11 A student attempts two tasks. The result of each task is either “Pass” or “No pass”.
The probability of the student passing the first task is 0.6. The probability of the student passing the second task is \(x\).
(a) Complete the tree diagram. [2]
(b) Write down the mathematical assumption that has been made about the two tasks. [1]
(c) The probability of the student passing just one of these two tasks is 0.528.
Work out the value of \(x\). [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
Correct tree diagram
2
B1 for 0.4 on missing branch in task 1 B1 for \(1 - x\) on both missing branches in task 2
If there is more than one answer on any branch, choose the one on the dotted line first
Mark scheme (b)
Answer
Marks
Part marks and guidance
They are independent oe
1
Pick the best comment, see appendix
Appendix: exemplar responses for Q11(b)
Response
Mark
The pass before does not affect the results after
1
They are independent
1
The first task results do not affect the second
1
Passing first task does not affect the second task
1
Passing the second does not rely on passing the first
1
They are not linked
1
Same probability to pass second on both times
0
The probabilities of the tests do not change
0
The probability of passing will always stay the same
0
Both second tests are the same
0
Mark scheme (c)
Answer
Marks
Part marks and guidance
0.36
4
M2 for \(0.6(1 - x) + 0.4x = 0.528\) or better or M1 for \(0.6(1 - x) + 0.4x\) or better M1 for rearranging their linear equation, \(kx + a = b\), to make \(kx\) the subject (\(0 \lt k\))
e.g. M2 for \(0.6 - 0.2x = 0.528\) e.g. M1 for \(0.6 - 0.2x\) e.g. \(0.6 - 0.528 = 0.2x\) or better allow similar equations involving 1–0.528 e.g \(0.6x + 0.4(1 - x) = 0.472\)
OR M2 for \(1 - 0.6x - 0.4(1 - x) = 0.528\) or better or M1 for \(1 - 0.6x - 0.4(1 - x)\) or better M1 for rearranging their linear equation, \(kx + a = b\), to make \(kx\) the subject (\(0 \lt k\))
e.g. \(0.6 - 0.528 = 0.2x\) or better
OR FTtheir(a) if not correct for up to 3 marks e.g. if in (a) \(1 - x\) is replaced with e.g. 0.4 then M2 for \(0.6 \times 0.4 + 0.4x = 0.528\) or better or M1 for \(0.6 \times 0.4 + 0.4x\) or better M1 for rearranging their linear equation, \(kx + a = b\), to make \(kx\) the subject (\(0 \lt k\))
e.g. \(0.4x = 0.528 - 0.6 \times 0.4\) or better
alternative method using trials M1 for each correct trial up to M3 and the correct answer from trials scores 4 marks, see appendix
If 0 scored SC1 for any 2 branches correctly written down and added e.g. \(0.6(1 - x) + 0.4(1 - x)\)
