Venn/tree diagrams; combined/mutually exclusive; conditional; independence practice questions.

Events A and B satisfy P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.2. Calculate P(A ∪ B).

• 0.8
• 0.6
• 0.2
• 1.0

Answer: 0.6

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.5 + 0.3 − 0.2 = 0.6; forgetting to subtract the intersection gives the wrong value 0.8.

Given P(A ∩ B) = 0.12 and P(B) = 0.4, determine P(A | B).

• 0.048
• 0.12
• 0.3
• 0.52

Answer: 0.3

P(A | B) = P(A ∩ B) / P(B) = 0.12 / 0.4 = 0.3; multiplying instead of dividing gives the distractor 0.048.

Events A and B satisfy P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.3. State which of the following is correct.

• A and B are independent
• A and B are mutually exclusive
• A and B are exhaustive
• A and B are dependent

Answer: A and B are independent

Since P(A) × P(B) = 0.6 × 0.5 = 0.3 = P(A ∩ B), the independence condition holds; they are not mutually exclusive because P(A ∩ B) ≠ 0.

In a class of 30 students, 18 study French, 14 study Spanish and 6 study both languages. Determine how many students study neither language.

• 2
• 6
• 10
• 4

Answer: 4

By a Venn diagram, French or Spanish = 18 + 14 − 6 = 26, so neither = 30 − 26 = 4.

A bag contains 3 red and 2 blue marbles. Two marbles are drawn one after the other without replacement. Determine the probability that both are red.

• 3/10
• 9/25
• 3/5
• 1/10

Answer: 3/10

On a tree diagram, P(RR) = 3/5 × 2/4 = 6/20 = 3/10; the distractor 9/25 comes from wrongly assuming replacement.

Events A and B are mutually exclusive with P(A) = 0.25 and P(B) = 0.45. State the value of P(A ∩ B).

• 0.1125
• 0.7
• 0
• 0.2

Answer: 0

Mutually exclusive events cannot occur together, so P(A ∩ B) = 0; multiplying the probabilities (0.1125) wrongly assumes independence.