Probability worksheet with problems on experimental and theoretical probability involving card selection and eye color distribution.

Let’s go step by step through each question. We’ll use the data from Amanda’s card experiment and Dale’s eye color survey to find probabilities.

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Amanda’s Card Experiment:

She drew cards 30 times total (7 Diamonds + 9 Hearts + 11 Spades + 3 Clubs = 30).

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Question 1: Experimental probability of selecting a heart?

Experimental probability = (number of times heart was picked) ÷ (total number of trials)

→ 9 hearts out of 30 draws → 9/30

We can simplify that: divide numerator and denominator by 3 → 3/10

✔ Answer: 3/10

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Question 2: Theoretical probability of selecting a heart?

In a standard deck, there are 52 cards, and 13 of them are hearts.

So theoretical probability = 13/52

Simplify: divide top and bottom by 13 → 1/4

✔ Answer: 1/4

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Question 3: Experimental probability of diamond OR spade?

Diamonds: 7
Spades: 11
Total for either: 7 + 11 = 18
Total trials: 30

→ 18/30

Simplify: divide by 6 → 3/5

✔ Answer: 3/5

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Question 4: Theoretical probability of diamond OR spade?

Diamonds: 13 cards
Spades: 13 cards
Total favorable: 13 + 13 = 26
Total cards: 52

→ 26/52 = 1/2

✔ Answer: 1/2

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Question 5: Compare results and describe findings.

Experimental vs Theoretical:

- Heart: Exp = 3/10 (0.3), Theo = 1/4 (0.25) → close but not same
- Diamond or Spade: Exp = 3/5 (0.6), Theo = 1/2 (0.5) → also close

They’re not exactly the same — which is normal in experiments because you’re only doing 30 tries. But they’re within about 10% of each other, so it’s reasonable.

✔ Answer: The experimental and theoretical probabilities are not the same, but they are close — within about 10% of each other. This makes sense because with more trials, experimental probability usually gets closer to theoretical.

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Dale’s Eye Color Survey:

Total students surveyed: 12 (Blue) + 58 (Brown) + 2 (Green) + 8 (Hazel) = 80

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Question 6a: Experimental probability distribution for each eye color

P(blue) = 12/80 → simplify: divide by 4 → 3/20

P(brown) = 58/80 → divide by 2 → 29/40

P(green) = 2/80 → divide by 2 → 1/40

P(hazel) = 8/80 → divide by 8 → 1/10

✔ Answers:
- P(blue) = 3/20
- P(brown) = 29/40
- P(green) = 1/40
- P(hazel) = 1/10

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Question 6b: Probability student has blue OR green eyes?

Add P(blue) + P(green) = 12/80 + 2/80 = 14/80

Simplify: divide by 2 → 7/40

✔ Answer: 7/40

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Question 6c: Probability student does NOT have brown OR hazel eyes?

That means they have BLUE or GREEN.

Same as above! Because “not brown or hazel” = blue or green.

So again: 12 + 2 = 14 → 14/80 = 7/40

Wait — let me double-check.

“Not have brown or hazel” → exclude brown (58) and hazel (8) → left with blue (12) + green (2) = 14 → yes, 14/80 = 7/40

But wait — the handwritten answer says 7/8? That must be wrong.

Hold on — maybe I misread.

Original question: “does not have brown or hazel eyes?”

Yes — so we want students who have NEITHER brown NOR hazel → so only blue or green → 12 + 2 = 14 → 14/80 = 7/40.

But the handwritten note says: 80 - 10 = 70 → 70/80 = 7/8.

Ah — that’s if they meant “does not have brown AND does not have hazel” — but that’s the same thing.

Wait — no. If you subtract brown AND hazel, you get 58 + 8 = 66 → 80 - 66 = 14 → still 14/80.

Unless... did they mean “does not have (brown or hazel)” — which is correct — 14/80.

But 70/80 would be if they subtracted only 10 — which doesn’t make sense.

Wait — looking back at the image: the handwritten part says “80 - 10 = 70” — where does 10 come from?

Maybe they added blue and green? 12 + 2 = 14 — not 10.

This seems like an error in the handwritten work.

Actually — perhaps they meant “does not have brown” — then it would be 80 - 58 = 22 → 22/80 = 11/40.

Or “does not have hazel” — 80 - 8 = 72 → 72/80 = 9/10.

But the question says: “does not have brown or hazel eyes”

In math, “or” in this context usually means union — so excluding both.

So correct answer should be 14/80 = 7/40.

But let’s check the original problem again.

Looking at the image: Question 6c says:

> c. Based on the survey, what is the experimental probability that a student in Dale’s class does not have brown or hazel eyes?

And the handwritten answer says: 80 - 10 = 70 → 70/80 = 7/8.

Where does 10 come from? Maybe they thought “brown or hazel” is 58 + 8 = 66, but wrote 10 by mistake? Or maybe they meant something else.

Wait — another interpretation: sometimes people say “does not have brown or hazel” meaning “has neither”, which is correct — 14/80.

But 7/8 is 70/80 — which would be if only 10 students had brown or hazel — but that’s not true.

I think the handwritten answer is incorrect.

Let me recalculate:

Total students: 80

Students with brown OR hazel: 58 + 8 = 66

Students without brown OR hazel: 80 - 66 = 14

Probability: 14/80 = 7/40

✔ Correct Answer: 7/40

*(Note: The handwritten answer of 7/8 appears to be a calculation error.)*

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Question 6d: If distribution is similar, how many of 360 students expected to have brown eyes?

From survey: P(brown) = 58/80

Apply to 360 students: (58/80) × 360

First, simplify 58/80 = 29/40

Now: (29/40) × 360

= 29 × (360 ÷ 40) = 29 × 9 = 261

✔ Answer: 261 students

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Final Answers Summary:

1. 3/10
2. 1/4
3. 3/5
4. 1/2
5. The experimental and theoretical probabilities are not the same, but they are close — within about 10% of each other. This makes sense because with more trials, experimental probability usually gets closer to theoretical.
6a. P(blue)=3/20, P(brown)=29/40, P(green)=1/40, P(hazel)=1/10
6b. 7/40
6c. 7/40 *(corrected from handwritten 7/8)*
6d. 261 students

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Final Answer:
1. \(\frac{3}{10}\)
2. \(\frac{1}{4}\)
3. \(\frac{3}{5}\)
4. \(\frac{1}{2}\)
5. The experimental and theoretical probabilities are not the same, but they are close — within about 10% of each other. This makes sense because with more trials, experimental probability usually gets closer to theoretical.
6a. P(blue) = \(\frac{3}{20}\), P(brown) = \(\frac{29}{40}\), P(green) = \(\frac{1}{40}\), P(hazel) = \(\frac{1}{10}\)
6b. \(\frac{7}{40}\)
6c. \(\frac{7}{40}\)
6d. 261