Bayes' theorem (max 3 events) practice questions.

A factory has two machines. Machine A produces 60% of all items and machine B produces the remaining 40%. Of the items from A, 2% are defective; of the items from B, 5% are defective. An item chosen at random is found to be defective. Calculate the probability that it was produced by machine A.

• 3/8
• 5/8
• 0.012
• 0.6

Answer: 3/8

P(A|D) = (0.6 × 0.02) / (0.6 × 0.02 + 0.4 × 0.05) = 0.012/0.032 = 3/8. Bayes' theorem divides the branch through A by the total probability of a defective item.

Three machines P, Q and R produce 50%, 30% and 20% of a factory's output, with defect rates 1%, 2% and 5% respectively. A randomly selected item is defective. Determine the probability that it came from machine R.

• 5/21
• 2/7
• 10/21
• 1/5

Answer: 10/21

P(R|D) = (0.2 × 0.05)/(0.5 × 0.01 + 0.3 × 0.02 + 0.2 × 0.05) = 0.010/0.021 = 10/21. The three-branch total probability forms the denominator.

Bag P contains 3 red and 2 blue marbles; bag Q contains 1 red and 4 blue marbles. A fair coin decides which bag is used, and one marble is drawn. Given that the marble is red, what is the probability that bag P was used?

• 3/5
• 3/4
• 1/2
• 3/8

Answer: 3/4

P(P|red) = (1/2 × 3/5)/(1/2 × 3/5 + 1/2 × 1/5) = (3/10)/(4/10) = 3/4. A common error is to answer 3/5, which is P(red|P), the reversed conditional.

On any day the probability of rain is 0.3. A forecaster predicts rain with probability 0.8 when it will rain, and with probability 0.1 when it will not. Given that rain is forecast, determine the probability that it actually rains.

• 0.24
• 24/31
• 0.8
• 7/31

Answer: 24/31

P(rain|forecast) = (0.3 × 0.8)/(0.3 × 0.8 + 0.7 × 0.1) = 0.24/0.31 = 24/31. Choosing 0.24 gives only the joint probability, not the conditional one.

For events A and B with P(A) > 0 and P(B) > 0, which expression is equal to P(A|B)?

• P(B|A)P(B)/P(A)
• P(A)P(B)/P(B|A)
• P(B|A)/P(B)
• P(B|A)P(A)/P(B)

Answer: P(B|A)P(A)/P(B)

Both conditionals share the numerator P(A ∩ B) = P(B|A)P(A), so P(A|B) = P(B|A)P(A)/P(B). The other options scramble which marginal multiplies and which divides.

One person in ten in a population has a condition. A test detects the condition in 90% of those who have it, but also gives a positive result for 20% of those who do not. Determine the probability that a person who tests positive actually has the condition.

• 9/11
• 0.9
• 1/3
• 0.09

Answer: 1/3

P(D|+) = (0.1 × 0.9)/(0.1 × 0.9 + 0.9 × 0.2) = 0.09/0.27 = 1/3. Answering 9/11 comes from forgetting to weight each branch by its prior probability.