Conditional Probability

Use the size of the group specified by the condition.

1. "Given that the pupil is a music-club member" means we only look at the 12 club members: they are the new total.
2. Of these 12 members, 7 play the piano.
3. So the probability is \frac{7}{12}.

Use only the rice row as the sample space.

1. There are 20 diners who chose rice.
2. Of these, 12 chose tea.
3. The probability is \frac{12}{20}=\frac{3}{5}.

List only scores consistent with the given information.

1. The condition leaves the equally likely scores 4, 5 and 6.
2. Two of these, 4 and 6, are even.
3. The probability is \frac{2}{3}.

The condition determines the denominator.

1. "Given that the student is in Year 11" means we only look at Year 11 students.
2. So the denominator is the total number in Year 11, and the numerator is the number in Year 11 who study art.
3. The total number in the school would be the denominator for P(\text{art}) with no condition, which is a different probability.

Update the contents after the known first draw.

1. After removing a red counter, 7 counters remain.
2. All 3 blue counters are still in the bag.
3. The required probability is \frac{3}{7}.

Multiply probabilities along the required path.

1. To find the probability of one outcome followed by another, multiply the probabilities along the branches.
2. Follow the branch to A, then the branch from A to S: 0.3\times0.8=0.24.
3. The probability is 0.24 (or \frac{6}{25}).

Add the two different paths that lead to undamaged contents.

1. Undamaged contents can arrive on time or late. These cases do not overlap.
2. On-time and undamaged: 0.6\times0.9=0.54. Late and undamaged: 0.4\times0.75=0.30.
3. Follow the two routes to undamaged contents.

4. Add the path probabilities: 0.54+0.30=0.84=\frac{21}{25}.

The condition selects a column rather than a row.

1. The Drama column contains 32 pupils.
2. There are 20 Year 9 pupils in this column.
3. The probability is \frac{20}{32}=\frac{5}{8}.

Discard outcomes that fail the condition before counting.

1. The four equally likely outcomes are HH, HT, TH and TT.
2. "Exactly one head" rules out HH (two heads) and TT (no heads), leaving HT and TH.
3. Only HT has a head on the first coin, so the probability is \frac{1}{2}.

Draw a Venn diagram and let x be the number who study both languages.

1. Draw a Venn diagram with circles F and S, and let x students study both.
2. 60-8=52 students study at least one language.
3. Adding 34+25 counts the x students in both circles twice, so 34+25-x=52, giving x=7.
4. Complete the Venn diagram before restricting the group.

5. "Given that the student studies French" means we only look at the 34 French students. 7 of them also study Spanish.
6. So the probability is \frac{7}{34}.

Keep the known first score fixed.

1. The second die can still show any score 1 to 6 with equal probability.
2. To make a total of at least 10, it must show 4, 5 or 6.
3. The probability is \frac{3}{6}=\frac{1}{2}.

There are two different paths that end at L.

1. The bus can be late in two ways: rain and late (R then L), or dry and late (D then L). Find each by multiplying along the branches.
2. Rain and late: \frac{1}{4}\times\frac{3}{5}=\frac{3}{20}.
3. Dry and late: \frac{3}{4}\times\frac{1}{10}=\frac{3}{40}.
4. The two ways cannot both happen, so add them: \frac{3}{20}+\frac{3}{40}=\frac{6}{40}+\frac{3}{40}=\frac{9}{40}.

Update the bag after the first counter, then remember that different colours can happen in two orders.

1. The first counter is blue, so the bag now holds 2 blue and 2 red counters: 4 counters.
2. Different colours means blue then red, or red then blue.
3. Blue then red: \frac{2}{4}\times\frac{2}{3}=\frac{4}{12}. Red then blue: \frac{2}{4}\times\frac{2}{3}=\frac{4}{12}.
4. Track the next two draws after the known first blue.

5. Add the two orders: \frac{4}{12}+\frac{4}{12}=\frac{8}{12}=\frac{2}{3}.

Compare the conditional probability with the original probability.

1. Two events are independent when knowing that one happened does not change the probability of the other.
2. Here P(A\mid B)=0.4 and P(A)=0.4: knowing B happened leaves the probability of A unchanged, so A and B are independent.
3. They are not mutually exclusive: that would mean A and B cannot happen together, but P(A\text{ and }B)=0.4\times0.3=0.12, which is not 0.

Use a convenient total to compare faulty items from the two sources.

1. Think of 10000 items: A makes 7000 and B makes 3000.
2. There would be 140 faulty items from A and 150 from B.
3. The faulty group contains 290 items in total.
4. Organise the imagined 10 000 items in a table.

5. The required probability is \frac{150}{290}=\frac{15}{29}.

Find the probability of two blue and of two red; "same colour" is either of these.

1. Two blue: \frac{4}{6}\times\frac{3}{5}=\frac{12}{30}.
2. Two red: \frac{2}{6}\times\frac{1}{5}=\frac{2}{30}.
3. Find the two paths that give matching colours.

4. Same colour (two blue or two red): \frac{12}{30}+\frac{2}{30}=\frac{14}{30}.
5. "Given that both are the same colour" means the \frac{14}{30} is now the whole. Two blue is \frac{12}{30} of it.
6. Compare the blue path with the whole same-colour group.

7. So the probability is \frac{12}{30}\div\frac{14}{30}=\frac{12}{14}=\frac{6}{7}. (Not \frac{12}{30}: that ignores the information given.)

Draw a three-circle Venn diagram and fill it in from the centre outwards.

1. Draw a Venn diagram with circles for swimming, tennis and yoga, and start in the centre: 10 do all three.
2. Swimming and yoga is 15 in total, so 15-10=5 do swimming and yoga only. Tennis and yoga is 12, so 12-10=2 do tennis and yoga only. 8 do swimming and tennis only.
3. Swimming only: 52-10-8-5=29. Yoga only: 6.
4. Tennis only: 100-(10+8+5+2+29+6)=100-60=40.
5. Fill the three-circle Venn diagram from the centre outwards.

6. Members who do tennis: 40+8+10+2=60. Of these, 40+2=42 do not do swimming.
7. "Given that the member does tennis" means the 60 tennis players are the total, so the probability is \frac{42}{60}=\frac{7}{10}.

Compare the A-and-white path with both paths producing white.

1. The A-and-white probability is \frac{1}{2}\times\frac{2}{5}=\frac{1}{5}.
2. The B-and-white probability is \frac{1}{2}\times\frac{4}{6}=\frac{1}{3}.
3. Follow both routes to a white ball.

4. The total white probability is \frac{1}{5}+\frac{1}{3}=\frac{8}{15}.
5. Compare the A-and-white path with all white outcomes.

6. So the probability of A given white is \frac{1/5}{8/15}=\frac{3}{8}.

Count the restricted sample space using the complement of an even product.

1. There are 6\times5=30 equally likely ordered selections.
2. An odd product needs two odd numbers, giving 3\times2=6 selections. Thus 24 selections have an even product.
3. Two even numbers also give 3\times2=6 selections.
4. Mark the allowed ordered selections in a grid.

5. Within the restricted sample space, the probability is \frac{6}{24}=\frac{1}{4}.

Work out the probability of at least one faulty bulb and of exactly one faulty bulb first.

1. Use a tree diagram (or list the orders). There are 4 sound and 2 faulty bulbs, and three are taken without replacement.
2. No faulty bulbs: \frac{4}{6}\times\frac{3}{5}\times\frac{2}{4}=\frac{24}{120}=\frac{1}{5}, so at least one faulty is 1-\frac{1}{5}=\frac{4}{5}.
3. Exactly one faulty, faulty bulb first: \frac{2}{6}\times\frac{4}{5}\times\frac{3}{4}=\frac{24}{120}=\frac{1}{5}.
4. The faulty bulb could be first, second or third, and each order has the same probability, so exactly one faulty is 3\times\frac{1}{5}=\frac{3}{5}.
5. List the orders used in the two probability calculations.

6. "Given that at least one is faulty" means \frac{4}{5} is now the whole, and "exactly one" is part of it.
7. Compare exactly one faulty with at least one faulty.

8. So the probability is \frac{3}{5}\div\frac{4}{5}=\frac{3}{4}.