Question 2 OR (g)
Probability that the contractor gets both contracts
6 marks
- Time to write
- About 12 minutes
Question
Question 2 OR (g) · 6 marksA contractor has two potential contracts, namely an electrification contract and a plumbing contract. The probability of the contractor getting the electrification contract is 4/7. The probability of the contractor not getting the plumbing contract is 7/10. The probability of the contractor getting at least one of the contracts is 7/9. Find the probability of the contractor getting both contracts.
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- FindFormula first, then the working, then the answer with its unit (₹, %, units).
What the paper leaves open
The three probabilities are printed as stacked fractions; they are written here as a/b.
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Events and the probability of the plumbing contract 1½ marks 2 min
- 2 Addition theorem and P both contracts 3½ marks 6 min
- 3 Check through the four outcomes 1 mark 2 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Short on time? Step 2 carries 3½ of the 6 marks: guard its minutes.
Try it first. About 12 minutes for 6 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Events. Let E be the event that the contractor gets the electrification contract and P the event of getting the plumbing contract. Given:
- P(E) = 4/7;
- P(not P) = 7/10, so P(P) = 1 − 7/10 = 3/10, because an event and its complement have probabilities adding to 1;
- P(E ∪ P) = 7/9 (at least one contract).
Probability of getting both contracts
The addition theorem for any two events, P(E ∪ P) = P(E) + P(P) − P(E ∩ P), gives P(E ∩ P) = P(E) + P(P) − P(E ∪ P):
P(E ∩ P) = 4/7 + 3/10 − 7/9.
With the common denominator 630 (= 7 × 10 × 9):
P(E ∩ P) = 360/630 + 189/630 − 490/630 = 59/630 ≈ 0.0937.
The probability that the contractor gets both contracts is 59/630, a little under 1 in 10. The figure is a valid probability: it is positive and smaller than both P(E) = 360/630 and P(P) = 189/630.
Check through all the outcomes.
| Outcome | Working (in 630ths) | Probability |
|---|---|---|
| Both contracts | 360 + 189 − 490 = 59 | 59/630 |
| Electrification only | 360 − 59 = 301 | 301/630 |
| Plumbing only | 189 − 59 = 130 | 130/630 |
| Neither contract | 630 − 490 = 140 | 140/630 |
| Total | 630/630 = 1 |
Where the 6 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Events and the probability of the plumbing contract | 1.5 |
| 2 | Addition theorem and P(both contracts) | 3.5 |
| 3 | Check through the four outcomes | 1 |
| Total | 6 |
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