Probabilities Summing to 1
Question 11 mark
A spinner is biased.
The probability that the spinner lands on red is 0.36
Work out the probability that the spinner does not land on red.
Hint
The spinner either lands on red or it does not, so the two probabilities add up to 1.
Worked solution
A spinner is biased.
The probability that the spinner lands on red is 0.36
Work out the probability that the spinner does not land on red.
- P(not red) =1- P(red).
- 1-0.36=0.64.
Question 21 mark
The probability that Amir wins a game is \frac{3}{8}
Work out the probability that Amir does not win the game.
Hint
Write 1 as a fraction with the same denominator.
Worked solution
The probability that Amir wins a game is \frac{3}{8}
Work out the probability that Amir does not win the game.
- P(does not win) =1-\frac{3}{8}.
- 1=\frac{8}{8}, so \frac{8}{8}-\frac{3}{8}=\frac{5}{8}.
Question 31 mark
The probability that it will rain tomorrow is 35\%
What is the probability that it will not rain tomorrow? Give your answer as a percentage.
Hint
As percentages, the probabilities add up to 100%.
Worked solution
The probability that it will rain tomorrow is 35\%
What is the probability that it will not rain tomorrow? Give your answer as a percentage.
- It will either rain or not rain, so the two probabilities add up to 100\%.
- 100\%-35\%=65\%.
Question 42 marks
A bag contains only red, blue and green counters.
Ali takes a counter at random from the bag. The table shows the probability that the counter is red and the probability that it is blue.
Work out the probability that the counter is green.
Hint
The probabilities for the three colours must add up to 1.
Worked solution
A bag contains only red, blue and green counters.
Ali takes a counter at random from the bag. The table shows the probability that the counter is red and the probability that it is blue.
Work out the probability that the counter is green.
- Red and blue together: 0.2+0.45=0.65.
- Green: 1-0.65=0.35.
Question 52 marks
Here is a biased 4-sided spinner numbered 1, 2, 3 and 4.
The table shows the probabilities that the spinner will land on 1, on 2 and on 3.
Work out the probability that the spinner will land on 4.
Hint
Write all the fractions with the same denominator before you add them.
Worked solution
Here is a biased 4-sided spinner numbered 1, 2, 3 and 4.
The table shows the probabilities that the spinner will land on 1, on 2 and on 3.
Work out the probability that the spinner will land on 4.
- Write \frac{1}{4} as \frac{2}{8}.
- Landing on 1, 2 or 3: \frac{1}{8}+\frac{3}{8}+\frac{2}{8}=\frac{6}{8}.
- Landing on 4: 1-\frac{6}{8}=\frac{2}{8}=\frac{1}{4}.
Question 62 marks
Each student in a class travels to school in only one way.
A student from the class is picked at random.
The probability that the student walks to school is 0.38
The probability that the student cycles to school is 0.17
Work out the probability that the student does not walk and does not cycle to school.
Hint
Add the two probabilities you are given, then take the total away from 1.
Worked solution
Each student in a class travels to school in only one way.
A student from the class is picked at random.
The probability that the student walks to school is 0.38
The probability that the student cycles to school is 0.17
Work out the probability that the student does not walk and does not cycle to school.
- Walks or cycles: 0.38+0.17=0.55.
- Neither: 1-0.55=0.45.
Question 72 marks
Every morning a bus can be early, on time, late or cancelled.
The table shows the probability that the bus is early, the probability that it is on time and the probability that it is late.
Work out the probability that the bus is cancelled.
Hint
The four probabilities add up to 1. Line up the decimal points when you add.
Worked solution
Every morning a bus can be early, on time, late or cancelled.
The table shows the probability that the bus is early, the probability that it is on time and the probability that it is late.
Work out the probability that the bus is cancelled.
- Early, on time or late: 0.08+0.65+0.24=0.97.
- Cancelled: 1-0.97=0.03.
Question 83 marks
A 6-sided dice is biased.
The table shows the probability that the dice will land on each number, except 4
Work out the probability that the dice will land on 4
2 marks
The dice is rolled once. Work out the probability that it lands on a number greater than 3
1 mark
Hint
The probabilities for all six numbers add up to 1.
Worked solution
A 6-sided dice is biased.
The table shows the probability that the dice will land on each number, except 4
Part (a)
Work out the probability that the dice will land on 4
- The other five probabilities: 0.1+0.2+0.15+0.25+0.1=0.8.
- Landing on 4: 1-0.8=0.2.
Part (b)
The dice is rolled once. Work out the probability that it lands on a number greater than 3
- The numbers greater than 3 are 4, 5 and 6.
- 0.2+0.25+0.1=0.55.
Question 91 mark
A spinner can land on A, on B or on C.
Four students each write down what they think the probabilities are.
Which list could be correct?
Hint
The probabilities must add up to exactly 1, and no probability can be less than 0.
Worked solution
A spinner can land on A, on B or on C.
Four students each write down what they think the probabilities are.
Which list could be correct?
- A: 0.3, B: 0.4, C: 0.4
- A: 0.25, B: 0.25, C: 0.25
- A: 0.5, B: 0.2, C: 0.3
- A: 0.6, B: 0.5, C: -0.1
- 0.3+0.4+0.4=1.1, which is more than 1.
- 0.25+0.25+0.25=0.75, which is less than 1.
- 0.5+0.2+0.3=1, and none of them is negative, so this list could be correct.
- A probability cannot be negative, so -0.1 is not possible.
Question 101 mark
A student in a class is picked at random.
The probability that the student has a dog is 0.4
The probability that the student has a cat is 0.3
Jack says, "The probability that the student has a dog or a cat is 0.4+0.3=0.7"
Why might Jack be wrong?
Hint
Think about a student who has both pets.
Worked solution
A student in a class is picked at random.
The probability that the student has a dog is 0.4
The probability that the student has a cat is 0.3
Jack says, "The probability that the student has a dog or a cat is 0.4+0.3=0.7"
Why might Jack be wrong?
- You should multiply the probabilities, not add them.
- Some students might have a dog and a cat, so they are counted twice.
- 0.7 is more than 0.5, so it is too big to be a probability.
- The probabilities must add up to 1, so the answer should be 0.3
- Having a dog and having a cat can both happen to the same student.
- Those students are included in the 0.4 and in the 0.3, so adding counts them twice.
- You can only add probabilities like this when the two things cannot happen together.
Question 113 marks
A spinner is biased. It can land on red, blue, yellow or green.
The probability that it lands on red is 0.35
The probability that it lands on blue is 0.17
The probability that it lands on yellow is the same as the probability that it lands on green.
Work out the probability that the spinner lands on yellow.
Hint
Find what is left for yellow and green together, then share it equally.
Worked solution
A spinner is biased. It can land on red, blue, yellow or green.
The probability that it lands on red is 0.35
The probability that it lands on blue is 0.17
The probability that it lands on yellow is the same as the probability that it lands on green.
Work out the probability that the spinner lands on yellow.
- Red or blue: 0.35+0.17=0.52.
- Yellow and green together: 1-0.52=0.48.
- Yellow: 0.48\div2=0.24.
Question 123 marks
In a game at a school fair, every player wins one prize: a pen, a badge, a sweet or a bookmark.
The probability of winning a pen is 0.2
The probability of winning a badge is 0.35
The probability of winning a sweet is twice the probability of winning a bookmark.
Work out the probability of winning a sweet.
Hint
Find the probability left for a sweet or a bookmark. A sweet gets 2 shares of it and a bookmark gets 1.
Worked solution
In a game at a school fair, every player wins one prize: a pen, a badge, a sweet or a bookmark.
The probability of winning a pen is 0.2
The probability of winning a badge is 0.35
The probability of winning a sweet is twice the probability of winning a bookmark.
Work out the probability of winning a sweet.
- Pen or badge: 0.2+0.35=0.55.
- Sweet or bookmark: 1-0.55=0.45.
- Split 0.45 into 3 equal shares: 0.45\div3=0.15 each.
- Bookmark: 1 share =0.15. Sweet: 2 shares =0.3.
Question 132 marks
A bag contains only red, blue, green and white beads.
A bead is taken at random from the bag. The probability that it is red is \frac{2}{5}
The probabilities that the bead is blue, green or white are all the same.
Work out the probability that the bead is green. Give your answer as a fraction.
Hint
Find the probability that the bead is not red, then share it equally between three colours.
Worked solution
A bag contains only red, blue, green and white beads.
A bead is taken at random from the bag. The probability that it is red is \frac{2}{5}
The probabilities that the bead is blue, green or white are all the same.
Work out the probability that the bead is green. Give your answer as a fraction.
- Not red: 1-\frac{2}{5}=\frac{3}{5}.
- Blue, green and white share \frac{3}{5} equally.
- Green: \frac{3}{5}\div3=\frac{1}{5}.
Question 143 marks
When Zak takes a shot in basketball, he can score 3 points, score 2 points or miss.
The probability that he scores 3 points is 0.15
The probability that he scores 2 points is 0.45
Work out the probability that Zak misses.
1 mark
Zak takes 60 shots. Work out an estimate for the number of times he misses.
2 marks
Hint
For (b), multiply the probability of a miss by the number of shots.
Worked solution
When Zak takes a shot in basketball, he can score 3 points, score 2 points or miss.
The probability that he scores 3 points is 0.15
The probability that he scores 2 points is 0.45
Part (a)
Work out the probability that Zak misses.
- Scores: 0.15+0.45=0.6.
- Misses: 1-0.6=0.4.
Part (b)
Zak takes 60 shots. Work out an estimate for the number of times he misses.
- Estimate = probability \times number of shots.
- 0.4\times60=24.
Question 153 marks
A box contains only milk chocolates, dark chocolates and white chocolates.
A chocolate is taken at random from the box.
The probability that it is a milk chocolate is 0.45
The probability that it is a dark chocolate is 0.3
There are 14 white chocolates in the box. Work out the total number of chocolates in the box.
Hint
First find the probability of taking a white chocolate.
Worked solution
A box contains only milk chocolates, dark chocolates and white chocolates.
A chocolate is taken at random from the box.
The probability that it is a milk chocolate is 0.45
The probability that it is a dark chocolate is 0.3
There are 14 white chocolates in the box. Work out the total number of chocolates in the box.
- White: 1-(0.45+0.3)=1-0.75=0.25.
- So 0.25 of the chocolates are white, and 0.25=\frac{1}{4}.
- 14 is \frac{1}{4} of the total, so the total is 14\times4=56 chocolates.
Question 16Challenge5 marks
Every student at a school has one type of lunch: a hot meal, a sandwich, a salad or a packed lunch from home.
A student is picked at random.
The probability that the student has a hot meal is 0.28
The probability that the student has a sandwich is 0.12
number of students who have a salad : number of students who have a packed lunch =3:2
Work out the probability that the student has a salad.
3 marks
70 students have a hot meal. How many students have a packed lunch?
2 marks
Hint
Find the probability left for a salad or a packed lunch, then share it in the ratio 3 : 2.
Worked solution
Every student at a school has one type of lunch: a hot meal, a sandwich, a salad or a packed lunch from home.
A student is picked at random.
The probability that the student has a hot meal is 0.28
The probability that the student has a sandwich is 0.12
number of students who have a salad : number of students who have a packed lunch =3:2
Part (a)
Work out the probability that the student has a salad.
- Hot meal or sandwich: 0.28+0.12=0.4.
- Salad or packed lunch: 1-0.4=0.6.
- Share 0.6 in the ratio 3:2: 3+2=5 parts, 0.6\div5=0.12 for each part.
- Salad: 3\times0.12=0.36.
Part (b)
70 students have a hot meal. How many students have a packed lunch?
- Packed lunch: 2\times0.12=0.24.
- 0.28 of the students is 70, so the total is 70\div0.28=250 students.
- Packed lunch: 0.24\times250=60 students.
Question 17Challenge4 marks
A football team plays a match. The team can win, draw or lose.
The probability that the team wins is 0.45
The probability that the team loses is 0.15 more than the probability that the team draws.
Work out the probability that the team loses.
3 marks
Work out the probability that the team does not draw.
1 mark
Hint
Draw and lose together have probability 1 − 0.45. Try taking the difference away first, then share what is left equally.
Worked solution
A football team plays a match. The team can win, draw or lose.
The probability that the team wins is 0.45
The probability that the team loses is 0.15 more than the probability that the team draws.
Part (a)
Work out the probability that the team loses.
- Draw or lose: 1-0.45=0.55.
- Take away the difference: 0.55-0.15=0.4, then share equally: 0.4\div2=0.2. So the draw is 0.2.
- Lose: 0.2+0.15=0.35.
- Check: 0.45+0.2+0.35=1.
Part (b)
Work out the probability that the team does not draw.
- Draw: 0.2.
- Does not draw: 1-0.2=0.8.
Question 18Challenge3 marks
A spinner can land on A, B, C or D.
Ellie writes down the probabilities shown in the table. The probabilities for A, B and C are correct. The probability for D is wrong.
How can you tell that Ellie's table is wrong?
1 mark
Work out the correct probability for D. Give your answer as a decimal.
2 marks
Hint
Write all the probabilities in the same form, as decimals, before you add them.
Worked solution
A spinner can land on A, B, C or D.
Ellie writes down the probabilities shown in the table. The probabilities for A, B and C are correct. The probability for D is wrong.
Part (a)
How can you tell that Ellie's table is wrong?
- The probabilities are not all written in the same way.
- The probabilities add up to 0.7, which is less than 1
- The probabilities add up to 1.1, which is more than 1
- D has the biggest probability, so it is wrong.
- Write them all as decimals: \frac{1}{4}=0.25 and 15\%=0.15.
- 0.3+0.25+0.15+0.4=1.1.
- The probabilities should add up to 1, but they add up to 1.1.
Part (b)
Work out the correct probability for D. Give your answer as a decimal.
- A, B and C: 0.3+0.25+0.15=0.7.
- D: 1-0.7=0.3.
Question 19Challenge5 marks
Every customer at a café buys exactly one drink: tea, coffee, hot chocolate or juice.
The probability that a customer buys tea is 0.3
The probability that a customer buys coffee is 0.5
The café expects 240 customers tomorrow.
The manager first guesses that hot chocolate and juice are equally likely.
Using this guess, work out an estimate for the number of customers who will buy juice tomorrow.
3 marks
In fact, the probability that a customer buys hot chocolate is 3 times the probability that a customer buys juice.
Work out a new estimate for the number of customers who will buy juice tomorrow.
2 marks
Hint
Find the probability left for hot chocolate and juice together first.
Worked solution
Every customer at a café buys exactly one drink: tea, coffee, hot chocolate or juice.
The probability that a customer buys tea is 0.3
The probability that a customer buys coffee is 0.5
The café expects 240 customers tomorrow.
Part (a)
The manager first guesses that hot chocolate and juice are equally likely.
Using this guess, work out an estimate for the number of customers who will buy juice tomorrow.
- Tea or coffee: 0.3+0.5=0.8.
- Hot chocolate or juice: 1-0.8=0.2, so each is 0.2\div2=0.1.
- Estimate: 0.1\times240=24 customers.
Part (b)
In fact, the probability that a customer buys hot chocolate is 3 times the probability that a customer buys juice.
Work out a new estimate for the number of customers who will buy juice tomorrow.
- Share 0.2 so that hot chocolate gets 3 shares and juice gets 1 share: 3+1=4 shares.
- Juice: 0.2\div4=0.05.
- Estimate: 0.05\times240=12 customers.
Question 20Challenge4 marks
A jar contains only lemon sweets, orange sweets and lime sweets.
A sweet is taken at random from the jar.
The probability that it is an orange sweet is 0.55
The probability that it is a lime sweet is twice the probability that it is a lemon sweet.
There are 12 lime sweets in the jar. Work out the total number of sweets in the jar.
Hint
Work out the probability of taking a lime sweet first.
Worked solution
A jar contains only lemon sweets, orange sweets and lime sweets.
A sweet is taken at random from the jar.
The probability that it is an orange sweet is 0.55
The probability that it is a lime sweet is twice the probability that it is a lemon sweet.
There are 12 lime sweets in the jar. Work out the total number of sweets in the jar.
- Lemon or lime: 1-0.55=0.45.
- Lime gets 2 shares and lemon gets 1 share: 0.45\div3=0.15, so lemon is 0.15 and lime is 2\times0.15=0.3.
- 0.3=\frac{3}{10}, so 12 lime sweets are \frac{3}{10} of the sweets.
- \frac{1}{10} of the sweets is 12\div3=4, so the total is 4\times10=40 sweets.