Unit 8
9 marks · NormalThis question connects the definition of a function (Algebra 8) and basic probability with tree diagrams (Handling Data 5).
Consider the following mapping diagrams:
| Input | M Output | N Output | P Output |
|---|---|---|---|
| 1 | 2 | 3 | 3 |
| 2 | 3 | 2 | 2 |
| 3 | 4 | 1 | 1 |
| 4 | 5 | 1 | 1 |
Which of the mappings M, N, P are functions? Write down all that are functions, giving a reason for each choice.
For one mapping that is not a function, explain why.
In a probability experiment, a spinner shows either 1, 2, 3 or 4 with equal probability. The outcome is used as the input for mapping N above.
If you spin the spinner twice, each time applying the mapping N to the number spun, let A be the event 'both outputs are the same'. Draw a probability tree diagram to represent this experiment and calculate the probability of event A.
Draw or describe a probability tree diagram for this experiment, labelling all outcomes with both spinner result and mapped output.
Work out the probability that both mapped outputs are the same, showing your steps.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a(i)) M and N are functions.
(a(ii)) P is not a function since input 1 maps to two different outputs (3 and 1), which is not allowed for functions.
(b(i)) A tree diagram with two stages: first and second spin, each branch labelled 1,2,3,4, with all outputs mapped using N: (1→3, 2→2, 3→1, 4→1). Each outcome labelled with the mapped value for that stage. Each branch has probability 1/4. All possible pairs of mapped outputs considered.
(b(ii)) The probability that both mapped outputs are the same is 7/16.
Common mistakes: Not checking one-to-one output in mapping diagrams; Errors in number of outcomes on the tree diagram; Ignoring equal probabilities
Show mark scheme and worked solution
Marking points
B1 (a(i)) · 1 mark · Correctly identifies that mapping M is a function, with valid reason.
B2 (a(i)) · 1 mark · Correctly identifies that mapping N is a function, with valid reason.
B3 (a(ii)) · 1 mark · Correctly explains (with example) why mapping P is not a function.
B4 (b(i)) · 1 mark · Draws or describes a correct probability tree with all paths/stages for two spins, mapping outputs correctly.
B5 (b(i)) · 1 mark · Labels all outcomes with both spinner result and mapped output.
B6 (b(ii)) · 2 marks · Correctly identifies all pairs where both mapped outputs are the same and counts them, or clearly sets out calculation method.
A1 (b(ii)) · 2 marks · Correct probability (7/16) found with clear final answer. (follow-through)
Worked solution
a(i) Mapping M: Each input (1,2,3,4) maps to a single unique output (2,3,4,5). Hence, M is a function.
Mapping N: Each input maps to only one output (1↦3, 2↦2, 3↦1, 4↦1). N is a function.
Mapping P: Input 1 maps to both 3 and 1 in the outputs, so P is not a function (a function assigns only one output to each input).
a(ii) Example: P is not a function because if you input 1, it maps to both 3 and 1, so it fails the definition of a function.
b(i) Tree diagram: First spin has outcomes 1,2,3,4 (each prob. 1/4). For each, the mapped output (by N) is 3,2,1,1. Second spin: again, 1,2,3,4 (each prob. 1/4), mapped as above. Thus, every pair of mapped outputs is possible: (3,3), (3,2), (3,1), etc.
All possible pairs: 16 outcomes in total. List all mapped pairs (map N to both spins): First spin: 1→3, 2→2, 3→1, 4→1 Second spin (same mapping for each): 1→3, 2→2, 3→1, 4→1 List all pairs where both mapped outputs are the same.
Pairs where both mapped outputs are the same: - (1,1): mapped pair (3,3) - (2,2): (2,2) - (3,3): (1,1) - (3,4): (1,1) - (4,3): (1,1) - (4,4): (1,1) Also: - (1,3): (3,1) (not same) - (1,4): (3,1) (not same) - (2,1): (2,3) (not same) - (2,3): (2,1) (not same) - etc. Count only matching pairs: For each first mapped output (M1): 3 (for 1), 2 (for 2), 1 (for 3 or 4). For each, how many times does the second mapped output match? See matrix below:
| First mapped output | How many second outcomes match |
|---|---|
| 3 (from 1st spin=1) | 1 (when second spin=1) |
| 2 (from 1st spin=2) | 1 (when second spin=2) |
| 1 (from 1st spin=3) | 2 (when second spin=3 or 4) |
| 1 (from 1st spin=4) | 2 (when second spin=3 or 4) |
Count them all: (1×1)+(1×1)+(1×2)+(1×2)=6. But actually, write out all possible pairings where outputs match:
(First Spin, Second Spin): (1,1): (3,3) match (2,2): (2,2) match (3,3): (1,1) match (3,4): (1,1) match (4,3): (1,1) match (4,4): (1,1) match Total = 6 matching pairs, but (3,3) and (4,3), (3,4), (4,4) are separate events. From the matrix, matches are: (1,1); (2,2); (3,3); (3,4); (4,3); (4,4) = 6 pairs.
Each combination has probability (1/4)×(1/4)=1/16, so probability of A = 6×(1/16)=6/16 = 3/8. But let's explicitly count for every value: First mapped: 3 (spin 1), Second mapped: 3 (spin 1) → (3,3) First mapped: 2 (spin 2), Second mapped: 2 (spin 2) → (2,2) First mapped: 1 (spin 3 or 4), Second mapped: 1 (spin 3 or 4) → (1,1); for these, 2×2=4 pairs (3,3;3,4;4,3;4,4) So total: (1)+ (1) + (2×2=4) = 6 pairs out of 16 possibilities.
Therefore, probability both outputs are the same is .