SME
Grade 10 Math · Problem type
A tree with conditional branches
Use this for “P(Rain) = 0.3, P(Late | Rain) = 0.4, P(Late | Dry) = 0.1. Find P(Late) and P(Rain | Late).”
Diagram
Each node’s branches add to 1: 0.3 + 0.7 = 1
P(Rain and Late) = P(Rain) × P(Late | Rain) = 0.3 × 0.4 = 0.12
P(Late) = P(Rain and Late) + P(Dry and Late) = 0.12 + 0.07 = 0.19
So
P(Rain
|
Late)
=
0.12
0.19
=
12
19
≈
0.63
This is an example. Type your own values to start a new problem.
P(Late)
=
P(R)
×
P(L | R)
+
(1
−
P(R)
)
×
P(L | D)
=
P(L)
P(R and L)
P(Rain and Late)
worked out (not typed)
P(R | L)
P(Rain | Late)
worked out (not typed)
Formulas
P(R and L) = P(R) × (P(L | R))
0.12 = 0.3 × 0.4
P(L) = P(R and L) + (1 − P(R)) × (P(L | D))
0.19 = 0.12 + (1 − 0.3) × 0.1
P(R
|
L)
=
P(R and L)
P(L)
12
19
=
0.12
0.19
Assumptions
•
Each second branch is conditional: P(Late | Rain) is the chance of being late when it rains.
•
Multiply along a path; add the paths that end in Late for P(Late).
•
Then P(Rain | Late) is the rain path’s share of P(Late).
Step-by-step
WE KNOW
P(R) = 0.3, P(L | R) = 0.4, P(L | D) = 0.1
FIND
P(Rain and Late) (P(R and L)), P(Late) (P(L)), P(Rain | Late) (P(R | L))
Step 1 · Find P(Rain and Late): P(R and L)
Use P(R and L) = P(R) × (P(L | R))
Multiply along the path: rain, then late.
P(R and L) = 0.3 × 0.4
P(R and L) = 0.12
Step 2 · Find P(Late): P(L)
Use P(L) = P(R and L) + (1 − P(R)) × (P(L | D))
Add the two paths that end in Late: rain then late, and dry then late.
P(L) = 0.12 + (1 − 0.3) × 0.1
P(L) = 0.12 + 0.7 × 0.1
P(L) = 0.12 + 0.07
P(L) = 0.19
Step 3 · Find P(Rain | Late): P(R | L)
Use P(R | L) = (P(R and L)) ÷ P(L)
The rain path’s share of all the ways to be late.
P(R
|
L)
=
0.12
0.19
P(R
|
L)
=
12
19
Check
Put the numbers back in. Both sides should match.
0.12 = 0.3 × 0.4 ✓
0.19 = 0.12 + (1 − 0.3) × 0.1 ✓
12
19
=
0.12
0.19
✓
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