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Worked solution

Condition on a full tank after a fuel-type split

At a petrol station,30%of customers buy leaded petrol,60%buy unleaded petrol and the remainder buy diesel. A leaded-petrol customer has a 25%chance of filling the tank; an unleaded-petrol customer has a 20%chance; a diesel customer has a 30%chance.

(i) Draw a probability tree.

(ii) Find the probability that a car leaving does not have a full tank.

(iii) Given that it has a full tank, find the probability that it contains unleaded petrol.

CJC · Lecture · Example 8

(i) Define the events and complete the first split

Let L,U,D denote leaded, unleaded, diesel purchases and F denote filling the tank. The three fuel events partition the customers: P(L)=0.3P(L)=0.3, P(U)=0.6P(U)=0.6, P(D)=1−0.3−0.6=0.1P(D)=1-0.3-0.6=0.1.

How did this part go?

(i) Complete the conditional full/not-full branches

Given L, the F and F′ branches have probabilities 0.25 and 0.75. Given U, they have 0.2 and 0.8. Given D, they have 0.3 and 0.7. The diagram incorporates every Teacher-completed probability; each pair sums to 1.

How did this part go?

(ii) Add the three not-full paths

P(F′)=P(L∩F′)+P(U∩F′)+P(D∩F′)=(0.3)(0.75)+(0.6)(0.8)+(0.1)(0.7)=0.775P(F')=P(L\cap F')+P(U\cap F')+P(D\cap F')=(0.3)(0.75)+(0.6)(0.8)+(0.1)(0.7)=0.775. The fuel cases are mutually exclusive.

How did this part go?

(iii) Use full-tank probability as the denominator

P(U∣F)=P(U∩F)/P(F)=(0.6×0.2)/(1−0.775)=0.12/0.225=8/15P(U\mid F)=P(U\cap F)/P(F)=(0.6\times0.2)/(1-0.775)=0.12/0.225=8/15. The numerator is only the unleaded-and-full path, whereas the denominator includes full tanks from all three fuels.

How did this part go?

Compare your sketch

Fuel type, then tank filling

Each full/not-full probability is conditional on the fuel type.

Start · probability 1

Open a branch to follow its outcomes. Outcome percentages are for complete paths. Open an outcome to see its calculation.

  • L · Leaded0.3 · 2 outcomes

    L · Leaded

    • F · FullPath 7.5%
      L · Leaded → F · Full
      Branch weights
      0.3 × 0.25
      Complete path probability
      3/40= 0.075 · 7.5%
    • F′ · Not fullPath 22.5%
      L · Leaded → F′ · Not full
      Branch weights
      0.3 × 0.75
      Complete path probability
      9/40= 0.225 · 22.5%
  • U · Unleaded0.6 · 2 outcomes

    U · Unleaded

    • F · FullPath 12%
      U · Unleaded → F · Full
      Branch weights
      0.6 × 0.2
      Complete path probability
      3/25= 0.12 · 12%
    • F′ · Not fullPath 48%
      U · Unleaded → F′ · Not full
      Branch weights
      0.6 × 0.8
      Complete path probability
      12/25= 0.48 · 48%
  • D · Diesel0.1 · 2 outcomes

    D · Diesel

    • F · FullPath 3%
      D · Diesel → F · Full
      Branch weights
      0.1 × 0.3
      Complete path probability
      3/100= 0.03 · 3%
    • F′ · Not fullPath 7%
      D · Diesel → F′ · Not full
      Branch weights
      0.1 × 0.7
      Complete path probability
      7/100= 0.07 · 7%
Compare outcome probabilities
Multiply along each complete path. All outcomes together have probability 1.
OutcomeProbability
L · Leaded → F · Full3/40= 0.075 · 7.5%
L · Leaded → F′ · Not full9/40= 0.225 · 22.5%
U · Unleaded → F · Full3/25= 0.12 · 12%
U · Unleaded → F′ · Not full12/25= 0.48 · 48%
D · Diesel → F · Full3/100= 0.03 · 3%
D · Diesel → F′ · Not full7/100= 0.07 · 7%
All 6 outcomes1= 1 · 100%
Tree assumptions and source notes

L, U and D partition customers by fuel type; F means a full tank.

Multiply along a fuel/fullness path and add disjoint fuel paths. P(F | U) is not P(U | F).

Open original probability tree (new tab)
Source tree: the first split is leaded L with probability 0.3, unleaded U with probability 0.6, or diesel D with probability 0.1. Under L, full F has probability 0.25 and not full has probability 0.75. Under U, these probabilities are 0.2 and 0.8; under D, they are 0.3 and 0.7. Initial probabilities sum to one, as does each fuel-specific second-stage pair. Multiply along a fuel/fullness path and add disjoint fuel paths.

Common mistakes

Do not use the 0.2 branch probability as P(U given F): it is P(F given U). The fuel categories form disjoint exhaustive cases, while fullness depends on the fuel type in this model.

Read the hint