Worksheet on probability concepts including relative frequency and expected outcomes with real-world scenarios.

Let’s go through each problem one by one, step by step. I’ll solve them carefully and check my work before giving the final answers.

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Problem 1:

The relative frequency of tails is 0.18 → that means heads must be 1 - 0.18 = 0.82 (since only two outcomes: heads or tails).

If flipped 50 times, expected number of heads = 50 × 0.82 = 41

✔ Check: 50 × 0.82 = 41 → correct.

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Problem 2:

Spinner spun 48 times. The spinner has 8 equal sections:

- B appears in 3 sections → probability = 3/8
- G appears in 3 sections → probability = 3/8
- Y appears in 2 sections → probability = 2/8 = 1/4

Expected outcome = total spins × probability for each color.

So:

- B: 48 × 3/8 = 18
- G: 48 × 3/8 = 18
- Y: 48 × 2/8 = 12

✔ Check: 18 + 18 + 12 = 48 → matches total spins.

Table:

| Colour | B | G | Y |
|--------|----|----|----|
| Expected Outcome | 18 | 18 | 12 |

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Problem 3:

Ben rolled a die 50 times. Results:

a) Ben says “die must be biased because only one 3 in 50 rolls” — Is he correct?

→ Not necessarily. Even with a fair die, you might get low numbers sometimes due to chance. You need more evidence (like many trials showing consistent bias). So no, Ben is not correct — one low count doesn’t prove bias.

b) Complete table with relative frequencies:

Relative frequency = frequency ÷ total rolls (50)

- Number 1: 12 ÷ 50 = 0.24
- Number 2: 8 ÷ 50 = 0.16
- Number 3: 1 ÷ 50 = 0.02
- Number 4: 15 ÷ 50 = 0.30
- Number 5: 5 ÷ 50 = 0.10
- Number 6: 6 ÷ 50 = 0.12

✔ Check sum: 0.24+0.16+0.02+0.30+0.10+0.12 = 0.94? Wait — that’s not 1.00!

Wait — let me recalculate frequencies:

Given frequencies: 12, 8, 1, 15, 5, 6 → sum = 12+8=20; +1=21; +15=36; +5=41; +6=47 → Oh! Total is 47, not 50? But problem says “rolled 50 times”.

Wait — maybe typo? Or did I misread?

Looking back: “Ben rolls a die 50 times, here are his results” — then table shows frequencies adding to 47? That can’t be right.

Wait — perhaps it’s a mistake in the problem? Or maybe I missed something.

Actually, looking again: the table lists numbers 1 to 6, frequencies: 12,8,1,15,5,6 → sum = 47. But problem says 50 rolls. Hmm.

Possibility: Maybe there’s an error in the problem, or perhaps we’re supposed to use 50 as denominator anyway? But that would be wrong.

Alternatively — maybe the table is incomplete? No, it lists all 6 numbers.

Wait — perhaps it’s intentional? Let me re-read: “here are his results” — maybe he didn’t record 3 rolls? Unlikely.

But since the problem says “rolled 50 times”, and gives these frequencies, perhaps we should assume the frequencies add to 50? But they don’t.

Wait — 12+8+1+15+5+6 = 47. Missing 3 rolls? That doesn’t make sense.

Perhaps it’s a trick? Or maybe I miscalculated.

12 + 8 = 20
20 + 1 = 21
21 + 15 = 36
36 + 5 = 41
41 + 6 = 47 → yes, 47.

But problem says 50 rolls. This is confusing.

Wait — maybe the table is correct and total is 47? But problem says 50. Perhaps typo in problem?

To proceed, I think we have to use the given frequencies and divide by 50 as stated, even if sum is 47 — but that would give relative frequencies that don’t add to 1.

That doesn’t make sense.

Alternative idea: Maybe “50 times” is approximate? Or perhaps the student made a recording error? But for math purposes, we should use the actual total from data.

I think the best approach is to use the sum of frequencies as the total for relative frequency calculation — so total = 47.

But the problem says “rolled 50 times”, so probably expects us to use 50.

This is ambiguous.

Wait — let me look at part c: “if Ben rolled the dice 80 times...” — so likely, the 50 is fixed.

Perhaps the frequencies are correct and sum to 50? Let me double-check addition:

Number 1: 12
Number 2: 8 → 20
Number 3: 1 → 21
Number 4: 15 → 36
Number 5: 5 → 41
Number 6: 6 → 47

Still 47. Unless... is there a number missing? No.

Perhaps it's a mistake, and it should be 50. Maybe number 3 is 4? But it's written as 1.

I think for the sake of solving, I'll assume the total is 50 as stated, and calculate relative frequencies using 50, even though frequencies sum to 47 — but that would mean some rolls are unaccounted for, which is odd.

Alternatively, perhaps the "50" includes something else? No.

Another thought: maybe the table is for the results, and total is 47, but problem says 50 — contradiction.

I recall that in some problems, they give frequencies that don't add up, but here it's clear.

Perhaps I should use 50 as denominator for relative frequency, as per problem statement.

Let me do that, and note the discrepancy.

So:

Relative frequency = frequency / 50

- 1: 12/50 = 0.24
- 2: 8/50 = 0.16
- 3: 1/50 = 0.02
- 4: 15/50 = 0.30
- 5: 5/50 = 0.10
- 6: 6/50 = 0.12

Sum: 0.24+0.16=0.4; +0.02=0.42; +0.30=0.72; +0.10=0.82; +0.12=0.94 — not 1.00.

That's problematic.

Perhaps the total is 47, and "50" is a typo? Let me check if 47 makes sense.

If total is 47, then:

- 1: 12/47 ≈ 0.255
- 2: 8/47 ≈ 0.170
- 3: 1/47 ≈ 0.021
- 4: 15/47 ≈ 0.319
- 5: 5/47 ≈ 0.106
- 6: 6/47 ≈ 0.128

Sum ≈ 1.000 — good.

But the problem says "50 times", so likely expects 50.

I think there might be an error in the problem, but for consistency, I'll use 50 as stated, and proceed. In real life, we'd question the data, but for homework, follow instructions.

So for b), relative frequencies with denominator 50:

| Number | Frequency | Relative Frequency |
|--------|-----------|---------------------|
| 1 | 12 | 0.24 |
| 2 | 8 | 0.16 |
| 3 | 1 | 0.02 |
| 4 | 15 | 0.30 |
| 5 | 5 | 0.10 |
| 6 | 6 | 0.12 |

c) If rolled 80 times, how many even numbers expected?

Even numbers on die: 2,4,6

From data, frequency of even numbers: 8 (for 2) + 15 (for 4) + 6 (for 6) = 29 out of 50 rolls.

So relative frequency of even = 29/50 = 0.58

For 80 rolls, expected even numbers = 80 × 0.58 = 46.4 → but since we can't have fraction, probably round to nearest whole number? Or keep as decimal? Usually in expected value, we can have decimal.

But let's see: 80 × 29/50 = (80/50)*29 = 1.6 * 29 = 46.4

So 46.4, but perhaps they want integer? Or leave as is.

In context, "how many" might expect integer, but expected value can be decimal.

I'll put 46.4, but let's confirm.

Since it's expected outcome, 46.4 is fine.

But let's calculate exactly: 80 * 29 / 50 = (80/50)*29 = 8/5 * 29 = 232/5 = 46.4

Yes.

Now, back to a): Ben is not correct — one low count doesn't prove bias; could be random variation.

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Problem 4:

Lewis and Gracie roll biased 4-sided die.

Lewis: 20 rolls, relative frequency of rolling 4 is 0.15 → so he got 4: 20 * 0.15 = 3 times.

Gracie: 50 rolls, relative frequency of 4 is 0.2 → so she got 4: 50 * 0.2 = 10 times.

a) Who has better estimate for probability of rolling 4?

Generally, larger sample size gives better estimate. Gracie has 50 rolls vs Lewis's 20, so Gracie's estimate is more reliable.

Also, her relative frequency is based on more data.

So Gracie has better estimate.

b) Using all their data, calculate amount of 4s expected from 200 rolls.

First, combine their data:

Total rolls: 20 + 50 = 70

Total 4s: Lewis: 3, Gracie: 10 → total 13

So combined relative frequency of 4 = 13/70

For 200 rolls, expected number of 4s = 200 * (13/70) = (200/70)*13 = (20/7)*13 = 260/7 ≈ 37.142...

So approximately 37.14, but since it's expected value, we can leave as fraction or decimal.

260/7 = 37 and 1/7, or 37.142...

Probably fine as 260/7 or decimal.

But let's compute: 200 * 13 / 70 = 2600 / 70 = 260 / 7 ≈ 37.1429

I'll keep it as fraction or decimal? Probably decimal is fine.

Note: 260 ÷ 7 = 37.142857..., so about 37.14

But in context, perhaps they want exact fraction or rounded.

I think 37.14 is acceptable, but let's see.

Since it's "amount expected", and rolls are discrete, but expected value can be fractional.

So 260/7 or 37.14

I'll write 37.14 for now.

But let's confirm calculation: 13/70 * 200 = 13 * 200 / 70 = 13 * 20 / 7 = 260 / 7 ≈ 37.1429

Yes.

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Problem 5:

Student rolls biased die. Table:

| Number | Frequency | Relative Frequency |
|--------|-----------|---------------------|
| 1 | ? | ? |
| 2 | ? | 0.25 |
| 3 | 2 | ? |
| 4 | ? | 0.2 |
| 5 | 4 | 0.1 |
| 6 | 12 | ? |

We need to find missing values.

First, from number 5: frequency 4, relative frequency 0.1 → so total rolls = frequency / relative frequency = 4 / 0.1 = 40

Check with others.

Number 6: frequency 12, so relative frequency = 12/40 = 0.3

Number 2: relative frequency 0.25, so frequency = 0.25 * 40 = 10

Number 4: relative frequency 0.2, so frequency = 0.2 * 40 = 8

Number 3: frequency 2, so relative frequency = 2/40 = 0.05

Number 1: total frequency should be 40.

Sum of known frequencies: num2:10, num3:2, num4:8, num5:4, num6:12 → 10+2=12; +8=20; +4=24; +12=36

So num1 frequency = 40 - 36 = 4

Relative frequency for num1 = 4/40 = 0.1

Now check relative frequencies sum: num1:0.1, num2:0.25, num3:0.05, num4:0.2, num5:0.1, num6:0.3 → 0.1+0.25=0.35; +0.05=0.4; +0.2=0.6; +0.1=0.7; +0.3=1.0 → good.

So completed table:

| Number | Frequency | Relative Frequency |
|--------|-----------|---------------------|
| 1 | 4 | 0.1 |
| 2 | 10 | 0.25 |
| 3 | 2 | 0.05 |
| 4 | 8 | 0.2 |
| 5 | 4 | 0.1 |
| 6 | 12 | 0.3 |

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Problem 6:

Tom and Sally flip biased coin.

Tom says they flipped 36 more heads than tails.

Sally says they flipped heads 4 times as much as tails.

Let T = number of tails, H = number of heads.

From Tom: H = T + 36

From Sally: H = 4T

Set equal: 4T = T + 36 → 3T = 36 → T = 12

Then H = 4*12 = 48

Total flips = H + T = 48 + 12 = 60

a) Relative frequency of flipping head = H / total = 48 / 60 = 0.8

b) Total flips = 60

✔ Check: H - T = 48 - 12 = 36, yes. H = 4*T = 48, yes.

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Problem 7:

Neda played game numerous times. Initial relative frequency of winning is 0.6.

She plays one additional game and wins. New relative frequency is 0.625.

How many games including recent win?

Let n = number of games before the last one.

Wins before last game: 0.6n

After winning one more, total wins = 0.6n + 1

Total games = n + 1

New relative frequency = (0.6n + 1) / (n + 1) = 0.625

So equation:

(0.6n + 1) / (n + 1) = 0.625

Multiply both sides by (n + 1):

0.6n + 1 = 0.625(n + 1)

Expand right side: 0.625n + 0.625

So:

0.6n + 1 = 0.625n + 0.625

Bring like terms together:

1 - 0.625 = 0.625n - 0.6n

0.375 = 0.025n

So n = 0.375 / 0.025 = 375 / 25 = 15

Calculate: 0.375 ÷ 0.025

0.025 * 15 = 0.375, yes.

So n = 15 games before.

Total games including recent win = 15 + 1 = 16

Check: initial wins = 0.6 * 15 = 9

After win: wins = 10, total games = 16, relative frequency = 10/16 = 5/8 = 0.625, yes.

Perfect.

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Now, compiling all final answers.

For Problem 1: 41 heads

Problem 2: B:18, G:18, Y:12

Problem 3a: No, Ben is not correct. (Explanation needed, but since options are false, just answer)

But the user said ExplainAnswer:false, etc., so only final answers.

But for multiple parts, need to specify.

Similarly, Problem 3b: relative frequencies as calculated.

But since it's a table, perhaps list them.

To save space, I'll summarize.

Final Answers:

1. 41

2. B: 18, G: 18, Y: 12

3a. No

3b. Relative frequencies: 1:0.24, 2:0.16, 3:0.02, 4:0.30, 5:0.10, 6:0.12 (using total 50)

But earlier I was unsure about total. Given the problem says 50 rolls, I'll stick with that.

3c. 46.4

4a. Gracie

4b. 260/7 or approximately 37.14, but let's write as fraction or decimal? Since others are decimals, perhaps 37.14, but better exact.

260/7 is exact, but maybe they want decimal.

In context, I'll put 37.14, but note it's approximate.

Actually, 260 ÷ 7 = 37.142857..., so perhaps 37.1 or 37.14.

But let's see the other problems; they have decimals.

I'll put 37.14

5. Completed table as above.

But for final answer, perhaps list the missing values.

Since it's "complete the table", but for answer, maybe state the values.

To be concise:

For 5: Freq:1:4,2:10,4:8; RelFreq:1:0.1,3:0.05,6:0.3

But perhaps not necessary to list all.

Since the question is to complete, but for final answer box, maybe skip or summarize.

I think for clarity, I'll provide the answers per part.

6a. 0.8

6b. 60

7. 16

Now for Problem 3a, it's explanation, but since ExplainAnswer is false, perhaps just say "No" or include brief.

The instruction is to provide final answer after reasoning, and for multiple parts, list them.

Also, for Problem 3b, the relative frequencies.

But to avoid clutter, I'll list numerical answers.

Let me organize:

Final Answer:

1. 41

2. B: 18, G: 18, Y: 12

3a. No

3b. Relative frequencies: 0.24, 0.16, 0.02, 0.30, 0.10, 0.12 for numbers 1 to 6 respectively

3c. 46.4

4a. Gracie

4b. \frac{260}{7} or 37.14 (but let's use fraction for accuracy)

In school, often decimal is fine, but 260/7 is exact.

I'll put 37.14, assuming rounding.

But 260/7 = 37.142857..., so to two decimals, 37.14

5. For the table:
- Number 1: Frequency 4, Relative Frequency 0.1
- Number 2: Frequency 10, Relative Frequency 0.25 (given)
- Number 3: Frequency 2 (given), Relative Frequency 0.05
- Number 4: Frequency 8, Relative Frequency 0.2 (given)
- Number 5: Frequency 4 (given), Relative Frequency 0.1 (given)
- Number 6: Frequency 12 (given), Relative Frequency 0.3

But since some are given, perhaps only missing ones.

For final answer, maybe state the completed values.

To simplify, since it's a bit messy, and the user might expect key answers.

6a. 0.8

6b. 60

7. 16

For Problem 4b, I'll put 37.14

But let's confirm if it should be integer. Expected value can be decimal.

Similarly for 3c, 46.4

Now for Problem 3a, since it's "explain", but ExplainAnswer is false, so perhaps just "No" is sufficient.

I think that's ok.

So compiling:

Final Answer:
1. 41
2. B: 18, G: 18, Y: 12
3a. No
3b. 0.24, 0.16, 0.02, 0.30, 0.10, 0.12
3c. 46.4
4a. Gracie
4b. 37.14
5. Frequencies: 1:4, 2:10, 4:8; Relative Frequencies: 1:0.1, 3:0.05, 6:0.3
6a. 0.8
6b. 60
7. 16

For 5, perhaps list the missing relative frequencies or something.

Since the table has blanks, but in answer, maybe specify.

To make it clean, for 5, the relative frequency for number 3 is 0.05, for number 6 is 0.3, and frequencies for 1,2,4 are 4,10,8 respectively.

But I think it's understood.

Now, I think this is comprehensive.

One thing: in Problem 3b, I used total 50, but frequencies sum to 47, which is inconsistent. But as per problem statement, I'll keep it.

Perhaps the problem intends total 50, and frequencies are correct, so relative frequencies are as calculated.

So I'll proceed.

Final Answer below.