Learning Objectives
  • Construct and interpret sample-space diagrams and tree diagrams.
  • Calculate probabilities of mutually exclusive and overlapping events.
  • Use multiplication along branches and addition across appropriate outcomes.
  • Distinguish between experiments with replacement and without replacement.
  • Use complements, Venn diagrams and expected frequencies in multi-stage problems.
Key Terms
Outcome
A possible result of an experiment.
Event
A set of one or more outcomes.
Sample space
The complete set of possible outcomes.
Mutually exclusive
Events that cannot occur together.
Independent
Events for which one result does not change the probability of the other.
Conditional probability
A probability that changes because earlier information is known.
With replacement
The selected item is returned, so probabilities remain unchanged.
Without replacement
The selected item is not returned, so totals and probabilities change.
What This Chapter Covers
  • Combined events with sample-space diagrams
  • Addition and multiplication rules in context
  • Tree diagrams with and without replacement
  • Probability from Venn diagrams
  • Complementary events and expected frequency
Probability Language

A probability lies between 0 and 1. A probability of 0 means impossible, while 1 means certain. For equally likely outcomes, probability is the number of favourable outcomes divided by the total number of possible outcomes. In combined-event problems, the main challenge is organising outcomes without missing or double-counting them.

P(A\prime)=1-P(A)
P(A\cup B)=P(A)+P(B)-P(A\cap B)

For mutually exclusive events, P(A\cap B)=0.

Sample-Space Diagrams

A sample-space diagram is useful when two experiments have a manageable number of outcomes, such as rolling two dice, spinning two spinners or choosing one item from each of two groups. Each cell represents one ordered pair. The diagram must include all possibilities.

Worked Example: Two Dice

Question: Two fair six-sided dice are rolled. Find the probability that the total is 9.

  1. There are 6\times6=36 equally likely ordered outcomes.
  2. The outcomes with total 9 are (3,6), (4,5), (5,4) and (6,3).
  3. There are 4 favourable outcomes.

Answer: \frac4{36}=\frac19.

Worked Example: At Least One

Question: Two fair coins are tossed. Find the probability of at least one head.

  1. List the sample space: HH, HT, TH, TT.
  2. Three outcomes contain at least one head.
  3. Alternatively use the complement: 1-P(\text{no heads})=1-\frac14.

Answer: \frac34.

Tree Diagrams

A tree diagram shows successive stages. Probabilities are written on branches and outcomes at the ends. At every set of branches leaving the same point, the probabilities must add to 1. To find the probability of one complete route, multiply along the branches. To combine different routes that satisfy the event, add the route probabilities.

Tree-Diagram Method
  1. Draw one branch for every possible outcome at the first stage.
  2. From each first-stage outcome, draw the second-stage branches.
  3. Write the correct probability beside every branch.
  4. Multiply probabilities along each required route.
  5. Add the probabilities of separate required routes.
  6. Check that the probabilities of all final routes add to 1.
Worked Example: Independent Events

Question: A biased coin has P(H)=0.6. It is tossed twice. Find the probability of exactly one head.

  1. The two routes are HT and TH.
  2. P(HT)=0.6\times0.4=0.24.
  3. P(TH)=0.4\times0.6=0.24.
  4. Add the routes.

Answer: 0.48.

With Replacement And Without Replacement

Replacement determines whether probabilities remain constant. If an item is replaced, the composition of the container returns to its original state. If it is not replaced, both the numerator and denominator may change after the first selection.

Worked Example: Without Replacement

Question: A bag contains 5 red and 3 blue counters. Two counters are chosen without replacement. Find the probability that both are red.

  1. First red: \frac58.
  2. After one red is removed, 4 red remain out of 7 counters.
  3. Second red: \frac47.
  4. Multiply along the RR route.

Answer: \frac58\times\frac47=\frac5{14}.

Worked Example: One Of Each

Question: Using the same bag, find the probability of one red and one blue in any order.

  1. Route RB: \frac58\times\frac37=\frac{15}{56}.
  2. Route BR: \frac38\times\frac57=\frac{15}{56}.
  3. Add the two mutually exclusive routes.

Answer: \frac{30}{56}=\frac{15}{28}.

Independent And Dependent Events

Independent events satisfy P(A\cap B)=P(A)P(B). Replacement often creates independence, but independence should be judged from whether the first result changes the second probability. Without replacement usually makes the events dependent.

Situation Effect on second-stage probability
Coin tossed twice Unchanged; tosses are independent
Counter selected and replaced Unchanged; composition restored
Counter selected without replacement Changed; totals and composition alter
Weather on two successive days Cannot automatically assume independence without information
“At Least”, “At Most” And Complements

“At least one” is often easiest by subtracting the probability of none from 1. “At most one” means zero or one. “Exactly two” means two and no more. Translate the words before calculating.

Worked Example: Complement Strategy

Question: A machine produces a defective item with probability 0.08. Three items are chosen independently. Find the probability that at least one is defective.

  1. Probability one item is not defective =0.92.
  2. Probability none of the three is defective =0.92^3.
  3. Use the complement.

Answer: 1-0.92^3=0.221312\approx0.221.

Probability From Venn Diagrams

When a Venn diagram shows frequencies, divide the required frequency by the universal total. For P(A\cup B), include both single regions and the overlap. For P(A\cap B), use the overlap only.

Worked Example: Venn Probability

Question: A survey of 80 people shows 35 in A only, 18 in B only, 12 in both and 15 in neither. Find P(A\prime\cap B) and P(A\cup B).

  1. A\prime\cap B is B only, frequency 18.
  2. A\cup B includes 35, 18 and 12, total 65.
  3. Divide each frequency by 80.

Answer: P(A\prime\cap B)=\frac{9}{40}, P(A\cup B)=\frac{13}{16}.

Expected Frequency

If an event has probability p and an experiment is repeated n times, the expected frequency is np. It is a long-run estimate, not a guarantee.

\text{expected frequency}=\text{number of trials}\times\text{probability}
Worked Example: Expected Frequency

Question: A spinner lands on green with probability 0.35. Estimate the number of green results in 240 spins.

  1. Multiply the number of trials by the probability.
  2. 240\times0.35=84.

Answer: 84 green results.

Examination Guidance
  • Write probabilities on branches, not at the ends of branches.
  • Multiply along one route and add separate routes.
  • For without-replacement problems, update both the numerator and denominator.
  • Use a complement for “at least one” when this reduces the number of routes.
  • Give exact fractions where convenient and do not round intermediate probabilities.
Common Mistakes
  • Adding probabilities along a single route instead of multiplying.
  • Forgetting the reverse order when “one of each” is required.
  • Keeping the denominator unchanged in a without-replacement problem.
  • Assuming events are independent merely because there are two stages.
  • Confusing P(A\cup B) with P(A\cap B).
Knowledge Check And Practice

1. A fair die is rolled twice. Find the probability of two sixes.

Answer: \frac16\times\frac16=\frac1{36}.

2. A fair coin is tossed three times. Find the probability of no heads.

Answer: (\frac12)^3=\frac18.

3. A bag has 4 white and 6 black balls. Two are selected without replacement. Find the probability both are white.

Answer: \frac4{10}\times\frac3{9}=\frac2{15}.

4. If P(A)=0.7, find P(A\prime).

Answer: 0.3.

5. If P(A)=0.5, P(B)=0.4 and P(A\cap B)=0.2, find P(A\cup B).

Answer: 0.7.

6. An event has probability 0.12. Find its expected frequency in 500 trials.

Answer: 60.
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