Sample 2022-23 Class 12 Maths Q38 Solution: Case study: two antiaircraft guns fire at a plane

The question

Case-Study 3: Read the following passage and answer the questions given below.

There are two antiaircraft guns, named as A and B. The probabilities that the shell fired from them hits an airplane are 0.3 and 0.2 respectively. Both of them fired one shell at an airplane at the same time.

(i) What is the probability that the shell fired from exactly one of them hit the plane? (2 marks)

(ii) If it is known that the shell fired from exactly one of them hit the plane, then what is the probability that it was fired from B? (2 marks)

Exam answer

Let $H_A$ be the event that the shell fired from A hits the plane, and $H_B$ the event that the shell fired from B hits the plane. The guns operate independently, so $H_A$ and $H_B$ are independent.

Why?

Whether one shell hits does not change the chance that the other shell hits, so the events are independent. For independent events,

P(H_AH_B) = P(H_A)\,P(H_B)

and the same holds with either event replaced by its complement.

\begin{aligned} P(H_A) &= 0.3,\ P(\overline{H_A}) = 0.7 \\ P(H_B) &= 0.2,\ P(\overline{H_B}) = 0.8 \end{aligned}

Why?

The probability that a shell misses is 1 minus the probability that it hits: $1 - 0.3 = 0.7$ and $1 - 0.2 = 0.8$ ⁠.

Let E be the event that the shell fired from exactly one of them hits the plane. Then $E = H_A\overline{H_B} \cup \overline{H_A}H_B$ ⁠, and these two events are mutually exclusive.

(i)

\begin{aligned} P(E) &= P(H_A\overline{H_B}) + P(\overline{H_A}H_B) \\ &= 0.3 \times 0.8 + 0.7 \times 0.2 \\ &= 0.24 + 0.14 = 0.38 \end{aligned}

Why?

Exactly one hit happens in two ways: A hits and B misses, or A misses and B hits. Both cannot happen at once, so their probabilities add.

Each way is an "and" of independent events, so its probability is a product, for example

\begin{aligned} P(H_A\overline{H_B}) &= 0.3 \times 0.8 \\ &= 0.24 \end{aligned}

and the other way:

\begin{aligned} P(\overline{H_A}H_B) &= 0.7 \times 0.2 \\ &= 0.14 \\[4pt] 0.24 + 0.14 &= 0.38 \end{aligned}

$\therefore$ The required probability is 0.38.

(ii) The hit was from B exactly when only the shell from B hits, i.e. $\overline{H_A}H_B$ ⁠, and $P(E \mid \overline{H_A}H_B) = 1$ ⁠. By Bayes' theorem,

Why?

We know that exactly one shell hit, and we want the chance that it was B's. Given exactly one hit, the hit came from B only in the case "A misses, B hits".

If A misses and B hits, then exactly one shell hits for certain, so that conditional probability is 1.

\begin{aligned} &P(\overline{H_A}H_B \mid E) \\ &= \frac{P(\overline{H_A}H_B)\,P(E \mid \overline{H_A}H_B)}{P(E)} \\ &= \frac{0.14 \times 1}{0.38} = \frac{0.14}{0.38} \\ &= \frac{14}{38} \end{aligned}

Why?

Bayes' theorem turns a probability given the cause into a probability of the cause given what was seen:

P(F \mid E) = \frac{P(F)\,P(E \mid F)}{P(E)}

Here $F = \overline{H_A}H_B$ ⁠, with $P(F) = 0.14$ from (i), $P(E \mid F) = 1$ and $P(E) = 0.38$ ⁠. Multiply top and bottom by 100, then divide both by 2: $\frac{14}{38} = \frac{7}{19}$ ⁠.

\begin{aligned} \frac{0.14}{0.38} &= \frac{0.14 \times 100}{0.38 \times 100} = \frac{14}{38} \\ &= \frac{14 \div 2}{38 \div 2} = \frac{7}{19} \end{aligned}

$\therefore$ The required probability is $\dfrac{7}{19}$ ⁠.

Where this comes from

Chapter 13, Probability: Independent events; Bayes' theorem.

Question type: (i) At least one, or exactly one; (ii) Bayes' theorem

How to spot it: click to see

(i) Type 6. At least one, or exactly one: think "not" first

Spot it: independent events, with "at least one", "the problem is solved", "neither" or "exactly one".

Here: The guns fire independently, and the question says "exactly one". Add A hits and B misses, $0.3 \times 0.8$ ⁠, to A misses and B hits, $0.7 \times 0.2$ ⁠: 0.38.

(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: Exactly one hit has already happened, and you want the chance it came from B. The B-only route is 0.14 out of the total 0.38 from (i), so Bayes gives $\frac{14}{38} = \frac{7}{19}$ ⁠.

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Q38: Case study: two antiaircraft guns fire at a plane