Questions on conditional probability

Question 1

P(A ∩ B) = 0.18 and P(B) = 0.6. Find P(A|B), and state what P(A) would have to equal for A and B to be independent.

Accepted Answer

P(A|B) = P(A ∩ B)/P(B) = 0.18/0.6 = 0.3, the overlap measured as a share of B's world rather than of the whole. Independence means conditioning on B changes nothing, so P(A) would have to be 0.3 as well. M1 for the conditional formula, A1 for 0.3, B1 for the value P(A) would need under independence. Dividing P(A ∩ B) = P(A)P(B) through by P(B) turns the product form of independence into this conditional form, so the two definitions say one thing.

Question 2

P(B) = 0.4 and P(A|B) = 0.25. Find P(A ∩ B).

Accepted Answer

Multiply along the branch. P(A ∩ B) = P(B) × P(A|B) = 0.4 × 0.25 = 0.1. M1 for the product along the branch, A1 for 0.1. This is the conditional formula rearranged, and it is what every tree diagram does. The second-stage probabilities on a tree are already conditional on the first stage, so multiplying along a path gives the probability of the whole path.

Question 3

P(A ∩ B) = 0.12 and P(A|B) = 0.3. Find P(B).

Accepted Answer

Rearrange the conditional formula rather than substituting into it. From P(A|B) = P(A ∩ B)/P(B) comes P(B) = P(A ∩ B)/P(A|B) = 0.12/0.3 = 0.4. M1 for rearranging the formula, A1 for 0.4. The unknown here is the denominator, so write the formula down before any numbers go in. Dividing the other way gives 0.3/0.12 = 2.5, which is not a probability, and that settles at once that the division has gone the wrong way round.

Question 4

Of 40 pupils, 22 walk to school and 18 do not. Nine of those who walk are in Year 7, and six of those who do not walk are in Year 7. Find P(Year 7 | walks) and P(walks | Year 7).

Accepted Answer

Along the walking row, 9 of the 22 who walk are in Year 7, so P(Year 7 | walks) = 9/22. For the other one, count the Year 7 pupils first: 9 + 6 = 15, and 9 of them walk, so P(walks | Year 7) = 9/15 = 3/5. M1 for the 15 pupils in Year 7, A1 for 9/22, A1 for 3/5. The same nine pupils form the numerator both times. What changes is the denominator, and the event written after the bar is what fixes it.

Question 5

A bag holds 4 red and 3 blue counters. Two are drawn without replacement. Find the probability that the second counter is blue, and the probability that both are red.

Accepted Answer

Two routes end in a blue second draw. Red then blue is 4/7 × 3/6 = 12/42, and blue then blue is 3/7 × 2/6 = 6/42. Adding them gives 18/42 = 3/7. Both red is a single route, 4/7 × 3/6 = 12/42 = 2/7. The second branch carries a denominator of 6 because a counter has already gone, and that shift is what 'without replacement' means. M1 for the products along the branches, M1 for adding the two routes, A1 for the first probability, A1 for the second.

Question 6

A bag holds 4 red and 3 blue counters, and two are drawn without replacement. Find the probability that the first counter was red, given that the second is blue.

Accepted Answer

P(first red | second blue) = P(red then blue)/P(second blue) = (12/42)/(18/42) = 12/18 = 2/3. Conditioning on the later event to ask about the earlier one feels backwards, but the formula has no sense of time. It compares an overlap with a whole and nothing more. M1 for the conditional formula, A1 for the correct pair of probabilities, A1 for 2/3. The denominator has to be P(second blue). Putting P(first red) underneath answers a different question.

Question 7

Of 100 gym members, 60 are adults and 40 are juniors. 45 adults and 10 juniors attend classes. Find P(attends | adult) and P(adult | attends).

Accepted Answer

Reading along the adult row, P(attends | adult) = 45/60 = 3/4. The attenders number 45 + 10 = 55, so P(adult | attends) = 45/55 = 9/11. B1 for the first probability, M1 for the 55 attenders, A1 for the second.

Question 8

Of 80 students, 50 study French and 28 study German, and 20 study both. Show that studying French and studying German are not independent events.

Accepted Answer

Compare a conditional with the plain probability. P(German) = 28/80 = 0.35, while P(German | French) = 20/50 = 0.4. M1 for computing both probabilities from the table, A1 for the pair of values. Independence requires P(German | French) = P(German), and 0.4 ≠ 0.35, so the two are not independent: knowing a student takes French raises the chance they take German. A1 for the comparison finished with a sentence. Comparing P(F ∩ G) = 0.25 with P(F)P(G) = 0.625 × 0.35 = 0.21875 is an equivalent test, provided the comparison is written out.

Question 9

A bag holds 5 green and 4 yellow beads. Two beads are taken at random without replacement. Find the probability that the two beads are the same colour.

Accepted Answer

Two routes end in a matching pair. Green then green is 5/9 × 4/8 = 20/72, and yellow then yellow is 4/9 × 3/8 = 12/72. Adding them gives 32/72 = 4/9. M1 for a product along one branch, M1 for adding the two matching routes, A1 for 4/9. On the second draw the denominator falls to 8 because a bead has gone, and the numerator falls by one as well, because the bead that went was of the colour being matched.

Question 10

In a college, 60% of the students take music. Of the students who take music, 25% play the piano; of those who do not take music, 10% play the piano. A student is chosen at random. Find the probability that the student plays the piano, and the probability that the student takes music given that the student plays the piano.

Accepted Answer

Multiply along each branch of the tree, then add. P(music and piano) = 0.6 × 0.25 = 0.15 and P(no music and piano) = 0.4 × 0.10 = 0.04, so P(piano) = 0.15 + 0.04 = 0.19. M1 for the two products, A1 for 0.19. Then P(music | piano) = 0.15/0.19 = 15/19 = 0.789 to 3 significant figures. M1 for dividing the right product by that total, A1 for 0.789. Music students are 60% of the college and 79% of its pianists, because the piano rate inside the music group is two and a half times the rate outside it, and conditioning on the piano reweights the two branches.

Question 11

P(A) = 0.3, P(B) = 0.5 and P(A ∪ B) = 0.65. Determine whether A and B are independent.

Accepted Answer

Recover the overlap from the addition rule first: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.3 + 0.5 − 0.65 = 0.15. M1 for rearranging the addition rule, A1 for 0.15. Now the test in conditional form: P(A|B) = P(A ∩ B)/P(B) = 0.15/0.5 = 0.3, which is exactly P(A) = 0.3, so A and B are independent: being told that B has happened leaves the probability of A where it was. B1 for the comparison, with both values quoted and a conclusion written out. The product form reaches the same verdict: P(A)P(B) = 0.3 × 0.5 = 0.15, which is P(A ∩ B).

Question 12

A factory has two machines. Machine A makes 60% of the components and 3% of them are faulty. Machine B makes the rest, and 8% of those are faulty. A component is chosen at random and found to be faulty. Find the probability that it came from machine B.

Accepted Answer

Draw the tree with the machine first, then condition. P(A and faulty) = 0.6 × 0.03 = 0.018 and P(B and faulty) = 0.4 × 0.08 = 0.032, so P(faulty) = 0.018 + 0.032 = 0.05. Then P(B | faulty) = 0.032/0.05 = 0.64. M1 A1 for the two products, M1 for adding them to reach the total probability of a fault, M1 for dividing the right product by that total, A1 for 0.64. Look at what the answer says. Machine B makes only 40% of the components yet accounts for 64% of the faulty ones, because the evidence has reweighted the branches. A patient who reads a 99% accurate positive test as a 99% chance of illness has run the same calculation backwards, quoting P(positive | ill) when the question asks for P(ill | positive).

Question 13

A weather app treats successive days as independent, each wet with probability 0.2, and multiplies to give 0.2 × 0.2 = 0.04 as the probability that both days of a weekend are wet. Explain why the independence assumption is doubtful here, and state which way the app's answer is likely to err.

Accepted Answer

The assumption doing the work is that Sunday's weather ignores Saturday's. B1 for naming it. Weather systems last for days, so rainy days cluster: given a wet Saturday, the chance of a wet Sunday is above 0.2, which is exactly what P(wet | wet) > P(wet) says. B1 for the criticism made through the conditional. Multiplying the unconditional 0.2 twice therefore understates the probability of a wet weekend, and the true value is likely to exceed 0.04. B1 for the direction.

Question 14

A bag contains 6 red counters and n blue counters. Two counters are taken at random without replacement. The probability that both are red is 1/3. Find the value of n.

Accepted Answer

Multiply along the red branch, keeping n in the denominators. P(both red) = 6/(6 + n) × 5/(5 + n) = 30/((6 + n)(5 + n)). M1 for the product along the branch, A1 for that expression. Setting it equal to 1/3 gives 90 = (6 + n)(5 + n), so n² + 11n + 30 = 90 and n² + 11n − 60 = 0. M1 for forming the quadratic. It factorises as (n + 15)(n − 4) = 0, so n = 4 or n = −15. A1 for both roots. A bag cannot hold −15 counters, so n = 4. A1 for n = 4 with the negative root rejected in a sentence. Check it forwards: 6/10 × 5/9 = 30/90 = 1/3, as the question stated.

Question 15

A and B are events with P(A) = 0.4, P(B|A) = 0.25 and P(B|A′) = 0.5. Find P(B), P(A|B) and P(A′|B′).

Accepted Answer

Draw the tree with A first, then multiply along its branches. P(A ∩ B) = 0.4 × 0.25 = 0.1 and P(A′ ∩ B) = 0.6 × 0.5 = 0.3, so P(B) = 0.1 + 0.3 = 0.4. M1 for the two products, A1 for P(B) = 0.4. Then P(A|B) = P(A ∩ B)/P(B) = 0.1/0.4 = 0.25. M1 for the conditional formula, A1 for 0.25. For the last one, work in complements: P(B′|A′) = 1 − 0.5 = 0.5, so P(A′ ∩ B′) = 0.6 × 0.5 = 0.3, while P(B′) = 1 − 0.4 = 0.6, giving P(A′|B′) = 0.3/0.6 = 0.5. M1 for a complement taken on a branch, A1 for 0.5. The pair of probabilities leaving any one branch of a tree sums to 1, which is what turns P(B|A′) = 0.5 into P(B′|A′) = 0.5 with no fresh calculation. Note that P(A|B) and P(B|A) both come out as 0.25 here; that is arithmetic coincidence on these numbers and not a rule.