Edexcel GCSE Maths (1MA1), Higher tier · Probability
Practise Tree diagrams and combined events. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
A school hockey team plays a match on Saturday. A school netball team plays a match on Sunday. The probability that the hockey team wins is 0.6. The probability that the netball team wins is 0.3. The two results are independent. The probability tree diagram shows this information.
[object Object]
(a) Work out the probability that both teams win.[2]
(b) Work out the probability that neither team wins.[2]
Show the answer and mark scheme
(a)Answer: \(0.18\)
M1 for \(0.6 \times 0.3\)
A1 for \(0.18\) oe
Worked solution: \(0.6 \times 0.3 = 0.18\)
(b)Answer: \(0.28\)
M1 for \(0.4 \times 0.7\)
A1 for \(0.28\) oe
Worked solution: \(0.4 \times 0.7 = 0.28\)
Question 2Medium3 marks
A fair six-sided dice is rolled twice. Dan says, 'The probability of getting at least one 6 is \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\).'
(a) Explain why Dan's method is wrong.[1]
(b) Work out the correct probability of getting at least one 6.[2]
Show the answer and mark scheme
(a)Answer: A 6 on the first roll and a 6 on the second roll can both happen, so the events are not mutually exclusive: adding counts (6, 6) twice.
C1 for explaining that the two events are not mutually exclusive (both can happen), so the probabilities cannot simply be added / (6, 6) is counted twice
Worked solution: The outcome (6, 6) is included in both \(\frac{1}{6}\)s, so adding them counts it twice.
(b)Answer: \(\frac{11}{36}\)
M1 for \(1 - \left(\frac{5}{6}\right)^2\) or \(\frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{6}\) or 11 outcomes out of 36 listed
A1 for \(\frac{11}{36}\) oe
Worked solution: P(no 6) \(= \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}\), so P(at least one 6) \(= 1 - \frac{25}{36} = \frac{11}{36}\).
Question 3Hard4 marks
Sam has a biased coin. The probability that the coin lands on heads is \(\frac{4}{5}\). Sam throws the coin three times.
(a) Work out the probability that the coin lands on heads at least once.[3]
(b) State one assumption you made in your calculation.[1]
Show the answer and mark scheme
(a)Answer: \(\frac{124}{125}\)
M1 for the probability of not getting heads on one throw, \(\frac{1}{5}\)
M1 for \(1 - \left(\frac{1}{5}\right)^{3}\)
A1 for \(\frac{124}{125}\) oe
Worked solution: The only way not to get heads at least once is to miss it on all three throws: \(\left(\frac{1}{5}\right)^{3} = \frac{1}{125}\). So \(P(\text{at least once}) = 1 - \frac{1}{125} = \frac{124}{125}\).
(b)Answer: The three throws are independent (the probability is the same each time).
C1 for stating that the throws are independent, or that the probability of heads stays the same for every throw
Worked solution: Multiplying probabilities for the three throws assumes they are independent.