IP EMaths Notes (Upper Sec, Year 3-4): 14) Probability

Probability questions reward tidy organisation. Label events clearly, state whether they are mutually exclusive or independent, and keep notation consistent across tables and tree diagrams.

Quick reference

• \( P(A) \) is the probability of event A; \( P(A') = 1 - P(A) \) for the complement.
• Addition (mutually exclusive events): \( P(A \cup B) = P(A) + P(B) \).
• General addition: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
• Multiplication (independent events): \( P(A \cap B) = P(A) \times P(B) \).
• Conditional probability: \( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} \), provided \( P(B) > 0 \).
• When outcomes repeat, use tables or tree diagrams so that multiplication along a path and addition across paths are obvious.

Organising your work

• Frequency tables: when counts are given, arrange them in a two-way table, sum rows and columns, then divide by the total to convert to probabilities.
• Tree diagrams: useful for sequential events (with or without replacement). Multiply the branch probabilities, then add the relevant paths.
• Complement strategy: for “at least” questions, consider the complement (e.g. “no successes”) and subtract from 1.

Mutually exclusive vs independent

• Mutually exclusive events cannot occur together (their intersection is zero). For example, drawing a heart and drawing a spade on a single card draw are mutually exclusive.
• Independent events do not influence each other. Tossing a coin and rolling a die are independent because the outcome of one does not change the distribution of the other.
• Mutually exclusive events are not independent unless one probability is zero; removing a red marble from a bag changes the probability of drawing another red marble without replacement.

Conditional probability workflow

1. Identify the conditioning event (e.g. “given that the first die shows an even number”).
2. Restrict the sample space to outcomes that satisfy that event.
3. Count or compute the favourable outcomes within the restricted space.
4. Divide by the number of outcomes in the restricted space. Use \( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} \) as a consistency check.

Worked example - two draws without replacement

A bag contains five red, three blue and two green counters. Two counters are drawn at random without replacement. What is the probability that both counters are the same colour?

1. Total counters: \( 10 \).
2. Two reds: \( \dfrac{5}{10} \times \dfrac{4}{9} = \dfrac{20}{90} \).
3. Two blues: \( \dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{6}{90} \).
4. Two greens: \( \dfrac{2}{10} \times \dfrac{1}{9} = \dfrac{2}{90} \).
5. Add the mutually exclusive outcomes: \( \dfrac{20}{90} + \dfrac{6}{90} + \dfrac{2}{90} = \dfrac{28}{90} = \dfrac{14}{45} \).

So both counters share a colour with probability \( \dfrac{14}{45} \).

Worked example - conditional probability with a table

A survey of 120 students records whether they take Biology (B) and/or Chemistry (C). The results are summarised below.

A student is chosen at random. Find P(B | C) and P(C | B).

• \( P(C) = \dfrac{63}{120} \). The intersection \( B \cap C \) has 38 students, so \( P(B \mid C) = \dfrac{38}{120} \div \dfrac{63}{120} = \dfrac{38}{63} \).
• \( P(B) = \dfrac{60}{120} \). Therefore \( P(C \mid B) = \dfrac{38}{120} \div \dfrac{60}{120} = \dfrac{38}{60} = \dfrac{19}{30} \).

Always mention which total formed the denominator when quoting conditional probabilities.

Worked example - tree diagram with conditional branches

A factory machine produces 70% standard widgets (S) and 30% premium widgets (P). Among standard widgets, 4% are defective; among premium widgets, 9% are defective. A widget is selected at random.

1. Draw a tree with two stages: choose S or P, then good or defective.
2. Probability of a defective widget: \( (0.7 \times 0.04) + (0.3 \times 0.09) = 0.028 + 0.027 = 0.055 \).
3. Probability that a defective widget came from the premium line: \( \dfrac{0.3 \times 0.09}{0.055} \approx 0.4909 \).

Thus just under half of the defective widgets originated from the premium line even though premiums form only 30% of production.

Try this

1. A spinner is equally likely to show any integer from 1 to 8. Find \( P(X \ge 6 \mid X \text{is even}) \).
2. A box contains eight black, five white and three red marbles. Two marbles are drawn without replacement. Find \( P(\text{at least one red}) \).
3. In a game show, a contestant faces three doors. One door hides a prize. After the contestant chooses a door, the host opens a different door that never contains the prize. The contestant may stay or switch. Use a tree diagram to determine \( P(\text{win} \mid \text{switch}) \).