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AFCAT Numerical Ability · Probability

Probability Basics: Equally Likely Outcomes and Three Rules

When every outcome is equally likely, probability is favourable outcomes over total outcomes; three rules (not, or, and) build new probabilities from known ones.

Why this matters

Every question in this chapter is this formula applied to something: a coin, a die, a deck or a bag. The questions on this page pick one number, ticket or person from a set, or use the three rules directly. Marks are lost on the count itself, such as forgetting that 1 is a perfect square or that a range includes both ends.

Concept 1 of 2: Favourable over total for one pick

Pick one ticket from 20 tickets numbered 1 to 20. Every ticket is equally likely, so the chance of a multiple of 7 is the number of such tickets, 2 (7 and 14), out of 20: 220=110\tfrac{2}{20} = \tfrac{1}{10}.

Definition

  • The sample space is the list of every outcome that can happen. An event is the set of outcomes you want, the favourable ones.
  • When every outcome is equally likely, P(E)=favourable outcomestotal outcomesP(E) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}.
  • A probability lies between 0 and 1: 0 for an impossible event, 1 for a certain one.
  • Count with care: 1 is neither prime nor composite, 1 is a perfect square (1=121 = 1^2), and 'from 1 to 40' includes both ends.
  • Reduce the fraction at the end, but look at the options first: they may print the unreduced form.

Classical probability

P(E)=n(E)n(S)P(E) = \dfrac{n(E)}{n(S)}

Worked example

A number is chosen at random from 1 to 40. Find the probability that it is a multiple of 6.
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The same idea in a practice question:

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Example 1 · Probability · Probability Basics and Rules

A number is chosen at random from the set {2, 3, 4, 5, 6, 7, 8, 9, 10}. What is the probability that it is prime?

1 is a perfect square

From 1 to 30 the perfect squares are 1, 4, 9, 16 and 25: five numbers, so 530=16\tfrac{5}{30} = \tfrac{1}{6}. Starting the list at 4 gives 430\tfrac{4}{30}.

Read which ends are included

'From 1 to 40' has 40 numbers. 'Between 1 and 40' can leave out both ends, which leaves 38. The total changes, so the answer does too.

Concept 2 of 2: Three rules: not, or for exclusive events, and for independent events

Most probabilities are built from simpler ones. The chance that something does not happen is what is left of 1. Two events that cannot happen together add. Two events that do not affect each other multiply.

Definition

  • Complement: P(not A)=1−P(A)P(\text{not } A) = 1 - P(A).
  • Mutually exclusive events cannot happen together, so P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B).
  • If they can happen together, take the overlap off once: P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B). Cards use this often, later in the chapter.
  • Independent events do not affect each other, so P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B).
  • For independent events, the chance that neither happens is (1−P(A))(1−P(B))(1 - P(A))(1 - P(B)), and at least one happens is 1 minus that.

The three rules

P(not A)=1−P(A);P(A or B)=P(A)+P(B) (exclusive);P(A and B)=P(A) P(B) (independent)P(\text{not } A) = 1 - P(A); \quad P(A \text{ or } B) = P(A) + P(B) \ (\text{exclusive}); \quad P(A \text{ and } B) = P(A)\,P(B) \ (\text{independent})

Worked example

Two independent events have P(A)=0.3P(A) = 0.3 and P(B)=0.8P(B) = 0.8. Find the probability that both happen, that neither happens, and that at least one happens.
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The same idea in a practice question:

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Example 2 · Probability · Probability Basics and Rules

Two events A and B have probabilities P(A) = 0.6 and P(B) = 0.7. If they are independent, what is P(A and B)?

Adding when you should multiply

For independent events with probabilities 0.4 and 0.9, 'both' is 0.4×0.9=0.360.4 \times 0.9 = 0.36. Adding gives 1.3, which is more than 1 and cannot be a probability.

Exclusive is not the same as independent

Mutually exclusive events cannot both happen, so P(A and B)=0P(A \text{ and } B) = 0. Independent events can happen together. Multiplying the probabilities of exclusive events gives a number where the answer is 0.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Favourable over total for one pick

    Classical probability

    P(E)=n(E)n(S)P(E) = \dfrac{n(E)}{n(S)}
  • Three rules: not, or for exclusive events, and for independent events

    The three rules

    P(not A)=1−P(A);P(A or B)=P(A)+P(B) (exclusive);P(A and B)=P(A) P(B) (independent)P(\text{not } A) = 1 - P(A); \quad P(A \text{ or } B) = P(A) + P(B) \ (\text{exclusive}); \quad P(A \text{ and } B) = P(A)\,P(B) \ (\text{independent})

Watch out for (4)

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