Probability - Sample Spaces and Two-Way Tables - Higher

List ordered outcomes, keeping HT and TH separate.

1. The equally likely ordered outcomes are HH, HT, TH and TT.
2. Exactly one head occurs in HT and TH.
3. The probability is \frac{2}{4}=\frac{1}{2}.

Pair each main course with every dessert.

1. Each of the three main courses can be paired with either dessert.
2. There are 3\times2=6 combinations.

Use the joint cell, with all pupils as the denominator.

1. Use the cell satisfying both conditions: Year 10 and walking.
2. There are 12 such pupils out of 50.
3. The probability is \frac{12}{50}=\frac{6}{25}.

Count outcome cells, rather than different totals.

1. There are 3\times3=9 equally likely card-spinner outcomes.
2. A total of 5 occurs for card 2 with spinner 3, and card 4 with spinner 1.
3. The probability is \frac{2}{9}.

Use the total for the row containing the missing value.

1. The morning row has total 32.
2. So 14+x=32, giving x=18.

Remove the overlap from each group before adding.

1. Cycle-route only: 35-12=23. Footpath only: 29-12=17.
2. Exactly one route is used by 23+17=40 people.
3. The probability is \frac{40}{80}=\frac{1}{2}.

Count the ordered pairs for every total that qualifies.

1. There are 36 equally likely ordered outcomes.
2. Totals 10, 11 and 12 occur in 3, 2 and 1 ways respectively.
3. The probability is \frac{3+2+1}{36}=\frac{1}{6}.

Repeated labels still represent separate equally likely sectors.

1. Use the sectors as outcomes, even when labels repeat. There are 9 equally likely sector pairs.
2. A 2 from A and a 4 from B gives 2\times2=4 pairs. A 5 and a 1 gives one more.
3. The probability is \frac{5}{9}.

Find the intersection of the required row and column.

1. The Train and Veg cell contains 9 passengers.
2. The probability is \frac{9}{60}=\frac{3}{20}.

The first card determines whether the number can exceed the given boundary.

1. There are 4\times3=12 equally likely ordered choices.
2. The tens digit must be 5 or 7. Each allows three remaining units digits, giving 6 favourable choices.
3. The probability is \frac{6}{12}=\frac{1}{2}.

Consider which parity each score must have.

1. A product is odd only when both factors are odd.
2. Each die has three odd scores, so 3\times3=9 of the 36 ordered outcomes qualify.
3. The probability is \frac{9}{36}=\frac{1}{4}.

List all 20 outcomes in a sample space and count the wins. Then multiply the probability by the number of games.

1. There are 4\times5=20 equally likely outcomes (score on A, score on B).
2. A beats B when A is 2 and B is 1 (1 way); A is 3 and B is 1 or 2 (2 ways); A is 4 and B is 1, 2 or 3 (3 ways).
3. That is 1+2+3=6 winning outcomes, so the probability of a win is \frac{6}{20}=\frac{3}{10}.
4. Estimated number of wins =\frac{3}{10}\times200=60.

You only need the cell marked y. Use the Guide column total.

1. The visitors who came in the afternoon and bought a guide are the cell marked y.
2. Use the Guide column: 18+y=33, so y=15. (Check with the afternoon row: 15+25=40.)
3. There are 70 visitors in total, so the probability is \frac{15}{70}=\frac{3}{14}.

Compare how many ordered pairs produce different totals.

1. The individual ordered die pairs are equally likely.
2. For example, total 2 has only (1,1), whereas total 7 has six pairs.
3. The totals therefore do not have equal probabilities.
4. The dice are fair (each single score is equally likely); it is the totals that are not. There are 11 totals, from 2 to 12, and both 2 and 12 can happen, in one way each.

Find the union before taking its complement.

1. The number playing at least one sport is 22+18-9=31.
2. Subtract the overlap once because it was counted in both totals.
3. There are 40-31=9 playing neither, so the probability is \frac{9}{40}.

Check both conditions for each possible sum.

1. There are 9 equally likely sector pairs.
2. With A showing -2, B can show 4 or 6. With A showing 0, B can show 4 or 6. With A showing 3, B must show 1.
3. These give 5 qualifying pairs, so the probability is \frac{5}{9}.

List pairs systematically without counting the same unordered pair twice.

1. The 10 unordered pairs are equally likely, since each has two possible orders.
2. Prime sums occur for (1,2), (1,4), (2,3), (2,5) and (3,4).
3. There are 5 favourable pairs, so the probability is \frac{5}{10}=\frac{1}{2}.

Draw a two-way table (adult or child, chess or draughts) and fill it in. Only the members who play chess count.

1. Adults: \frac58\times120=75. Children: 120-75=45.
2. Adults who play chess: 40\% of 75 is 30.
3. Children who play chess: 45-30=15 (every child who doesn't play draughts).
4. Members who play chess: 30+15=45.
5. The member is chosen from the 45 chess players (not from all 120 members), and 30 of them are adults, so the probability is \frac{30}{45}=\frac23.

For each coin outcome, count how many die scores can match it.

1. There are 8 equally likely coin outcomes and 6 die scores, giving 48 equally likely combined outcomes.
2. The coin outcomes have 1, 2 or 3 heads in 3, 3 or 1 ways. Zero heads cannot match a die score.
3. For each qualifying coin outcome, exactly one die score matches.
4. The probability is \frac{3+3+1}{48}=\frac{7}{48}.

Find the possible score patterns, then count their distinct orders.

1. There are 6^3=216 equally likely ordered triples.
2. A total of 5 comes from scores 1,1,3 or 1,2,2.
3. Each pattern has three different orders, giving 6 favourable triples.
4. The probability is \frac{6}{216}=\frac{1}{36}.