Board 2024 Class 12 Maths Q37 Solution: Case study: air turbulence and late flights

The question

According to recent research, air turbulence has increased in various regions around the world due to climate change. Turbulence makes flights bumpy and often delays the flights.

Assume that, an airplane observes severe turbulence, moderate turbulence or light turbulence with equal probabilities. Further, the chance of an airplane reaching late to the destination are 55%, 37% and 17% due to severe, moderate and light turbulence respectively.

On the basis of the above information, answer the following questions :

(i) Find the probability that an airplane reached its destination late. (2 marks)

(ii) If the airplane reached its destination late, find the probability that it was due to moderate turbulence. (2 marks)

Exam answer

(i) Let $A$ ⁠: the airplane reaches its destination late, $E_1$ ⁠: severe turbulence, $E_2$ ⁠: moderate turbulence, $E_3$ ⁠: light turbulence.

Why?

Name the events first. Exactly one kind of turbulence happens on a flight, so $E_1, E_2, E_3$ are mutually exclusive and together cover every case. Lateness, A, can happen with any of them.

P(E_1) = P(E_2) = P(E_3) = \frac13

Why?

The three kinds are equally likely and their probabilities add up to 1, so each is $\frac13$ ⁠.

\begin{aligned} P(A|E_1) &= \frac{55}{100},\ P(A|E_2) = \frac{37}{100}, \\ P(A|E_3) &= \frac{17}{100} \end{aligned}

Why?

"The chance of reaching late due to severe turbulence" is the chance of A given that $E_1$ has happened, $P(A|E_1)$ ⁠. Each percentage is written as a fraction of 100.

By the theorem of total probability,

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &\quad + P(E_3)P(A|E_3) \\ &= \frac13\cdot\frac{55}{100} + \frac13\cdot\frac{37}{100} \\ &\quad + \frac13\cdot\frac{17}{100} \\ &= \frac13\left(\frac{109}{100}\right) \end{aligned}

Why?

A late arrival happens along one of three routes: severe and late, moderate and late, or light and late. Total probability adds the three routes, each one being $P(E_i)P(A|E_i)$ ⁠.

Take out the common $\frac13$ ⁠: $55 + 37 + 17 = 109$ ⁠.

\begin{aligned} &\frac13\Big(\frac{55}{100} + \frac{37}{100} + \frac{17}{100}\Big) \\ &= \frac13\Big(\frac{55 + 37 + 17}{100}\Big) \\[4pt] &55 + 37 = 92 \\ &92 + 17 = 109 \end{aligned}

Then multiply the fractions, top times top and bottom times bottom:

\frac13\cdot\frac{109}{100} = \frac{1 \times 109}{3 \times 100} = \frac{109}{300}

$\therefore$ Required probability $= \frac{109}{300}$ ⁠.

(ii) By Bayes' theorem,

Why?

Now we know the result, a late arrival, and ask for the cause. Reversing a conditional probability in this way is what Bayes' theorem does: one route divided by all the routes together.

\begin{aligned} P(E_2|A) &= \frac{P(E_2)P(A|E_2)}{P(A)} \\ &= \frac{\frac13\times\frac{37}{100}}{\frac{109}{300}} \\ &= \frac{37}{300}\times\frac{300}{109} \end{aligned}

Why?

The numerator is the moderate route only: $\frac13\times\frac{37}{100} = \frac{37}{300}$ ⁠. The denominator is $P(A)$ from part (i).

Dividing by $\frac{109}{300}$ is multiplying by $\frac{300}{109}$ ⁠, and the 300s cancel.

\begin{aligned} \frac13\times\frac{37}{100} &= \frac{1 \times 37}{3 \times 100} = \frac{37}{300} \\[4pt] \frac{37}{300} \div \frac{109}{300} &= \frac{37}{300}\times\frac{300}{109} \\ &= \frac{37 \times 300}{300 \times 109} \\ &= \frac{37}{109} \quad (\div 300) \end{aligned}

$\therefore$ Required probability $= \dfrac{37}{109}$ ⁠.

Where this comes from

Chapter 13, Probability: Total probability and Bayes' theorem.

Question type: (i) Two routes to one result; (ii) Bayes' theorem

How to spot it: click to see

(i) Type 7. Two routes to one result: total probability

Spot it: the result can happen in two (or more) ways, such as a ball from bag I or from bag II, and you want its overall chance.

Here: A late arrival can come through severe, moderate or light turbulence, each with chance $\frac13$ ⁠. Multiply each by its late chance (55%, 37%, 17%) and add: $\frac{109}{300}$ ⁠.

(ii) Type 8. Bayes' theorem: from the result back to the cause

Spot it: the result has already happened (a red ball, a faulty item, an A grade), and you are asked for the chance that it came from one cause. It often comes as a case study or a 5-mark question.

Here: The plane is already late, and you want the chance the cause was moderate turbulence. Put the moderate route, $\frac13\cdot\frac{37}{100}$ ⁠, over the total $\frac{109}{300}$ from (i).

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Q37: Case study: air turbulence and late flights