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Exercise 9.5 · Q2

Q.A courier service company sends 30% of its orders by air, 50% by combination of bus and local transport and remaining 20% by train. Past record shows the courier is delivered late 2%, 7% and 5% of the time when orders are sent by air, bus local transport and train respectively. Find

(i) the probability that the order will be delivered late
(ii) the probability that the parcel delivered to a customer is sent by train if it is delivered late.
Sikkim CbseNCERTSubjective· 5mImportance★★★★★est
71% · 15/21 Questions
✓ Free question

Total-probability gives P(late)=0.051P(\text{late})=0.051, and Bayes' theorem then gives P(train∣late)=1051≈0.196P(\text{train}\mid\text{late})=\dfrac{10}{51}\approx0.196.

Law of Total Probability: P(L)=∑iP(Mi)P(L∣Mi)P(L)=\sum_i P(M_i)P(L\mid M_i).

Bayes' theorem: P(Mi∣L)=P(Mi)P(L∣Mi)P(L)P(M_i\mid L)=\dfrac{P(M_i)P(L\mid M_i)}{P(L)}, where Mi∈{Air, Bus/local, Train}M_i\in\{\text{Air, Bus/local, Train}\} and L=L= "delivered late".

  1. Given data.

P(Air)=0.30, P(L∣Air)=0.02P(\text{Air})=0.30,\ P(L\mid\text{Air})=0.02

P(Bus)=0.50, P(L∣Bus)=0.07P(\text{Bus})=0.50,\ P(L\mid\text{Bus})=0.07

P(Train)=0.20, P(L∣Train)=0.05P(\text{Train})=0.20,\ P(L\mid\text{Train})=0.05

  1. Part (i): compute each joint term.

P(Air)⋅P(L∣Air)=0.30×0.02=0.006P(\text{Air})\cdot P(L\mid\text{Air})=0.30\times0.02=0.006

P(Bus)⋅P(L∣Bus)=0.50×0.07=0.035P(\text{Bus})\cdot P(L\mid\text{Bus})=0.50\times0.07=0.035

P(Train)⋅P(L∣Train)=0.20×0.05=0.010P(\text{Train})\cdot P(L\mid\text{Train})=0.20\times0.05=0.010

  1. Sum for total probability of a late delivery.

P(L)=0.006+0.035+0.010=0.051P(L)=0.006+0.035+0.010=0.051

  1. Part (ii): apply Bayes' theorem for Train given Late.

P(Train∣L)=P(Train)⋅P(L∣Train)P(L)=0.0100.051P(\text{Train}\mid L)=\frac{P(\text{Train})\cdot P(L\mid\text{Train})}{P(L)}=\frac{0.010}{0.051}

  1. Simplify.

0.0100.051=1051≈0.1961\frac{0.010}{0.051}=\frac{10}{51}\approx0.1961

Self-check: Adding all three posterior contributions should sum to 1: 0.0060.051+0.0350.051+0.0100.051=0.0510.051=1\frac{0.006}{0.051}+\frac{0.035}{0.051}+\frac{0.010}{0.051}=\frac{0.051}{0.051}=1 ✓.

✓Final answer

  1. P(order delivered late)=0.051P(\text{order delivered late})=0.051 (i.e. 5.1%5.1\%).
  2. P(sent by train∣delivered late)=1051≈0.196P(\text{sent by train}\mid\text{delivered late})=\dfrac{10}{51}\approx0.196 (about 19.6%19.6\%).

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