Year 11 conditional probability worksheet answers to download as PDF

These Year 11 conditional probability worksheet answers give a full worked solution to all 20 questions of the conditional probability worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. \(\dfrac{18+15}{60}=\dfrac{33}{60}=\) \(\dfrac{11}{20}\). Add the boys and girls who go by bus: \(18+15=33\) out of 60 pupils, then simplify by dividing by 3.
2. \(\dfrac{18}{60}=\) \(\dfrac{3}{10}\). The boys who go by bus are one cell of the table: 18 out of all 60 pupils.
3. Only the 30 girls matter: \(\dfrac{15}{30}=\) \(\dfrac12\). The condition 'a girl is chosen' means we only look at the girls' row, so the denominator is 30, not 60.
4. Only the 27 pupils who walk matter: \(\dfrac{15}{27}=\) \(\dfrac59\). The condition 'walks' means we only look at the walk column: \(12+15=27\) pupils, of whom 15 are girls.
Answer to Question 2
1. Tea only 14, both 6, coffee only 9, neither 11, and \(14+6+9+11=40\).
2. \(\dfrac{11}{40}\). 11 of the 40 people drink neither, so the probability is \(\tfrac{11}{40}\).
3. There are \(14+6=20\) tea drinkers and 6 of them also drink coffee: \(\dfrac{6}{20}=\) \(\dfrac{3}{10}\).
4. There are \(9+6=15\) coffee drinkers and 6 also drink tea: \(\dfrac{6}{15}=\) \(\dfrac25\).
Answer to Question 3
1. After a red counter is removed there are 4 red and 3 blue among 7. After a blue counter is removed there are 5 red and 2 blue among 7. The numbers on the branches are as given.
2. \(\dfrac58\times\dfrac47=\dfrac{20}{56}=\) \(\dfrac{5}{14}\).
3. \(\dfrac38\times\dfrac27=\dfrac{6}{56}=\) \(\dfrac{3}{28}\).
4. Red then blue: \(\dfrac58\times\dfrac37=\dfrac{15}{56}\). Blue then red: \(\dfrac38\times\dfrac57=\dfrac{15}{56}\). Total \(=\dfrac{30}{56}=\) \(\dfrac{15}{28}\).
Answer to Question 4
1. After a red counter is taken there are 4 red and 3 blue left: \(\dfrac47\). The first counter is not replaced, so there are only 7 counters left for the second draw.
2. \(1-P(\text{both red})=1-\dfrac{5}{14}=\) \(\dfrac{9}{14}\). 'At least one blue' is the complement of 'both red', so we subtract from 1.
3. \(P(\text{both blue}\mid\text{at least one blue})=\dfrac{3/28}{9/14}=\dfrac{3}{28}\times\dfrac{14}{9}=\dfrac{42}{252}=\) \(\dfrac16\). Both blue is part of 'at least one blue', so we divide the probability of both blue by the probability of at least one blue.
Answer to Question 5
1. \(P(A\mid B)=\dfrac{0.2}{0.4}=\) 0.5. Divide the probability of both events by the probability of the condition \(B\).
2. \(P(B\mid A)=\dfrac{0.2}{0.5}=\) 0.4. This time the condition is \(A\), so we divide by \(P(A)=0.5\).
3. \(P(A)\times P(B)=0.5\times0.4=0.2=P(A\cap B)\), so yes, they are independent. Also \(P(A\mid B)=P(A)\). For independent events, knowing that \(B\) happened does not change the probability of \(A\).
Answer to Question 6
1. \(0.6\times0.25=\) 0.15. Multiply the probability of \(A\) by the probability of \(B\) given \(A\). The answer is 0.15.
2. \(\dfrac{0.12}{0.3}=\) 0.4. Rearrange the formula: \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). The answer is 0.4.
3. \(\dfrac{0.18}{0.45}=\) 0.4. Rearrange the formula: \(P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}\). The answer is 0.4.
Answer to Question 7
1. First branches: rain 0.3, no rain 0.7. Second branches after rain: late 0.4, on time 0.6. After no rain: late 0.1, on time 0.9. The probabilities on each pair of branches add up to 1.
2. \(0.3\times0.4=\) 0.12. Multiply along the branches rain then late.
3. \(0.12+0.7\times0.1=0.12+0.07=\) 0.19. Maya can be late in two ways: rain and late, or no rain and late. Add the two products.
4. \(P(\text{rain}\mid\text{late})=\dfrac{0.12}{0.19}=\) \(\dfrac{12}{19}\) \(\approx0.63\). The condition is that Maya is late, so we divide the 'rain and late' probability by the total probability of being late.
Answer to Question 8
1. First branches: disease 0.02, no disease 0.98. Then positive 0.95 or negative 0.05 after disease; positive 0.04 or negative 0.96 after no disease.
2. \(0.02\times0.95+0.98\times0.04=0.019+0.0392=\) 0.0582.
3. \(P(\text{disease}\mid\text{positive})=\dfrac{0.019}{0.0582}=\) 0.326.
4. Most people are healthy, so even a small false positive rate produces many more false positives than true positives.
Answer to Question 9
1. \(\dfrac7{12}\times\dfrac6{11}=\dfrac{42}{132}=\) \(\dfrac{7}{22}\).
2. Mint then toffee: \(\dfrac7{12}\times\dfrac5{11}=\dfrac{35}{132}\). Toffee then mint: \(\dfrac5{12}\times\dfrac7{11}=\dfrac{35}{132}\). Total \(=\dfrac{70}{132}=\) \(\dfrac{35}{66}\).
3. After taking a mint: 6 mints and 5 toffees in 11, so \(\dfrac5{11}\).
4. By symmetry the two ways are equally likely: \(\dfrac{35/132}{70/132}=\) \(\dfrac12\).
Answer to Question 10
1. Both 15, drama only \(35-15=20\), music only \(40-15=25\), neither \(100-60=40\).
2. \(\dfrac{40}{100}=\) 0.4.
3. \(\dfrac{15}{35}=\) \(\dfrac37\). Only the 35 drama pupils count, and 15 of them also go to music.
4. \(\dfrac{15}{40}=\) \(\dfrac38\). Only the 40 music pupils count, and 15 of them also go to drama.
5. \(P(\text{drama})\times P(\text{music})=0.35\times0.4=0.14\), but \(P(\text{both})=0.15\). These are different, so the events are not independent.
Answer to Question 11
1. First branches: scores first 0.55, concedes first 0.45. Then win 0.6 / not win 0.4 after scoring first; win 0.2 / not win 0.8 after conceding first.
2. \(0.55\times0.6+0.45\times0.2=0.33+0.09=\) 0.42.
3. \(\dfrac{0.33}{0.42}=\) \(\dfrac{11}{14}\) \(\approx0.786\).
4. \(40\times0.42=\) 16.8, so about 17 wins.
Answer to Question 12
1. There are 5 even cards: \(\dfrac5{10}\times\dfrac49=\) \(\dfrac29\).
2. The sum is even when both are even or both are odd. Both odd is also \(\dfrac5{10}\times\dfrac49=\dfrac29\). Total \(=\dfrac29+\dfrac29=\) \(\dfrac49\).
3. \(\dfrac{2/9}{4/9}=\) \(\dfrac12\). Both odd is one of the two equally likely ways to get an even sum.
Answer to Question 13
1. \(0.6\times0.7=\) 0.42. Multiply along the branches green then green.
2. \(0.42+0.4\times0.3=0.42+0.12=\) 0.54. The second light can be green in two ways, so we add the two products: green then green, and red then green.
3. \(P(\text{second red})=1-0.54=0.46\). The first green and second red has probability \(0.6\times0.3=0.18\). So \(\dfrac{0.18}{0.46}=\) \(\dfrac{9}{23}\) \(\approx0.39\). Divide the probability of 'first green and second red' by the probability of 'second red'.
Answer to Question 14
1. \(\dfrac{60+45}{200}=\dfrac{105}{200}=\) 0.525. Add the two Year groups in the 'has a pet' column, then divide by 200 pupils.
2. \(\dfrac{45}{105}=\) \(\dfrac37\). The condition 'has a pet' means only the 105 pupils with a pet count, and 45 of them are in Year 11.
3. \(\dfrac{60}{100}=\) 0.6. The condition 'Year 10' means only the 100 Year 10 pupils count, and 60 of them have a pet.
4. \(P(\text{pet})=0.525\) but \(P(\text{pet}\mid\text{Year 10})=0.6\). They are different, so the events are not independent. For independent events the conditional probability would equal the overall probability.
Answer to Question 15
1. First branches: A 0.5, B 0.5. After A: red \(\tfrac35\), blue \(\tfrac25\). After B: red \(\tfrac15\), blue \(\tfrac45\). The probabilities on each pair of branches add up to 1.
2. \(0.5\times0.6+0.5\times0.2=0.3+0.1=\) 0.4. A red counter can come from Bag A or from Bag B, so we add the two products.
3. \(\dfrac{0.3}{0.4}=\) 0.75. Divide the probability of 'Bag A and red' by the total probability of red.
4. \(200\times0.4=\) 80. Multiply the number of trials by the probability of red. We expect about 80 red counters.
Answer to Question 16
1. \(\dfrac6{10}\times\dfrac59\times\dfrac48=\dfrac{120}{720}=\dfrac16\).
2. \(1-\dfrac16=\) \(\dfrac56\).
3. One white can be in three positions: WBB, BWB, BBW. WBB \(=\dfrac4{10}\times\dfrac69\times\dfrac58=\dfrac{120}{720}\). The other two orders have the same probability, so the total is \(3\times\dfrac{120}{720}=\dfrac{360}{720}=\) \(\dfrac12\).
4. \(\dfrac4{10}\times\dfrac39\times\dfrac28=\dfrac{24}{720}=\) \(\dfrac1{30}\). Multiply the probabilities of three whites, reducing the number of white discs and the total each time.
Answer to Question 17
1. \(P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.4+x-0.4x\) (independent so \(P(A\cap B)=0.4x\)).
2. \(0.4+0.6x=0.7\), so \(0.6x=0.3\) and \(x=0.5\).
3. \(P(A\cap B)=0.4\times0.5=\) 0.2. \(P(A\mid B)=\dfrac{0.2}{0.5}=\) 0.4 (equal to \(P(A)\), as expected).
4. Exactly one: \(P(A\cup B)-P(A\cap B)=0.7-0.2=\) 0.5.
Answer to Question 18
1. \(\dfrac n{10}\times\dfrac{n-1}9=\dfrac{n(n-1)}{90}=\dfrac1{15}\), so \(n(n-1)=\dfrac{90}{15}=6\). Multiply the probabilities along the branches, cross-multiply, then simplify.
2. \(n^2-n-6=0\), \((n-3)(n+2)=0\). As \(n\) is positive, \(n=3\). Rearrange to make one side zero and factorise. A number of sweets cannot be negative.
3. Orange then other: \(\dfrac3{10}\times\dfrac79=\dfrac{21}{90}\). Other then orange: \(\dfrac7{10}\times\dfrac39=\dfrac{21}{90}\). Total \(=\dfrac{42}{90}=\) \(\dfrac7{15}\). Exactly one orange can happen in two orders, so we add the two products.
Answer to Question 19
1. First branches: A 0.6, B 0.4. After A: faulty 0.02. After B: faulty 0.05. The probabilities on each pair of branches add up to 1, so the 'not faulty' branches are 0.98 and 0.95.
2. \(0.6\times0.02+0.4\times0.05=0.012+0.020=\) 0.032. A faulty item can come from Machine A or from Machine B, so we add the two products.
3. \(\dfrac{0.020}{0.032}=\) 0.625 (or \(\dfrac58\)). Divide the probability of 'Machine B and faulty' by the total probability of faulty.
4. \(1000\times0.032=\) 32 items. Multiply the number of items by the probability of a fault.
Answer to Question 20
1. BB, BG, GB, GG.
2. At least one boy removes only GG, leaving BB, BG, GB. Only BB is "both boys": \(\dfrac13\).
3. The older child is a boy leaves BB and BG. Only BB has two boys: \(\dfrac12\).
4. The two pieces of information are different. "At least one boy" keeps three outcomes; "the older is a boy" keeps only two, because it also tells you which child is the boy. More specific information changes the conditional probability.

How to use these Year 11 conditional probability worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the conditional probability lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 11 conditional probability worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 11 conditional probability worksheet answers show every step, so nothing is left unexplained.

Conditional probability – Year 11 GCSE maths worksheet answers to download as PDF

Answer to Question 1

Answer to Question 2

Answer to Question 3

Answer to Question 4

Answer to Question 5

Answer to Question 6

Answer to Question 7

Answer to Question 8

Answer to Question 9

Answer to Question 10

Answer to Question 11

Answer to Question 12

Answer to Question 13

Answer to Question 14

Answer to Question 15

Answer to Question 16

Answer to Question 17

Answer to Question 18

Answer to Question 19

Answer to Question 20

How to use these Year 11 conditional probability worksheet answers

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