white}) = \frac{8}{14} \times \frac{7}{13} = \frac{56}{182} = \frac{28}{91} = \frac{4}{13}
$$

✘ Mistake: Did not account for reduced number of white marbles after first draw (used 8/13 instead of 7/13).

✔ Correct Answer: $ \boxed{\frac{4}{13}} $

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Problem 6:

> A bag contains 4 red, 3 blue, and 5 green marbles. A marble is chosen at random, replaced, and then another is chosen at random. What is the probability that both marbles are red?

Student’s Solution:
$$
\frac{4}{12} \times \frac{3}{12} = \frac{12}{144} = \frac{1}{12}
$$

Error:
The student used $ \frac{3}{12} $ for the second draw, but since the marble is replaced, the probability of red on the second draw is still $ \frac{4}{12} $, not $ \frac{3}{12} $.

They probably thought that one red was removed, but replacement means the count stays the same.

✔ Correct Calculation:
$$
P(\text{red and red}) = \frac{4}{12} \times \frac{4}{12} = \frac{16}{144} = \frac{1}{9}
$$

✘ Mistake: Used $ \frac{3}{12} $ instead of $ \frac{4}{12} $ for the second draw, ignoring replacement.

✔ Correct Answer: $ \boxed{\frac{1}{9}} $

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Summary of Errors:

| Problem | Mistake | Correct Answer |
|--------|--------|----------------|
| 1 | Used no replacement when replacement was specified | $ \frac{7}{75} $ |
| 2 | Added probabilities instead of multiplying | $ \frac{1}{25} $ |
| 3 | Ignored condition that first person is a girl; computed unconditional probability | $ 0 $ |
| 4 | No mistake — student is correct | $ \frac{9}{64} $ |
| 5 | Used $ \frac{8}{13} $ instead of $ \frac{7}{13} $ for second draw | $ \frac{4}{13} $ |
| 6 | Used $ \frac{3}{12} $ instead of $ \frac{4}{12} $, ignoring replacement | $ \frac{1}{9} $ |

Wait — Problem 4 seems correct, but the worksheet says "find the mistake".

Possibility: Maybe the student wrote $ \frac{3}{8} \times \frac{3}{8} $, but the correct answer is different?

Or perhaps the wheel is not independent? But it is.

Another possibility: The student should have used tree diagram or conditional probability, but no — multiplication rule is fine.

Alternatively, maybe the problem says "at least one win", but it says "both times".

Wait — double-check the original text:

> "What is the probability that she wins both times?"

Yes — so correct.

But the worksheet says "find the mistake", so perhaps the student made a calculation error?

$ \frac{3}{8} \times \frac{3}{8} = \frac{9}{64} $ — correct.

Maybe the answer is simplified incorrectly? No — it's already simplified.

So I think Problem 4 has no mistake, but perhaps the intended mistake is that the student assumed independence, but maybe the wheel is biased?

But no — equal sections imply fairness.

So likely, Problem 4 is correct, and the worksheet has a typo.

But since the instruction says each problem has a mistake, perhaps in Problem 4, the student wrote the answer as 9/64, but the correct answer is 3/8 × 3/8, which is the same.

No.

Alternatively, maybe the student should have used addition for "or", but no — it's "both".

I think it's safe to conclude that Problem 4 is correct, and the others have clear errors.

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Final Answers:

1. Mistake: Didn't use replacement; used $ \frac{3}{14} $ instead of $ \frac{3}{15} $.
✔ Correct: $ \frac{7}{15} \times \frac{3}{15} = \frac{7}{75} $

2. Mistake: Added instead of multiplied.
✔ Correct: $ \frac{1}{5} \times \frac{1}{5} = \frac{1}{25} $

3. Mistake: Ignored the condition that the first person is a girl; computed P(boy then girl) despite contradiction.
✔ Correct: $ 0 $

4. Mistake: None — student is correct.
✔ Answer: $ \frac{9}{64} $ (but possibly a typo in worksheet)

5. Mistake: Used $ \frac{8}{13} $ instead of $ \frac{7}{13} $.
✔ Correct: $ \frac{8}{14} \times \frac{7}{13} = \frac{4}{13} $

6. Mistake: Used $ \frac{3}{12} $ instead of $ \frac{4}{12} $, ignoring replacement.
✔ Correct: $ \frac{4}{12} \times \frac{4}{12} = \frac{1}{9} $

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