Probability and risk: reason about uncertain outcomes

1. Define the experiment and event

A random experiment has an outcome that is not known beforehand. Its sample space is the set of possible outcomes. For one ordinary six-faced die, a simple model uses 1, 2, 3, 4, 5 and 6. An event is a specified collection of outcomes, such as “roll an even number”, which includes 2, 4 and 6. Write the event precisely before calculating.

Write P(A) for the probability of an event named A. A probability of 0 represents impossibility in the model, and 1 represents certainty. Values between them express degrees of chance: 0.25 = 1/4 = 25%. A probability cannot be negative or greater than 1. These bounds provide a quick check, but a number within the bounds can still come from a faulty model.

Sources: OpenStax: Introductory Statistics 2e: Probability Terminology ↗

2. Counting works when elementary outcomes have equal chances

For equally likely outcomes, probability is the number meeting the event divided by the total number. A fair die gives P(even) = 3/6 = 1/2. Fairness is an assumption about the die and the rolling process. Merely listing six labels does not prove equal chances. A spinner divided into unequal sectors cannot be analysed by counting sector labels as though every sector has the same area.

For two fair independent coin tosses, label heads H and tails T. The four equally likely ordered outcomes are HH, HT, TH and TT. “One head and one tail” contains two outcomes, HT and TH, so its probability is 2/4 = 1/2. The three descriptions “two heads”, “one of each” and “two tails” are not equally likely groups. Preserve the underlying outcomes while grouping them.

Sources: NCERT Class X Mathematics Exemplar: Statistics and Probability ↗ · OpenStax: Introductory Statistics 2e: Probability Terminology ↗

3. Solved example: replacement changes the next draw

An illustrative bag contains 3 red counters and 2 blue counters, identical except for colour. Mix them and draw without looking, assuming each counter has an equal chance. The first red probability is 3/5. If a red counter is drawn and kept out, 2 red and 2 blue remain, so the next red probability is 2/4. The probability of two reds in sequence is (3/5) × (2/4) = 3/10.

If instead the first counter is replaced and the bag is mixed again, the second red probability returns to 3/5. Under the independent-draw model, two reds have probability (3/5) × (3/5) = 9/25. The values 0.30 and 0.36 differ because the experiment differs. Multiplying probabilities is appropriate when the second factor is the correct probability after the first event, not automatically the original probability.

Sources: OpenStax: Introductory Statistics 2e: Independent and Mutually Exclusive Events ↗ · OpenStax: Introductory Statistics 2e: Two Basic Rules of Probability ↗

4. Solved example: use the opposite event

An event and its complement cover every possible outcome without overlapping. The complement of “at least one head in two tosses” is “no heads”, which is TT. For fair independent tosses, P(TT) = 1/2 × 1/2 = 1/4. Therefore P(at least one head) = 1 − 1/4 = 3/4. Listing HH, HT and TH gives the same answer and checks the reasoning.

For three such tosses, no heads has probability 1/8, so at least one head has probability 7/8. “At least one” includes two or three heads; it is not the same as “exactly one”. More opportunities can raise a probability without making an event certain. Even with probability 7/8, all three tosses can still be tails.

Sources: OpenStax: Introductory Statistics 2e: Probability Terminology ↗ · OpenStax: Introductory Statistics 2e: Two Basic Rules of Probability ↗

5. Avoid double counting when combining events

On one fair die, let A mean an even result and B mean a multiple of 3. A contains 2, 4, 6 and B contains 3, 6. “A or B” means at least one condition is met, including both. Adding 3/6 and 2/6 counts the 6 twice. Subtract that overlap once: 3/6 + 2/6 − 1/6 = 4/6 = 2/3. The qualifying results are 2, 3, 4 and 6.

Mutually exclusive events cannot happen together in the same trial, such as rolling 1 and rolling 2 on one die. Independent events are different: knowing one occurred does not change the other's probability. Two separate fair coin tosses can be independent, while heads and tails on the same toss are mutually exclusive. Do not use these terms interchangeably.

Sources: OpenStax: Introductory Statistics 2e: Independent and Mutually Exclusive Events ↗ · OpenStax: Introductory Statistics 2e: Two Basic Rules of Probability ↗

6. Compare a model with observed frequencies

Experimental probability estimates chance from repeated observations. If red appears 17 times in 40 replacement draws, the observed proportion is 17/40 = 0.425. It need not exactly equal the theoretical 0.4 for a bag with four red counters out of ten. Small samples vary. More trials can make a well-run estimate more stable, but they do not fix biased drawing or counters that are distinguishable by touch.

A sequence does not force compensation. After several tails, the next independent fair toss still has head probability 1/2. Over many trials, the proportion can move closer to half without the counts of heads and tails becoming exactly equal. Record all planned trials rather than stopping as soon as the results look convenient.

Sources: OpenStax: Introductory Statistics 2e: Probability Terminology ↗

7. Combine likelihood with consequence carefully

In an illustrative classroom model, a spill has probability 0.10 per session and damages 20 paper sheets if it occurs; otherwise no sheets are damaged. The expected damage is 0.10 × 20 + 0.90 × 0 = 2 sheets per session. This is an average over repeated comparable sessions, not a prediction that exactly two sheets will be damaged in the next session.

A cover that reduces the assumed spill-damage probability to 0.02 gives expected damage 0.4 sheet, saving 1.6 sheets per session on average. Fractional expected counts are meaningful averages even though damaged sheets are whole objects. Decisions may also depend on the worst possible outcome, the reliability of the probability estimate and the cost of prevention. Expected value is one comparison tool, not a complete decision rule.

Sources: OpenStax: Introductory Statistics 2e: Mean or Expected Value and Standard Deviation ↗

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