Probability of combined events Revision Notes

A sample space diagram lists all possible outcomes of an experiment, making it easier to identify the total number of outcomes and the favourable ones for combined events.

For example, when tossing two coins, the sample space is {HH, HT, TH, TT}. To find the probability of getting at least one head, count outcomes with one or more heads: HH, HT, TH (3 outcomes). The probability is therefore 3/4.

Sample space diagrams are especially useful when events are simple and outcomes can be listed exhaustively.

• List all possible outcomes clearly.
• Identify favourable outcomes for combined events.
• Calculate probability as favourable outcomes divided by total outcomes.

Venn diagrams visually represent events as overlapping circles within a sample space. They help calculate probabilities involving intersections (both events occur) and unions (at least one event occurs).

Notation: P(A ∩ B) is the probability that both events A and B occur; P(A ∪ B) is the probability that event A or event B or both occur.

For example, if P(A) = 0.5, P(B) = 0.4, and P(A ∩ B) = 0.2, then P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.5 + 0.4 − 0.2 = 0.7. This formula avoids double counting the intersection when adding probabilities.

• Draw overlapping circles for events A and B.
• Label probabilities inside each region.
• Use P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to find union probabilities.
• Use P(A ∩ B) to find intersection probabilities.

Tree diagrams show all possible outcomes of sequential events, with branches representing each possible outcome and probabilities written alongside.

Each path from the start to an end node represents a sequence of outcomes. Multiply probabilities along branches to find the probability of that sequence.

For example, drawing a card from a deck, replacing it, then drawing again: the probability of drawing an ace then a king is (4/52) × (4/52) = 1/169. The tree diagram has two levels, each with branches for possible outcomes and their probabilities.

• Draw branches for each possible outcome at each stage.
• Write probabilities on branches.
• Multiply probabilities along a path to find combined probability.
• Sum probabilities of relevant paths for combined events.

Replacement affects probabilities in sequential events. With replacement, the sample space remains the same for each event; without replacement, the sample space changes after each event.

For example, drawing two cards without replacement: probability first card is an ace is 4/52; probability second card is a king is 4/51 (since one card was removed). The combined probability is (4/52) × (4/51) = 16/2652 ≈ 0.00604.

Tree diagrams must reflect these changes by adjusting probabilities on branches accordingly.

• Identify if events are independent (with replacement) or dependent (without replacement).
• Adjust probabilities for each stage accordingly.
• Use tree diagrams to visualise and calculate combined probabilities.
• Multiply probabilities along branches for sequences.

Suppose a bag contains 3 red and 2 blue balls. Two balls are drawn one after the other without replacement. Find the probability that both balls are red.

Step 1: Total balls initially = 5. Probability first ball is red = 3/5.

Step 2: After removing one red ball, remaining balls = 4, red balls = 2. Probability second ball is red = 2/4 = 1/2 = 0.5. Step 3: Multiply probabilities: (3/5) × (1/2) = 3/10 = 0.3. Step 4: Conclusion: The probability both balls drawn are red is 0.3 or 30%. This can be represented on a tree diagram with branches labelled with these probabilities.

• Calculate probability at each stage considering replacement or not.
• Multiply probabilities along the sequence.
• Check that probabilities sum to 1 across all possible outcomes.