Probability rules: 7 exam questions solved step by step
What this lesson covers
- Probabilities are multiplied for "and" along a path, and added for "or" across outcomes that cannot happen together.
- A probability is always between 0 and 1, which is a check on every answer.
7 questions, three rules
- And (independent): multiply.
- Or (cannot both happen): add.
- Not, or at least one: 1 minus.
- Every answer is between 0 and 1.
Question 1 · Achieved, in context
- A fundraiser lucky dip: one prize a dip.
- P(chocolate) = 0.35, P(sticker) = 0.4,
- and every other prize is a pen.
- Find P(chocolate or sticker), and P(pen).
Q1: the lucky dip
- One prize: "or" is add.
- Pens are the rest.
- one prize: mutually exclusive
- P(choc or sticker) = 0.35 + 0.4
- = 0.75or: add
- P(pen) = 1 − 0.75the complement
- = 0.25
- check: 0.35 + 0.4 + 0.25 = 1
Question 2 · Achieved, in context
- At a kiwifruit packhouse, each tray fails
- inspection with probability 0.05,
- independently. Two trays are checked.
- Find P(both fail) and P(neither fails).
Q2: two trays
- "And": multiply.
- Pass: 1 − 0.05.
- P(both fail) = 0.05 × 0.05and: multiply
- = 0.0025
- 0.05 + 0.05 = 0.1"and" is not add
- P(pass) = 1 − 0.05 = 0.95complement
- P(neither) = 0.95 × 0.95and: multiply
- = 0.9025
Question 3 · Merit, in context
- A rugby goal kicker succeeds with
- probability 0.8 each kick, independently.
- She takes three kicks.
- Find P(she misses at least one).
Q3: at least one miss
- At least one: 1 − P(none).
- No misses: all three score.
- P(miss) = 1 − 0.8 = 0.2
- 0.2 + 0.2 + 0.2 = 0.6counts overlaps
- P(no misses) = 0.8 × 0.8 × 0.8
- = 0.512and: multiply
- P(at least one miss) = 1 − 0.512
- = 0.488
Question 4 · Merit, in context
- Two Wellington ferries sail each morning.
- P(A cancelled) = 0.1, P(B cancelled) = 0.15,
- independently.
- Find P(exactly one is cancelled).
Q4: exactly one cancelled
- Two separate ways: add.
- Each way: multiply.
- P(A runs) = 0.9, P(B runs) = 0.85
- A only: 0.1 × 0.85 = 0.085and: multiply
- B only: 0.9 × 0.15 = 0.135and: multiply
- exactly one: 0.085 + 0.135or: add
- = 0.22
- both: 0.1 × 0.15 = 0.015
- neither: 0.9 × 0.85 = 0.765
- 0.015 + 0.22 + 0.765 = 1
Question 5 · Merit, in context
- For an Auckland commuter:
- P(bus) = 0.4, P(podcast) = 0.25,
- P(bus and podcast) = 0.1.
- Independent? Mutually exclusive?
- Find P(bus or podcast). Pause and try it.
Q5: bus and podcast
- Test the product.
- Overlap: take it off once.
- P(bus) × P(podcast) = 0.4 × 0.25
- = 0.1 = P(bus and podcast)
- so independent
- P(both) = 0.1, not 0not mutually exclusive
- 0.4 + 0.25 = 0.65overlap counted twice
- P(bus or podcast)
- = 0.4 + 0.25 − 0.1take off the overlap
- = 0.55
- check: 1 − 0.6 × 0.75 = 0.55
Question 6 · Excellence, in context
- On a Bay of Islands boat trip,
- P(seeing dolphins) = 0.6, independently.
- How many trips are needed for at least
- a 0.99 chance of seeing dolphins?
Q6: how many trips?
- At least once: 1 − P(never).
- Never round before comparing.
- P(no dolphins) = 1 − 0.6 = 0.4
- P(none in n trips) = 0.4ⁿand: multiply
- 1 − 0.4ⁿ ≥ 0.99at least once
- n = 5: 1 − 0.4⁵ = 0.98976
- 0.98976 < 0.99not enough
- n = 6: 1 − 0.4⁶ = 0.995904
- 0.995904 ≥ 0.99
- 6 trips
- n ≥ log 0.01 ÷ log 0.4 = 5.03
Question 7 · Excellence, in context
- Two smoke alarms work independently.
- P(alarm A detects smoke) = 0.9.
- P(at least one detects it) = 0.98.
- Find P(alarm B detects smoke).
Q7: the second alarm
- Name the unknown.
- Work through "neither".
- let p = P(B detects)
- P(A fails) = 1 − 0.9 = 0.1
- P(neither) = 1 − 0.98 = 0.02complement
- P(neither) = 0.1 × (1 − p)and: multiply
- 0.1(1 − p) = 0.02
- 1 − p = 0.2÷ 0.1
- p = 0.8
- check: 1 − 0.1 × 0.2 = 0.98
The rules that solved all seven
- Read the words, pick the rule, check between 0 and 1.
| Step | How |
|---|---|
| And | multiply, if independent |
| Or | add, if they cannot overlap |
| Overlap | subtract P(A and B) once |
| Not | 1 minus |
| At least one | 1 − P(none) |
| Check | between 0 and 1; totals 1 |
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