Formal conditional probability; testing independence practice questions.

Events A and B are independent with P(A) = 0.5 and P(B) = 0.3. Calculate P(A ∪ B).

• 0.80
• 0.65
• 0.15
• 0.35

Answer: 0.65

For independent events P(A ∩ B) = P(A) × P(B) = 0.15, so P(A ∪ B) = 0.5 + 0.3 − 0.15 = 0.65. Adding to 0.8 forgets to subtract the intersection.

Two events A and B satisfy P(A) = 0.6 and P(B) = 0.4. Determine the value of P(A ∩ B) for which A and B are independent.

• 0
• 0.10
• 0.24
• 1.0

Answer: 0.24

Independence requires P(A ∩ B) = P(A) × P(B) = 0.6 × 0.4 = 0.24. An intersection of 0 would make them mutually exclusive, not independent.

In a group of 40 students, 24 play football, 20 play basketball and 10 play both sports. A student is chosen at random. Given that the student plays basketball, determine the probability that the student also plays football.

• 1/4
• 5/12
• 3/4
• 1/2

Answer: 1/2

P(football | basketball) = 10/20 = 1/2: restrict the sample space to the 20 basketball players. Using 10/24 (conditioning on football) or 10/40 (the joint probability) are the classic errors.

Events A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.5. State the value of P(A | B).

• 0
• 0.15
• 0.3
• 0.6

Answer: 0

Mutually exclusive events cannot occur together, so P(A ∩ B) = 0 and hence P(A | B) = 0 / 0.5 = 0. Answering 0.3 confuses mutual exclusivity with independence.

A bag contains 3 red and 5 blue balls. Two balls are drawn at random without replacement. Determine the probability that the second ball is red, given that the first ball drawn is red.

• 3/8
• 2/7
• 1/4
• 3/7

Answer: 2/7

After a red ball is removed, 2 red balls remain among 7, so the conditional probability is 2/7. Using 3/8 ignores that the draw is without replacement.

A factory has two machines. Machine X produces 60% of all items and 2% of its items are defective. Machine Y produces 40% of all items and 5% of its items are defective. An item is selected at random. Calculate the probability that it is defective.

• 0.032
• 0.07
• 0.035
• 0.02

Answer: 0.032

By the total probability rule: P(D) = 0.6 × 0.02 + 0.4 × 0.05 = 0.012 + 0.020 = 0.032. Simply adding 2% and 5%, or averaging them, ignores the production weights.