Probability: Combined Events

Grade 7 — Cambridge Lower Secondary Stage 7, Unit 13

The Key Rules

Mutually exclusive events cannot happen at the same time.

P(A or B) = P(A) + P(B)

Exhaustive events cover every possible outcome:

P(A) + P(not A) = 1  →  P(not A) = 1 − P(A)

Combined events (two-stage): multiply along tree branches.

P(A and B) = P(A) × P(B)

Probability Scale

0
Impossible 0.5
Even 1
Certain

Probabilities are always between 0 and 1 (or 0% and 100%).

Animated Tree Diagram: Flipping Two Coins

Watch the tree build itself — each branch appears one at a time.

Start H ½ Heads T ½ Tails H ½ HH = ¼ T ½ HT = ¼ H ½ TH = ¼ T ½ TT = ¼

Reading the Tree

Each branch shows one outcome and its probability.

Follow a path from root to end: multiply the probabilities along the way.

P(HH) = ½ × ½ = ¼

All end probabilities must add up to 1: ¼+¼+¼+¼ = 1 ✓

Sample Space: Two Dice

A sample space lists all possible outcomes. For two dice, use a grid.

Die 1 \ Die 2 123456
1(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
2(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
3(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
4(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
5(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
6(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

36 outcomes total. P(sum = 7) = 6/36 = 1/6

Experimental vs Theoretical Probability

Theoretical probability: what we expect mathematically. P(H) = 1/2

Experimental probability (relative frequency): what actually happens when you run an experiment.

Relative frequency = Number of times event occurs ÷ Total trials

As the number of trials increases, experimental probability gets closer to theoretical probability.

Worked Examples

Study these carefully before the exercises!

Example 1: Mutually Exclusive Events — Bag of Balls

Question: A bag contains red, blue, yellow and green balls. P(red) = 1/5, P(blue) = 3/10, P(yellow) = 1/4. Find P(red or blue) and P(not green).

Step 1: Check mutually exclusive — a ball can only be one colour, so yes.

Step 2: P(red or blue) = P(red) + P(blue) = 1/5 + 3/10 = 2/10 + 3/10 = 5/10 = 1/2

Step 3: All probabilities add to 1, so find P(green) first.
P(green) = 1 − P(red) − P(blue) − P(yellow) = 1 − 1/5 − 3/10 − 1/4

Common denominator 20: = 20/20 − 4/20 − 6/20 − 5/20 = 5/20 = 1/4

P(not green) = 1 − 1/4 = 3/4

Example 2: Sample Space Grid — Two Dice

Question: Two fair dice are rolled. Find P(sum > 9).

Step 1: Total outcomes = 6 × 6 = 36

Step 2: List outcomes with sum > 9:
(4,6),(5,5),(5,6),(6,4),(6,5),(6,6) — and (6,5),(5,6) already counted... let me list carefully:
Sum=10: (4,6),(5,5),(6,4) → 3 outcomes
Sum=11: (5,6),(6,5) → 2 outcomes
Sum=12: (6,6) → 1 outcome
Total favourable = 3+2+1 = 6

Step 3: P(sum > 9) = 6/36 = 1/6

Example 3: Tree Diagram — Two-Stage Event

Question: A bag has 3 red and 2 blue balls. Draw one, note colour, replace it, draw again. Find P(both same colour).

Step 1: P(R) = 3/5, P(B) = 2/5. With replacement, these stay the same.

Step 2: Tree paths:

  • R then R: 3/5 × 3/5 = 9/25
  • R then B: 3/5 × 2/5 = 6/25
  • B then R: 2/5 × 3/5 = 6/25
  • B then B: 2/5 × 2/5 = 4/25

Step 3: P(both same) = P(RR) + P(BB) = 9/25 + 4/25 = 13/25

Check: All four paths: 9/25+6/25+6/25+4/25 = 25/25 = 1 ✓

Example 4: Experimental vs Theoretical — Biased Coin

Question: A coin is flipped 200 times. Heads appears 130 times. Is the coin fair? What is the experimental probability of heads?

Step 1: Experimental P(H) = 130/200 = 0.65 (or 13/20)

Step 2: Theoretical P(H) for a fair coin = 0.5

Step 3: 0.65 ≠ 0.5, and with 200 trials this is a large sample, so the difference is significant.

Conclusion: The coin appears to be biased towards heads.

Note: More trials give a better estimate. With only 10 trials, getting 7 heads might just be chance.

Probability Simulator

Run experiments and watch experimental probability approach theoretical probability!

Choose an experiment:

Experimental Theoretical

Trials run: 0  | 

Ex 1: Mutually Exclusive Events

Use P(A or B) = P(A) + P(B). Type your answer as a fraction or decimal.

Ex 2: Complementary Events

Use P(not A) = 1 − P(A). Write answers as fractions or decimals.

Ex 3: Sample Space Grids

Click cells to highlight favourable outcomes, then type the probability as a fraction.

Problem 1: Two fair dice are rolled. Find P(sum = 5).

Click the cells that give a sum of 5, then type the probability:

P(sum = 5) =

Problem 2: Two fair dice are rolled. Find P(both numbers even).

Click the cells where both dice show even numbers:

P(both even) =

Problem 3: Two fair dice are rolled. Find P(at least one 6).

Click every cell where at least one die shows a 6:

P(at least one 6) =

Ex 4: Tree Diagram Builder

A bag has 3 Red and 2 Blue balls. Draw one ball, replace it, draw again.
Fill in all the branch probabilities and end probabilities.

Start R B R B R B RR = RB = BR = BB =

Hint: P(R) = 3/5, P(B) = 2/5. Multiply along branches for end probabilities.

Ex 5: Probability Challenge

Mixed problems. Enter each answer as a fraction (numerator / denominator).

Ex 6: Experimental vs Theoretical

A spinner with 4 equal sections (A, B, C, D) was spun 200 times. Here are the results:

OutcomeABCDTotal
Frequency62485535200
Rel. Freq.????1

Practice Questions

Work through these on paper. Show all working!

  1. A bag contains red, blue, and green balls. P(red) = 1/4, P(blue) = 1/3. Find P(red or blue).
  2. The probability of rain tomorrow is 0.35. What is the probability it does NOT rain?
  3. A fair die is rolled. Find P(even or greater than 4).
  4. P(A) = 3/8 and A and B are mutually exclusive. P(B) = 1/4. Find P(A or B).
  5. A spinner has 5 equal sections: 1, 2, 3, 4, 5. Find P(prime or even).
  6. Two fair coins are flipped. Using a sample space, find P(exactly one head).
  7. Two fair dice are rolled. Find P(sum = 8) using a sample space grid.
  8. Two fair dice are rolled. Find P(sum is a prime number).
  9. A bag has 4 red and 6 blue marbles. One is drawn, replaced, then another drawn. Find P(both red) using a tree diagram.
  10. Using the same bag as Q9, find P(one of each colour).
  11. A biased coin has P(H) = 0.6. It is flipped twice. Find P(both heads).
  12. A biased coin has P(H) = 0.6. It is flipped twice. Find P(at least one tail).
  13. A spinner is spun 50 times. It lands on red 18 times. What is the experimental P(red)?
  14. The theoretical P(red) for the spinner in Q13 is 1/4. Is the spinner likely biased? Explain.
  15. Events X and Y are mutually exclusive. P(X) = 0.3, P(Y) = 0.45. Find P(neither X nor Y).
  16. A card is drawn from a standard 52-card deck. Find P(heart or king).
  17. A number is chosen at random from 1 to 20. Find P(multiple of 3 or multiple of 5).
  18. In an experiment, a die is rolled 120 times. How many times would you expect to get a 6? If 6 appeared 25 times, calculate the relative frequency.
  19. Two bags each contain 3 red and 2 yellow beads. One bead is drawn from each bag. Draw a tree diagram and find P(same colour).
  20. A and B are exhaustive, mutually exclusive events. P(A) = 2x − 0.1 and P(B) = x + 0.4. Find x and hence P(A) and P(B).

Answers

  1. P(red or blue) = 1/4 + 1/3 = 3/12 + 4/12 = 7/12
  2. P(not rain) = 1 − 0.35 = 0.65
  3. Even: {2,4,6}; greater than 4: {5,6}; union (no overlap except 6): {2,4,5,6} = 4 outcomes. P = 4/6 = 2/3
  4. P(A or B) = 3/8 + 1/4 = 3/8 + 2/8 = 5/8
  5. Prime: {2,3,5}, Even: {2,4}. Union: {2,3,4,5} = 4 outcomes. P = 4/5
  6. Sample space: {HH, HT, TH, TT}. Exactly one head: {HT, TH} = 2. P = 2/4 = 1/2
  7. Sum=8: (2,6),(3,5),(4,4),(5,3),(6,2) = 5 outcomes. P = 5/36
  8. Prime sums (2,3,5,7,11): 1+2+4+6+2 = 15 outcomes. P = 15/36 = 5/12
  9. P(R)=4/10=2/5. P(RR) = 2/5 × 2/5 = 4/25
  10. P(RB) + P(BR) = (2/5)(3/5)+(3/5)(2/5) = 6/25+6/25 = 12/25
  11. P(HH) = 0.6 × 0.6 = 0.36
  12. P(at least one T) = 1 − P(HH) = 1 − 0.36 = 0.64
  13. Experimental P(red) = 18/50 = 9/25 = 0.36
  14. Theoretical = 0.25. Experimental = 0.36. Difference is noticeable. With 50 trials it could be chance, but the spinner may be slightly biased. More trials needed to be sure.
  15. P(X or Y) = 0.3+0.45=0.75. P(neither) = 1−0.75 = 0.25
  16. Hearts: 13, Kings: 4, King of hearts counted once. 13+4−1 = 16. P = 16/52 = 4/13
  17. Multiples of 3: {3,6,9,12,15,18}=6; Multiples of 5: {5,10,15,20}=4; Overlap {15}=1. 6+4−1=9. P = 9/20
  18. Expected 6s = 120 × 1/6 = 20. Relative frequency = 25/120 = 5/24 ≈ 0.208
  19. P(R)=3/5, P(Y)=2/5. P(same) = P(RR)+P(YY) = (3/5)(3/5)+(2/5)(2/5) = 9/25+4/25 = 13/25
  20. Since exhaustive and ME: P(A)+P(B)=1. (2x−0.1)+(x+0.4)=1 → 3x+0.3=1 → 3x=0.7 → x=7/30. P(A)=2(7/30)−0.1=14/30−3/30=11/30, P(B)=7/30+12/30=19/30

Challenge Problems

Hard problems — show all working. Some introduce new ideas!

  1. A bag contains 5 red, 3 blue, and 2 green balls. Two balls are drawn without replacement. Find P(both red). (Hint: probabilities change on second draw!)
  2. A biased spinner has 4 sections: P(1)=0.1, P(2)=0.3, P(3)=0.4, P(4)=x. Find x. The spinner is spun twice — find P(total score > 6).
  3. Three coins are flipped simultaneously. Using a systematic list of outcomes, find P(at least two heads).
  4. Two fair dice are rolled. Given that the sum is greater than 8, find the probability that the sum is exactly 10. (This is conditional probability.)
  5. In a class of 30 students, 18 play football, 12 play tennis, and 7 play both. A student is chosen at random. Find P(plays football but not tennis).
  6. A bag has r red and 4 blue balls. P(red) = 3/7. Find r. If two balls are drawn without replacement, find P(both blue).
  7. Events A and B are such that P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.2. Are A and B mutually exclusive? Find P(A or B).
  8. A factory tests components. P(faulty) = 0.04. Two components are tested. Find P(exactly one faulty). Draw a tree diagram.
  9. A game: roll two dice. Win £5 if the product of the dice is a perfect square; lose £1 otherwise. How many winning outcomes are there? Would you expect to win or lose money in the long run?
  10. Amira rolls a fair die. If she gets an odd number, she flips a coin. If she gets an even number, she rolls the die again. Draw the full tree diagram. Find P(she ends with a head OR a 6 on the second die roll).

Challenge Answers with Workings

  1. Without replacement: P(1st red) = 5/10 = 1/2. After drawing one red, 4 red remain from 9 balls. P(2nd red | 1st red) = 4/9.
    P(both red) = 1/2 × 4/9 = 4/18 = 2/9 ≈ 0.222
  2. P(1)+P(2)+P(3)+P(4)=1 → 0.1+0.3+0.4+x=1 → x=0.2.
    Total > 6 means sum = 7 or 8. Sum=7: (3,4),(4,3) = 0.4×0.2+0.2×0.4=0.08+0.08=0.16. Sum=8: (4,4)=0.2×0.2=0.04.
    P(total > 6) = 0.16+0.04 = 0.20
  3. 8 outcomes: HHH,HHT,HTH,HTT,THH,THT,TTH,TTT. At least 2 heads: HHH,HHT,HTH,THH = 4.
    P(at least 2 heads) = 4/8 = 1/2
  4. Outcomes with sum > 8: sum=9 (4 ways), sum=10 (3 ways), sum=11 (2 ways), sum=12 (1 way) = 10 total.
    Outcomes with sum=10: (4,6),(5,5),(6,4) = 3.
    P(sum=10 | sum>8) = 3/10 = 0.3
  5. Football only = 18−7=11. P(football but not tennis) = 11/30. P = 11/30 ≈ 0.367
  6. P(R) = r/(r+4) = 3/7 → 7r = 3(r+4) → 7r=3r+12 → 4r=12 → r=3. Bag has 3 red, 4 blue = 7 balls.
    P(both blue, no replacement) = 4/7 × 3/6 = 12/42 = 2/7
  7. P(A and B)=0.2 ≠ 0, so A and B are NOT mutually exclusive.
    P(A or B) = P(A)+P(B)−P(A and B) = 0.4+0.5−0.2 = 0.7
  8. Tree: Faulty (0.04) → F or NF; Not Faulty (0.96) → F or NF.
    P(exactly one faulty) = P(F,NF)+P(NF,F) = 0.04×0.96 + 0.96×0.04 = 0.0384+0.0384 = 0.0768
  9. Perfect square products from two dice (1×1=1, 1×4=4, 4×1=4, 2×2=4, 3×3=9, 4×4=16 — wait, max is 6×6=36).
    Products that are perfect squares: 1(1,1), 4(1,4),(2,2),(4,1), 9(3,3), 16(4,4),(2,8 no), 25(5,5),(1 only), 36(6,6).
    Valid: (1,1), (1,4),(2,2),(4,1), (3,3), (4,4 no—16 but dice go to 6), wait: 4×4=16 yes, (2,8) not possible, (4,4)=16 yes.
    Listing: (1,1)=1; (1,4)=4,(2,2)=4,(4,1)=4; (3,3)=9; (2,8) n/a; (5,5)=25; (6,6)=36; (4,4 wait 4×4=16 — 4 is on a die!) (4,4)=16.
    Win outcomes: (1,1),(1,4),(2,2),(4,1),(3,3),(4,4),(5,5),(6,6) = 8 outcomes.
    E(winnings per game) = 8/36 × £5 + 28/36 × (−£1) = 40/36 − 28/36 = 12/36 = £1/3 > 0.
    Expected to WIN money in the long run (expected gain ≈ £0.33 per game).
  10. Odd number (1,3,5): P=1/2 each. Even (2,4,6): P=1/2 each. On odd: flip coin (H or T, each 1/2). On even: roll die again (1-6 each 1/6).
    P(head) = P(odd) × P(H) = 1/2 × 1/2 = 1/4.
    P(6 on second roll) = P(even on first) × P(6 on second) = 1/2 × 1/6 = 1/12.
    These outcomes are mutually exclusive (one branch is coin, other is die), so:
    P(head OR 6) = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3
Created by Mr Josh