SME
Grade 10 Math · Problem type
Independent events: three stages
Use this for “Three trains are on time with chances 0.9, 0.8 and 0.7. What is the chance all three are? That at least one is late?”
Diagram
8 paths; the chances on them add to 1
P(On time, On time, On time) = 0.9 × 0.8 × 0.7 = 0.504
Each stage has the same chances on every branch: the stages are independent, so multiply.
This is an example. Type your own values to start a new problem.
P(all
on
time)
=
P(A)
×
P(B)
×
P(C)
=
P(all)
P(some late)
At least one late
worked out (not typed)
Formulas
P(all) = P(A) × P(B) × P(C)
0.504 = 0.9 × 0.8 × 0.7
P(some late) = 1 − P(all)
0.496 = 1 − 0.504
Assumptions
•
Each train is on time or late, whatever the others do: the stages are independent.
•
Along one path, multiply the chances: P(A and B and C) = P(A) × P(B) × P(C).
•
At least one late is every other path: 1 − P(all on time).
Step-by-step
WE KNOW
P(A) = 0.9, P(B) = 0.8, P(C) = 0.7
FIND
all three on time (P(all)), at least one late (P(some late))
Step 1 · Find all three on time (P(all))
Use P(all) = P(A) × P(B) × P(C)
Independent stages: multiply along the path.
P(all) = 0.9 × 0.8 × 0.7
P(all) = 0.72 × 0.7
P(all) = 0.504
Step 2 · Find at least one late (P(some late))
Use P(some late) = 1 − P(all)
Every path but the all-on-time one.
P(some late) = 1 − 0.504
P(some late) = 0.496
Check
Put the numbers back in. Both sides should match.
0.504 = 0.9 × 0.8 × 0.7 ✓
0.496 = 1 − 0.504 ✓
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