Probability: questions with answers

Every subset of the sample space is an event, however small: the empty set and a one-outcome collection both qualify. What disqualifies a collection is a member the experiment cannot produce, and HH is not one of the eight results of tossing three coins. So {HTH, HH} is not an event of this experiment.

The event is the subset of the eight outcomes that satisfy 'at least two heads appear', and counting them gives 4. One condition does not mean one outcome, and the answer is not the number of coins either: the count comes from the sample space.

The sure event is the whole sample space, and nothing less will do. 'the number shown is odd or even' holds for every one of 1, 2, 3, 4, 5 and 6. 'the number shown is not 6' holds for five of the six numbers, which is close but not certain: the single outcome 6 is enough to disqualify it.

The probability of showing 1 is the number of faces carrying 1 divided by 6, so the face count is recovered by multiplying the probability by 6, giving 3. Spreading the three numbers evenly over the six faces would put two faces on each, which is not what the given probability says, and the probability 1/2 is not itself a count of faces.

'A and B' is the intersection, so an outcome qualifies only if it satisfies BOTH descriptions. A is {1, 2, 3} and B is {2, 3, 5}, and the outcomes common to them are {2, 3}. Pooling the two lists instead would give {1, 2, 3, 5}, which is the event 'A or B'.

The union counts each outcome once, however many of the events hold it, so adding the three sizes over-counts and gives 8 instead of 5. The answer is not the number of events either, and it is not automatically 6: here 5 of the six numbers are covered.

Mutual exclusion is a statement about the INTERSECTION: A and B are mutually exclusive exactly when no outcome lies in both. The condition that every outcome lies in A or in B is about the UNION, and makes the pair exhaustive instead. A pair can have either property without the other.

The four outcome probabilities must total 1, so the last one is forced: it is 1 minus the 7/12 already used, which is 5/12. It is not free to be any number in range, and there is no rule making the outcomes equally likely -- an equal assignment would give every outcome 1/4, which does not fit these three values.

Because A and B overlap, the third axiom does not apply and the two probabilities may not simply be added: the sizes of A and B add to 5, which double-counts the 1 outcome the two events share. The event 'A or B' really holds 4 of the six equally likely numbers, so P(A or B) is 2/3.

The sample space lists what actually happens on each object, so it holds 4 equally likely outcomes. The 3 summary descriptions are a different list altogether: they group several outcomes together, so they cannot all be equally likely and they must not be used as the denominator of a probability.

Counting the two places in order gives each committee twice over, once for each order of its two members, so the ordered count must be halved: there are 28 committees. The number of people is not the number of committees, and doubling the group counts nothing at all.

Every student of the class is equally likely to be chosen, and the class holds 17 students. The students belonging to atleast one group are those in A only, in B only and in both, which is 14 students. Adding the size of A to the size of B instead would count the 2 students in both twice; leaving them out altogether would count them no times, and neither is what 'atleast one' means.

'Exactly one' leaves out the 5 students who belong to both groups, so the favourable students are the 4 in A alone together with the 9 in B alone, which is 13. Adding the two group sizes without removing the shared students counts them twice, and keeping them in the favourable count answers a different question -- the probability of belonging to atleast one group.

'Red' and 'not red' are complementary events, so P(red) = 1 - 7/12 = 5/12. The complement is found by subtracting the probability from 1 -- not by subtracting 1 from it, which gives a negative number, and not by inverting the fraction, which gives a number bigger than 1.

Subtracting the 'neither' probability from 1 gives the UNION, 2/5, because 'neither' is the complement of 'A or B'. Reading it as the complement of 'A and B' would make P(A and B) equal to 1 - 3/5 = 2/5 straight away, which is the probability that atleast one occurs, not the probability that both do. The addition rule then gives 1/20.