Questions on probability and Venn diagrams

Question 1

P(A) = 0.45 and P(B) = 0.35, and A and B are mutually exclusive. Write down P(A ∩ B), and find P(A ∪ B).

Accepted Answer

Mutually exclusive means the events cannot happen together, so P(A ∩ B) = 0. The addition rule then loses its correction term and P(A ∪ B) = 0.45 + 0.35 = 0.8. B1 for P(A ∩ B) = 0, B1 for P(A ∪ B) = 0.8. On a Venn diagram the two circles are drawn apart, with no region between them.

Question 2

P(A) = 0.55, P(B) = 0.4 and P(A ∪ B) = 0.75. Find P(A ∩ B), and determine whether A and B are independent.

Accepted Answer

Rearrange the addition rule: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.55 + 0.4 − 0.75 = 0.2. Now test independence. P(A) × P(B) = 0.55 × 0.4 = 0.22, which does not equal 0.2, so A and B are not independent. M1 for rearranging the addition rule, A1 for 0.2, B1 for the independence test with both values quoted. The same test in conditional form gives P(A|B) = 0.2/0.4 = 0.5 against P(A) = 0.55, and reaches the same answer.

Question 3

P(A) = 0.6, P(B) = 0.3 and P(A ∩ B) = 0.18. Find P(A ∪ B).

Accepted Answer

The addition rule adds the two circles and then takes the overlap off once, because adding has counted it twice: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.3 − 0.18 = 0.72. M1 for the rule with the subtraction present, A1 for 0.72. Adding and stopping there gives 0.9, which counts the 0.18 twice over. These two events happen also to be independent, since 0.6 × 0.3 = 0.18, but the addition rule holds whether or not they are.

Question 4

P(A) = 0.35, P(A ∪ B) = 0.8 and P(A ∩ B) = 0.15. Find P(B), and write down P(B′).

Accepted Answer

Run the addition rule backwards. From P(A ∪ B) = P(A) + P(B) − P(A ∩ B) comes P(B) = P(A ∪ B) − P(A) + P(A ∩ B) = 0.8 − 0.35 + 0.15 = 0.6. M1 for rearranging the rule, A1 for 0.6. Then P(B′) = 1 − P(B) = 1 − 0.6 = 0.4. B1. The overlap returns with a plus sign when the rule is rearranged this way; keeping it negative gives 0.3, and then 0.35 + 0.3 − 0.15 = 0.5, not the 0.8 given, which is the signal to check the rearrangement.

Question 5

Of 100 students, 55 study art, 40 study music, and 20 study both. Draw the Venn diagram in numbers: art only, music only, both, neither.

Accepted Answer

Put the overlap in first: 20 study both. Art only is then 55 − 20 = 35 and music only is 40 − 20 = 20. Those three regions hold 75, so 25 students sit outside both circles. B1 for the overlap, M1 for subtracting it, A1 for 35 and 20, B1 for the 25 outside. The figures 55 and 40 already include the 20, so subtracting the overlap before anything else stops it being counted twice.

Question 6

Using the diagram above, find the probability that a randomly chosen student studies exactly one of the subjects, and the probability that they study neither.

Accepted Answer

Exactly one means art only or music only, so (35 + 20)/100 = 0.55. Neither is the region outside both circles, 25/100 = 0.25. M1 for adding the two single regions, A1 for 0.55, B1 for 0.25. 'Exactly one' deliberately excludes the overlap, which is what makes it a different question from P(A ∪ B) = 0.75.

Question 7

Explain the difference between mutually exclusive events and independent events.

Accepted Answer

Mutually exclusive events cannot both happen, so their circles do not overlap and P(A ∩ B) = 0. Independent events are those where one happening tells you nothing about the other, so P(A ∩ B) = P(A)P(B). B1 for mutual exclusivity, B1 for independence, B1 for why they cannot hold together. The two ideas are distinct properties rather than opposites. For events of non-zero probability they cannot hold together, because mutual exclusivity is an extreme form of dependence, with knowing that A occurred ruling B out entirely, and a pair of events may easily be neither exclusive nor independent.

Question 8

A and B are independent events with P(A) = 0.6 and P(B) = 0.25. Find P(A ∪ B), and the probability that exactly one of A and B occurs.

Accepted Answer

Independence hands over the overlap as a product: P(A ∩ B) = 0.6 × 0.25 = 0.15. M1 for using the product, A1 for 0.15. The addition rule then gives P(A ∪ B) = 0.6 + 0.25 − 0.15 = 0.7. M1. Exactly one excludes the overlap, so it is the two single regions: (0.6 − 0.15) + (0.25 − 0.15) = 0.45 + 0.1 = 0.55, which is also 0.7 − 0.15. A1 for both values. The four regions 0.45, 0.15, 0.1 and 0.3 total 1.

Question 9

For two events A and B, P(A ∩ B′) = 0.25, P(A′ ∩ B) = 0.15 and P(A ∩ B) = 0.2. Find P(A), P(B) and P(A′ ∩ B′), and state which region of the Venn diagram P((A ∪ B)′) describes.

Accepted Answer

Each expression names a region, so read them off and add. P(A) is the whole A circle, which is A-only together with the overlap: 0.25 + 0.2 = 0.45. B1. P(B) is 0.15 + 0.2 = 0.35. B1. The three regions inside the circles total 0.25 + 0.2 + 0.15 = 0.6, so the region outside both holds 1 − 0.6 = 0.4, and that region is P(A′ ∩ B′). M1 for taking the inside from 1, A1 for 0.4. P((A ∪ B)′) is the same region, so it is 0.4 as well: being in neither circle and being outside the union say one thing.

Question 10

P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.1. By identifying the regions of a Venn diagram that each event covers, find P(A′ ∪ B′) and P(A ∪ B′).

Accepted Answer

Fill the diagram first: the overlap is 0.1, A only is 0.5 − 0.1 = 0.4, B only is 0.3 − 0.1 = 0.2, and the region outside both is 1 − (0.4 + 0.1 + 0.2) = 0.3. B1 for the four regions.
A′ ∪ B′ is the set of outcomes not in A or not in B, which is everything except the outcomes in both: it is the complement of A ∩ B. So P(A′ ∪ B′) = 1 − 0.1 = 0.9, and the three regions it covers, 0.4 + 0.2 + 0.3, add to the same. M1 for identifying the region as the complement of the overlap.
A ∪ B′ covers the whole A circle together with everything outside B, which leaves out only the B-only region: P(A ∪ B′) = 0.4 + 0.1 + 0.3 = 0.8, or 1 − 0.2. The addition rule agrees: P(A) + P(B′) − P(A ∩ B′) = 0.5 + 0.7 − 0.4 = 0.8. A1 for 0.8 with the region named.
A complement of a union is the intersection of the complements, and a complement of an intersection is the union of the complements, so A′ ∪ B′ and (A ∩ B)′ are one region under two names. Shading the diagram settles the question in a line, where working from the notation alone tends to turn 'not both' into 'neither'.

Question 11

A student states that P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.95. Show that these three values cannot all be right, and find the largest value P(A ∪ B) can take when P(A) = 0.5 and P(B) = 0.4.

Accepted Answer

Rearranging the addition rule gives P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.4 − 0.95 = −0.05. M1 for the rearrangement, A1 for −0.05. A probability cannot be negative, so no pair of events carries all three of those values. B1. The union is largest when the overlap is smallest, and the smallest an overlap can be is zero, which is the mutually exclusive case: then P(A ∪ B) = 0.5 + 0.4 = 0.9. B1 for 0.9 with mutual exclusivity named as the case.

Question 12

Of 60 people surveyed, 30 like tea, 25 like coffee and 22 like juice. 12 like tea and coffee, 10 like coffee and juice, 8 like tea and juice, and 5 like all three. Find the number who like none of the three drinks, and the probability that a person chosen at random likes exactly one drink.

Accepted Answer

Fill the three-circle diagram from the centre outwards. The middle holds 5. Tea and coffee but not juice is 12 − 5 = 7; coffee and juice but not tea is 10 − 5 = 5; tea and juice but not coffee is 8 − 5 = 3. Now the single regions. Tea only is 30 − 7 − 3 − 5 = 15, coffee only is 25 − 7 − 5 − 5 = 8, and juice only is 22 − 3 − 5 − 5 = 9. Those seven regions total 52, so 60 − 52 = 8 people like none of the three. Exactly one drink covers 15 + 8 + 9 = 32 people, a probability of 32/60 = 8/15, or 0.533. The inclusion-exclusion formula gives the same total: 30 + 25 + 22 − 12 − 10 − 8 + 5 = 52. B1 for the centre, M1 A1 for the three pairwise regions, A1 for the three single regions, A1 for the 8 who like none, M1 A1 for the probability.

Question 13

Given P(A) = 0.5, a student writes 0.5 into the A-only region of a Venn diagram. Explain the error, the order in which the regions of a diagram should be filled, and the check that catches this mistake before any probabilities are read off.

Accepted Answer

P(A) is the whole circle, overlap included, so the A-only region holds P(A) − P(A ∩ B), and writing 0.5 there counts the overlap twice as soon as the B circle is filled. B1 for P(A) already containing the overlap. The diagram is filled from the inside out: the overlap first, then each only region by subtraction, then the outside, and a three-circle diagram runs centre, pairwise slivers, singles, outside, so that every unknown met has one equation attached. B1 for the order. The regions partition the sample space, so they must total exactly 1; a total that comes out above 1 is a sign of this error. B1 for the total-1 check.

Question 14

For two events A and B, P(A) = 0.6, P(B) = x and P(A ∩ B) = 0.2x. Given that P(A ∪ B) = 0.9, find the value of x, and determine whether A and B are independent.

Accepted Answer

Put all three into the addition rule and keep x: 0.9 = 0.6 + x − 0.2x. M1 for the substitution, A1 for 0.9 = 0.6 + 0.8x. Then 0.8x = 0.3, so x = 0.375. A1 for x = 0.375. Now the test. P(A ∩ B) = 0.2 × 0.375 = 0.075, while P(A)P(B) = 0.6 × 0.375 = 0.225. M1 for computing both. They are not equal, so A and B are not independent. A1 for the conclusion with both values quoted. In conditional form the same verdict reads P(A|B) = 0.075/0.375 = 0.2 against P(A) = 0.6: B happening cuts the chance of A to a third of what it was.

Question 15

A and B are events with P(A) = a and P(B) = b, where a and b are both non-zero. Show that if A and B are mutually exclusive then they cannot be independent. Show also that if A and B are independent then P(A′ ∩ B′) = (1 − a)(1 − b), and hence write P(A ∪ B) in terms of a and b.

Accepted Answer

Mutually exclusive events satisfy P(A ∩ B) = 0, and independent events satisfy P(A ∩ B) = ab. M1 for writing both conditions down. If both held at once then ab = 0, so a or b would be zero, and neither is; the two properties therefore cannot hold together for these events. A1. For the second part, the region outside both circles is the complement of the union, so P(A′ ∩ B′) = 1 − P(A ∪ B) = 1 − (a + b − ab), using the addition rule with the independent overlap ab. M1 for the complement of the union, A1 for 1 − a − b + ab. That expression factorises as (1 − a)(1 − b), which is the printed result. A1. Rearranging it gives P(A ∪ B) = 1 − (1 − a)(1 − b) = a + b − ab. A1. The factorised form says that A′ and B′ are independent whenever A and B are, and that is the shortest route to a union: find the probability that neither event happens and take it from 1.