Sample space worksheet | KS3-4 maths | Teachit - Free Printable

Let's solve each problem step by step, explaining the reasoning and calculations clearly.

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Problem 1: Tom’s Pizza Toppings

Tom can choose two extra toppings from:
- Ham (H)
- Mushroom (M)
- Pineapple (P)
- Sweetcorn (S)
- Cheese (C)

The sample space is given as a table of all possible combinations (including repeats), where:
- The first letter is the first topping.
- The second letter is the second topping.

So the total number of outcomes = 5 × 5 = 25 (since repetition is allowed and order matters in the table).

#### a. Probability that Tom has mushrooms on his pizza

We want the probability that mushrooms appear at least once in the two toppings.

Let’s count how many combinations include M (either first or second or both):

Look at the table:

- First topping is M: MH, MM, MP, MS, MC → 5 outcomes
- Second topping is M: HM, MM, PM, SM, CM → 5 outcomes

But MM appears in both lists, so we must avoid double-counting.

So total with at least one M = 5 + 5 – 1 = 9

Alternatively, list them:
- MH, MM, MP, MS, MC
- HM, PM, SM, CM

Wait — let’s list all unique ones:
- MH, MM, MP, MS, MC
- HM, PM, SM, CM

That’s 9 distinct outcomes with M.

✔ So favorable outcomes = 9

Total outcomes = 25

So,
$$
P(\text{mushrooms}) = \frac{9}{25}
$$

> ✔ Answer to (a): $ \boxed{\frac{9}{25}} $

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#### b. Probability he chooses two of the same topping

These are the diagonal entries: HH, MM, PP, SS, CC → 5 outcomes

So,
$$
P(\text{same topping}) = \frac{5}{25} = \frac{1}{5}
$$

> ✔ Answer to (b): $ \boxed{\frac{1}{5}} $

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Problem 2: Mary rolls a die and tosses a coin

#### a. Draw a sample space diagram

A die has 6 outcomes: {1, 2, 3, 4, 5, 6}
A coin has 2 outcomes: {H, T}

So total outcomes = 6 × 2 = 12

Sample space (list of all outcomes):

| Die | Coin | Outcome |
|-----|------|--------|
| 1 | H | (1,H) |
| 1 | T | (1,T) |
| 2 | H | (2,H) |
| 2 | T | (2,T) |
| 3 | H | (3,H) |
| 3 | T | (3,T) |
| 4 | H | (4,H) |
| 4 | T | (4,T) |
| 5 | H | (5,H) |
| 5 | T | (5,T) |
| 6 | H | (6,H) |
| 6 | T | (6,T) |

You could also draw a grid with die on one axis and coin on the other.

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#### b. Conditions
- She goes for a run if she rolls a 4
- She has chips for tea if she gets a tail

We want the probability that:
- She goes for a run AND has chips for tea

This happens only if:
- Die shows 4 AND Coin shows Tail

From the sample space, this is just (4,T)

Only 1 outcome satisfies both conditions.

Total outcomes = 12

So,
$$
P(\text{run and chips}) = \frac{1}{12}
$$

> ✔ Answer to (c): $ \boxed{\frac{1}{12}} $

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Problem 3: Puppies (3 male, 2 female)

Let’s denote:
- M = Male
- F = Female

There are 5 puppies: M, M, M, F, F

A vet picks one puppy at random for dental, then another (possibly the same one) for blood test.

So with replacement? Yes — "it could be the same puppy twice" → independent selections.

So total possible outcomes = 5 × 5 = 25 (since each pick is from 5 puppies, and they can repeat)

But since we only care about sex, not which individual puppy, we can model this based on sex.

Let’s define the sample space based on gender of the two selected puppies.

Each selection has:
- P(Male) = 3/5
- P(Female) = 2/5

And selections are independent.

#### a. Sample space diagram

We can represent the outcomes as ordered pairs:

- (M, M)
- (M, F)
- (F, M)
- (F, F)

Now, probabilities:

- P(M,M) = (3/5)(3/5) = 9/25
- P(M,F) = (3/5)(2/5) = 6/25
- P(F,M) = (2/5)(3/5) = 6/25
- P(F,F) = (2/5)(2/5) = 4/25

You could draw a tree diagram or a grid.

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#### b. Probability both puppies are of the same sex

Same sex means:
- Both male: (M,M) → 9/25
- Both female: (F,F) → 4/25

So total:
$$
P(\text{same sex}) = \frac{9}{25} + \frac{4}{25} = \frac{13}{25}
$$

> ✔ Answer to (b): $ \boxed{\frac{13}{25}} $

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✔ Final Answers Summary

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Problem 1:
- a) $ \boxed{\frac{9}{25}} $
- b) $ \boxed{\frac{1}{5}} $

Problem 2:
- a) Sample space: 12 outcomes (all combinations of die roll 1–6 and coin H/T)
- c) $ \boxed{\frac{1}{12}} $

Problem 3:
- a) Sample space: (M,M), (M,F), (F,M), (F,F) with probabilities 9/25, 6/25, 6/25, 4/25
- b) $ \boxed{\frac{13}{25}} $

Let me know if you'd like diagrams drawn or formatted!