AQA GCSE Maths · 3.5

Probability

50 maths questions with worked answers with answers for 3.5 Probability in AQA GCSE Maths (spec 8300). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below. Not official AQA material.

All 50 questions and answers

Question 1 · Basic probability · P3
A fair six-sided die is rolled. What is P(rolling 4)?
Answer:
There is 1 favourable face out of 6 equally likely faces.
P(rolling 4) = 16.
Question 2 · Basic probability · P3
A fair die is rolled. What is P(an even number)?
Answer:
The even outcomes are 2, 4 and 6: three of six equally likely outcomes.
P(even) = 36 = 12.
Question 3 · Basic probability · P3
A fair coin is tossed. What is P(tails)?
Answer:
There is 1 tails outcome out of 2 equally likely outcomes.
P(tails) = 12.
Question 4 · Basic probability · P3
A bag has 3 red and 7 blue counters. One is selected at random. Find P(red).
Answer:
Total counters = 3 + 7 = 10.
P(red) = 310.
Question 5 · Basic probability · P3
What is the probability of an impossible event?
Answer:
An impossible event has no favourable outcomes.
Its probability is 0.
Question 6 · Basic probability · P3
What is the probability of a certain event?
Answer:
A certain event includes every possible outcome.
Its probability is 1.
Question 7 · Basic probability · P4
If P(rain) = 0.3, find P(no rain).
Answer:
Rain and no rain are complementary.
P(no rain) = 1 − 0.3 = 0.7.
Question 8 · Basic probability · P4
A spinner has only red, blue and green outcomes. P(red) = 0.2 and P(blue) = 0.5. Find P(green).
Answer:
The probabilities of all outcomes total 1.
P(green) = 1 − 0.2 − 0.5 = 0.3.
Question 9 · Basic probability · P3
Which is more likely: probability 25 or probability 0.45?
Answer:
Convert 25 to 0.4.
0.45 > 0.4, so 0.45 is more likely.
Question 10 · Basic probability · P3
A probability is 35%. Write it as a decimal.
Answer:
35% means 35100.
35 ÷ 100 = 0.35.
Question 11 · Frequency · P1
A coin lands heads 18 times in 40 tosses. Find the relative frequency of heads.
Answer:
Relative frequency = number of heads ÷ number of tosses.
1840 = 0.45.
Question 12 · Frequency · P2
A fair die is rolled 120 times. Find the expected number of sixes.
Answer:
The probability of a six is 16.
Expected number = 120 × 16 = 20.
Question 13 · Frequency · P2
A biased spinner has P(red) = 0.3. In 200 spins, how many reds are expected?
Answer:
Expected frequency = probability × number of spins.
0.3 × 200 = 60.
Question 14 · Frequency · P5
Why might 100 coin tosses not produce exactly 50 heads?
Answer:
The expected count for a fair coin is 100 × 12 = 50.
The actual result can vary by chance, so it need not be exactly 50.
Question 15 · Frequency · P5
Why can more independent, unbiased trials improve an estimate of probability?
Answer:
Calculate relative frequency as successes ÷ trials.
With many independent trials it tends to settle near the underlying probability, though a particular estimate can still differ.
Question 16 · Frequency · P1
A spinner gives 12 red results in 60 spins. Estimate P(red).
Answer:
Relative frequency = red results ÷ total spins.
1260 = 15 = 0.2.
Question 17 · Frequency · P1
Of 80 customers, 50 buy tea. How many do not buy tea?
Answer:
Subtract those who buy tea from the total.
80 − 50 = 30 customers.
Question 18 · Combined outcomes · P7
A fair coin is tossed twice. Find P(two heads).
Answer:
P(heads) = 12 on each independent toss.
P(two heads) = 12 × 12 = 14.
Question 19 · Combined outcomes · P7
A fair coin is tossed twice. Find P(exactly one head).
Answer:
The equally likely outcomes are HH, HT, TH and TT.
HT and TH have exactly one head, so P = 24 = 12.
Question 20 · Combined outcomes · P7
Two fair dice are rolled. Find P(total 7).
Answer:
There are 36 equally likely ordered outcomes.
Six total 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).
P = 636 = 16.
Question 21 · Combined outcomes · P7
Two fair dice are rolled. Find P(two sixes).
Answer:
There are 36 equally likely ordered outcomes; only (6, 6) has two sixes.
P = 136.
Question 22 · Combined outcomes · P7
A fair die and fair coin are used independently. Find P(4 and heads).
Answer:
P(4) = 16 and P(heads) = 12.
The events are independent, so multiply: 16 × 12 = 112.
Question 23 · Combined outcomes · P7
Two fair coins are tossed. List all equally likely outcomes.
Answer:
For first toss H the outcomes are HH and HT.
For first toss T the outcomes are TH and TT.
Outcomes: HH, HT, TH, TT.
Question 24 · Combined outcomes · P7
Two fair dice are rolled. Find P(total 3).
Answer:
There are 36 equally likely ordered outcomes.
The total 3 occurs for (1, 2) or (2, 1).
P = 236 = 118.
Question 25 · Combined outcomes · P7
Two fair dice are rolled. Find P(a double).
Answer:
There are 36 equally likely ordered outcomes and 6 doubles: (1, 1) to (6, 6).
P = 636 = 16.
Question 26 · Independent events · P8
Events A and B are independent, with P(A) = 13 and P(B) = 12. Find P(A and B).
Answer:
For independent events, multiply their probabilities.
P(A and B) = 13 × 12 = 16.
Question 27 · Independent events · P8
A fair coin is tossed three times. Find P(three heads).
Answer:
Each independent head has probability 12.
P(three heads) = 12 × 12 × 12 = 18.
Question 28 · Independent events · P8
The chance of a machine passing a test is 0.9 each time independently. Find P(passing both tests).
Answer:
The two tests are independent.
P(both pass) = 0.9 × 0.9 = 0.81.
Question 29 · Independent events · P8
A fair die is rolled twice. Find P(no six on either roll).
Answer:
P(no six on one roll) = 56.
The rolls are independent, so P = 56 × 56 = 2536.
Question 30 · Independent events · P8
A fair coin is tossed twice. Find P(at least one head).
Answer:
The complement of at least one head is no heads, which means TT.
P(TT) = 14, so P(at least one head) = 1 − 14 = 34.
Question 31 · Dependent events · P8
A bag has 3 red and 2 blue counters. Two are selected without replacement. Find P(two red).
Answer:
P(first red) = 35; after a red, P(second red) = 24.
Multiply: 35 × 24 = 310.
Question 32 · Dependent events · P8
A bag has 4 red and 1 blue counter. Two are selected without replacement. Find P(red then blue).
Answer:
P(first red) = 45; then 1 blue remains among 4 counters.
P(red then blue) = 45 × 14 = 15.
Question 33 · Dependent events · P8
A bag has 2 red and 3 blue counters. Two are selected without replacement. Find P(two blue).
Answer:
P(first blue) = 35; then 2 blue remain among 4 counters.
P(two blue) = 35 × 24 = 310.
Question 34 · Dependent events · P8
A bag has 2 red and 3 blue counters. Two are selected without replacement. Find P(one red and one blue).
Answer:
P(red then blue) = 25 × 34 = 310.
P(blue then red) = 35 × 24 = 310.
Add the separate orders: 310 + 310 = 35.
Question 35 · Dependent events · P8
A bag has 5 green and 5 yellow counters. Two are selected without replacement. Find P(two yellow).
Answer:
P(first yellow) = 510; then 4 yellow remain among 9 counters.
P(two yellow) = 510 × 49 = 29.
Question 36 · Dependent events · P8
A bag has 2 red and 3 blue counters. Two are selected with replacement. Find P(two red).
Answer:
Replacement restores the original bag before the second pick.
P(two red) = 25 × 25 = 425.
Question 37 · Mutual exclusivity · P4
On a fair die, find P(1 or 2).
Answer:
A roll cannot be both 1 and 2, so add their probabilities.
16 + 16 = 26 = 13.
Question 38 · Mutual exclusivity · P4
A spinner lands on red with probability 0.2 and blue with probability 0.35. Find P(red or blue).
Answer:
On a single spin the red and blue outcomes are mutually exclusive.
P(red or blue) = 0.2 + 0.35 = 0.55.
Question 39 · Mutual exclusivity · P4
Can mutually exclusive events both happen on the same trial?
Answer:
Mutually exclusive means that the events have no common outcomes.
They cannot both happen on the same trial.
Question 40 · Sets · P6
In a class of 30, 12 study art, 10 study music and 4 study both. How many study art or music?
Answer:
Adding 12 and 10 counts the 4 students studying both twice.
Subtract the overlap: 12 + 10 − 4 = 18.
Question 41 · Sets · P6
In a class of 30, 18 students study art or music. How many study neither?
Answer:
Everyone is either in the art-or-music group or neither.
30 − 18 = 12 study neither.
Question 42 · Sets · P6
A set A has 8 members and B has 6; they share 3. How many are in A union B?
Answer:
Add the set sizes and subtract the 3 counted twice.
8 + 6 − 3 = 11.
Question 43 · Conditional · P9 · Higher only
A group has 12 girls and 8 boys. Five girls wear glasses. A girl is chosen at random. Find P(glasses).
Answer:
The condition restricts the sample to the 12 girls.
Five of those 12 wear glasses, so P = 512.
Question 44 · Conditional · P9 · Higher only
Of 40 students, 24 take music; 6 of the music students also take drama. Find P(drama given music).
Answer:
The condition restricts the sample to the 24 music students.
Six also take drama, so P = 624 = 14.
Question 45 · Conditional · P9 · Higher only
A bag has 3 red and 2 blue counters. Given that the first pick was red, find P(red on the second pick without replacement).
Answer:
After the first red is removed, 2 red remain among 4 counters.
P(second red given first red) = 24 = 12.
Question 46 · Conditional · P9 · Higher only
A survey finds 18 adults, including 12 drivers. Given the person is an adult, find P(driver).
Answer:
The condition restricts the sample to the 18 adults.
12 are drivers, so P = 1218 = 23.
Question 47 · Conditional · P9 · Higher only
A box has 4 faulty and 16 good bulbs. Two are chosen without replacement. Given the first was good, find P(the second is faulty).
Answer:
After a good bulb is removed, 19 bulbs remain, including all 4 faulty bulbs.
P(second faulty given first good) = 419.
Question 48 · Error check · P8
A student adds 12 + 12 for P(two heads on two independent fair tosses). Correct it.
Answer:
Both heads must happen, so multiply independent probabilities.
12 × 12 = 14.
Question 49 · Error check · P8
Why do probabilities change after a counter is removed without replacement?
Answer:
Removing a counter reduces the total by one.
If it is a favourable counter, that count also falls; the fraction used for the next probability can therefore change.
Question 50 · Error check · P3
A student says a probability of 1.3 is possible. Correct them.
Answer:
Every probability lies between 0 and 1 inclusive.
1.3 > 1, so it cannot be a probability.

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