Unit 10
10 marks · HardA manufacturer fits security systems in homes. The company records the number of alarms installed each month. The number of alarms, n, is related to the cost per alarm, c, by the equation:
On one occasion:
Given that 8 alarms were installed that month, solve the equation above to find the cost per alarm, c.
A follow-up promotion means that the number of alarms in a month is now a random variable with the following probabilities:
| Number of alarms, n | Probability |
|---|---|
| 8 | 0.2 |
| 12 | 0.3 |
| 16 | 0.4 |
| 20 | 0.1 |
A supervisor picks a month at random from the coming year and uses the value of n from that month in the same cost equation as above, with c = 12. (i) For each possible value of n, state if the equation has a solution for n > 0. Briefly explain your answer.
The supervisor now wants to know, "What is the probability that the equation has a valid solution for n > 0, based on the probability distribution in part b?"
Complete a probability tree diagram for randomly selecting a value of n and checking if the equation can be solved for n > 0. Clearly label the branches with numbers and probabilities.
Use your tree diagram to find the probability that a randomly selected month leads to a valid value of n from part (c).
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) c = 12
(b) n = 8, 12, 16: solution exists for n > 0; n = 20: no solution for n > 0.
(c) Tree diagram with branches for n = 8, 12, 16 (solution exists, with probabilities 0.2, 0.3, 0.4) and n = 20 (no solution, 0.1) · must show Branches for all n values, Correct probabilities, Correct indication of valid/invalid solutions
(d) Probability = 0.2 + 0.3 + 0.4 = 0.9
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Rearranges and correctly substitutes n = 8 into the equation.
A1 (a) · 2 marks · Accurately solves for c = 12, with correct calculations.
B1 (b) · 1 mark · Considers each value of n from the table and substitutes into the equation with c = 12.
B2 (b) · 1 mark · Correctly identifies which values of n allow a solution for n > 0 with explanation.
M2 (c) · 1 mark · Draws tree diagram with at least two branches representing n values and solution/no solution outcomes.
A2 (c) · 2 marks · Labels branches with correct probabilities and identifies the valid solution branches.
M3 (d) · 1 mark · Selects branches which correspond to valid n based on the tree diagram. (follow-through)
A3 (d) · 1 mark · Correctly calculates the probability as 0.2+0.3+0.4 = 0.9. (follow-through)
Worked solution
a) Substitute n = 8 in the equation:
Subtract from both sides:
Solve for c:
But checking: 1/8 + 2/16 = 1/8 + 1/8 = 2/8 = 1/4 is correct. The calculation above was correct; alternative approach:
So, the cost per alarm, c, is 16.
b) For c = 12, investigate solutions for each n:
| n | \frac{1}{n} + \frac{2}{12} | Equals 1/4? |
|---|---|---|
| 8 | 1/8 + 1/6 = 7/24 | Yes (since 7/24 < 1/4) |
| 12 | 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4 | Yes |
| 16 | 1/16 + 1/6 = (3/48 + 8/48) = 11/48 ≈ 0.229 | No |
| 20 | 1/20 + 1/6 = (3/60 + 10/60) = 13/60 ≈ 0.217 | No |
Actually, for n=12, equation holds exactly; for n=8, not exactly equal. So only n = 12 gives the solution for 1/n + 1/6 = 1/4. Re-compute for all to confirm. Let's set up for n = 12:
So, valid solution when n = 12 only.
c) Tree diagram structure: - First branch: pick n (probability for each value) - Second branch: 'solution exists' (only for n=12) vs 'no solution' (all others) Tree diagram must show: - n = 8: 0.2, no solution - n = 12: 0.3, solution exists - n = 16: 0.4, no solution - n = 20: 0.1, no solution
d) The probability that a randomly selected n gives a valid solution (i.e., solution exists) is probability for n = 12: P(solution exists) = 0.3