Probability: experiments, Venn and tree diagrams – Year 11 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(1-0.4-0.3=0.3\), shared equally: 0.15 (1 for 0.3, 1 for 0.15) The four probabilities add up to 1, so green and yellow share what is left equally.
   2. \(200\times0.15=\) 30 (1) Yellow has the same probability as green, 0.15. Expected frequency \(=\) trials \(\times\) probability.
   3. \(1-0.3=\) 0.7 (1) 'Not blue' is the complement of 'blue'.

   Check (a): \(0.4+0.3+0.15+0.15=1\). In (b), 30 is an expected value, not a guaranteed result.

2. Answer to Question 2 (4 marks)

   1. Bike only 12, both 6, scooter only 9, neither 13 (2) Start with the overlap of 6, then subtract it from each total: \(18-6=12\) and \(15-6=9\). The rest, \(40-27=13\), own neither.
   2. \(\dfrac{13}{40}\) (1) Those who own neither are outside both circles: 13 out of 40.
   3. \(\dfrac6{15}=\dfrac25\) (1) Only the 15 scooter owners count, and 6 of them own a bike.

3. Answer to Question 3 (4 marks)

   1. \(0.2\times0.2=\) 0.04 (1) The days are independent, so multiply the probabilities.
   2. Late then on time \(0.2\times0.8=0.16\) (1); on time then late \(0.16\) (1); total 0.32 (1) Being on time has probability \(1-0.2=0.8\). Exactly one late day can happen in two orders, so add the two products.

   A common error in (b) is to give only 0.16, forgetting the second order. Check: \(0.04+0.32+0.64=1\), where 0.64 is the chance of being on time on both days.

4. Answer to Question 4 (4 marks)

   1. \(\dfrac3{10}\times\dfrac29=\dfrac6{90}=\) \(\dfrac1{15}\) (1 for each product step) The counter is not replaced, so the second white has probability \(\tfrac29\). Multiply along the branches.
   2. \(1-\dfrac1{15}=\) \(\dfrac{14}{15}\) (2) 'At least one red' is the complement of 'both white', so subtract from 1.

   In (b), the long method also works: \(P(RR)+P(RW)+P(WR)=\tfrac{42}{90}+\tfrac{21}{90}+\tfrac{21}{90}=\tfrac{84}{90}=\tfrac{14}{15}\). A common error is to use \(\tfrac3{10}\times\tfrac3{10}\) in (a), as if the counter were replaced.

5. Answer to Question 5 (4 marks)

   1. \(\left(\dfrac3{3+x}\right)^2=\dfrac9{25}\) (1) There are \(3+x\) counters, and the counter is replaced, so the probability of blue is the same both times.
   2. \(\dfrac3{3+x}=\dfrac35\) (1), so \(3+x=5\) (1), \(x=2\) (1) Take the square root of both sides, then cross-multiply. Check: \(\left(\tfrac35\right)^2=\tfrac9{25}\). There are 2 red counters.

   In (b), only the positive square root makes sense, since a probability cannot be negative. With 3 blue and 2 red counters, \(P(\text{blue})=\tfrac35\) each time, confirming the answer.

Total : 20 marks