S3.1Mutually exclusive and independent events; Venn and tree diagrams
Edexcel A level Maths (9MA0) · Statistics › Probability
Practise Mutually exclusive and independent events; Venn and tree diagrams. 1 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 1Easy5 marks
A survey of 40 members of a gym records whether they use the swimming pool and whether they take fitness classes. \(P\) is the event that a randomly chosen member uses the swimming pool and \(C\) is the event that a randomly chosen member takes fitness classes. The Venn diagram shows the number of members in each region.
[object Object]
(a) Find \(P(P)\).[1]
(b) Find \(P(P \cup C)\).[2]
(c) Find the probability that a randomly chosen member is in exactly one of the events \(P\) and \(C\).[2]
Show the answer and mark scheme
(a)Answer: \(\frac{17}{40} = 0.425\)
B1 for \(\frac{17}{40}\) (or 0.425)
(b)Answer: \(\frac{33}{40} = 0.825\)
M1 for \(\dfrac{4 + 13 + 16}{40}\) or \(1 - \dfrac{7}{40}\)
A1 for \(\frac{33}{40}\) (or 0.825)
(c)Answer: \(\frac{1}{2} = 0.5\)
M1 for \(\dfrac{4 + 16}{40}\)
A1 for \(\frac{1}{2}\) (or 0.5)
Question 2Medium8 marks
For the events \(A\) and \(B\), \(P(A) = 0.4\), \(P(B) = 0.5\) and \(P(A \cup B) = 0.7\).
(a) Show that \(A\) and \(B\) are independent.[2]
(b) Find \(P(A \mid B')\).[2]
(c) Prove that, for any two independent events \(C\) and \(D\), the events \(C'\) and \(D'\) are also independent.[3]
(d) A student says: “\(A\) and \(B\) are independent, so they cannot both happen. This means they are mutually exclusive.” Explain why the student is wrong.[1]
M1 for \(P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.2\)
A1* for \(0.2 = 0.4 \times 0.5 = P(A)P(B)\), so independent
Worked solution: \(P(A \cap B) = 0.4 + 0.5 - 0.7 = 0.2\), and \(P(A)P(B) = 0.4 \times 0.5 = 0.2\). These are equal, so \(A\) and \(B\) are independent.
(b)Answer: 0.4
M1 for \(\frac{P(A \cap B')}{P(B')} = \frac{0.4 - 0.2}{0.5}\)
A1 for 0.4
Worked solution: \(P(A \cap B') = 0.4 - 0.2 = 0.2\), so \(P(A \mid B') = \frac{0.2}{0.5} = 0.4\) (equal to \(P(A)\), as expected for independent events).
(d)Answer: \(P(A \cap B) = 0.2 \ne 0\), so \(A\) and \(B\) can happen together; independence means one event does not change the probability of the other.
B1 for \(P(A \cap B) = 0.2 \ne 0\) so they are not mutually exclusive (independent events with non-zero probabilities can occur together)
Worked solution: Mutually exclusive events have \(P(A \cap B) = 0\), but here \(P(A \cap B) = 0.2\). Independence means that knowing \(B\) has happened does not change the probability of \(A\); it does not mean they cannot happen together.
Question 3Hard7 marks
A bag contains 11 balls. Each ball is either yellow or white, and there are fewer yellow balls than white balls. Two balls are taken at random from the bag, without replacement. The probability that the two balls are different colours is \(\frac{24}{55}\).
(a) Find the number of yellow balls in the bag.[5]
(b) Find the probability that both balls taken are yellow.[2]
Show the answer and mark scheme
(a)Answer: 3 yellow balls
M1 for \(2 \times \dfrac{x}{11} \times \dfrac{11 - x}{10}\) (two orders), with \(x\) the number of yellow balls
A1 for \(\dfrac{2x(11 - x)}{110} = \frac{24}{55}\)
M1 for rearranging to a 3-term quadratic, \(x^2 - 11x + 24 = 0\)
A1 for \(x = 3\) or \(x = 8\)
A1 for \(x = 3\) only, since there are fewer yellow balls than white balls