Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow:
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?
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 $B$ ?
iii. What is the probability that the shell was fired from $A$ ?
OR
How many hypotheses are possible before the trial, with the guns operating independently? Write the conditions of these hypotheses.

Answer

i. Let $P$ be the event that the shell fired from $A$ hits the plane and $Q$ be the event that the shell fired from $B$ hits the plane.
The following four hypotheses are possible before the trial, with the guns operating independently:
$E _1= PQ , E _2=\bar{P} \bar{Q}, E _3=\bar{P} Q, E _4=P \bar{Q}$
Let $E = $ The shell fired from exactly one of them hits the plane.
$ P \left( E _1\right)=0.3 \times 0.2=0.06, P $
$\left( E _2\right)=0.7 \times 0.8=0.56, P$
$ \left( E _3\right)=0.7 \times 0.2=0.14, P$
$ \left( E _4\right)=0.3 \times 0.8=0.24$
$P\left(\frac{E}{E_1}\right)=0, P\left(\frac{E}{E_2}\right)=0, P\left(\frac{E}{E_3}\right)=1, P\left(\frac{E}{E_4}\right)=1$
$P ( E )= P \left( E _1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right)+P\left(E_4\right) \cdot P\left(\frac{E}{E_4}\right)$
$=0.14+0.24+=0.38$
ii. By Bayes' Theorem,
$P \left(\frac{E_3}{E}\right)=\frac{P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right)+P\left(E_4\right) \cdot P\left(\frac{E}{E_4}\right)}$
$=\frac{0.14}{0.38}=\frac{7}{19}$
$\text{NOTE}$: The four hypotheses form the partition of the sample space and it can be seen that the sum of their probabilities is $1$ .
The hypotheses $E _1$ and $E _2$ are actually eliminated as $P\left(\frac{E}{E_1}\right)=P\left(\frac{E}{E_2}\right)=0$
$iii$. By Bayes' Theorem $, P \left(\frac{E_4}{E}\right)=\frac{P\left(E_4\right) \cdot P\left(\frac{E}{E_4}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right)+P\left(E_4\right) \cdot P\left(\frac{E}{E_4}\right)}$
$=\frac{0.24}{0.38}=\frac{12}{19}$
OR
Let $P$ be the event that the shell fired from $A$ hits the plane and $Q$ be the event that the shell fired from $B$ hits the plane.
The following four hypotheses are possible before the trial, with the guns operating independently:
$E _1= PQ , E _2=\bar{P} \bar{Q}, E _3=\bar{P} Q, E _4=P \bar{Q}$
Let $E =$ The shell fired from exactly one of them hits the plane.
$P\left(E_1\right)=0.3 \times 0.2=0.06,$
$P\left(E_2\right)=0.7 \times 0.8=0.56,$
$P\left(E_3\right)=0.7 \times 0.2=0.14,$
$P\left(E_4\right)=0.3 \times 0.8=0.24$

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Question 24 Marks

Read the following text carefully and answer the questions that follow :
In a street two lamp posts are $600$ feet apart. The light intensity at a distance d from the first $($stronger$)$ lamp post. $\frac{1000}{d^2}$ the light intensity at distance d from the second $($weaker$)$ lamp post is $\frac{125}{d^2} \ ($in both cases the light intensity is inversely proportional to the square of the distance to the light source$)$. The combined light intensity is the sum of the two light intensities coming from both lamp posts.

i. If $l\ (x)$ denotes the combined light intensity, then find the value of $x$ so that $I\ (x)$ is minimum.
ii. Find the darkest spot between the two lights.
iii. If you are in between the lamp posts, at distance $x$ feet from the stronger light, then write the combined light intensity coming from both lamp posts as function of $x$ .
OR
Find the minimum combined light intensity ?

Answer

i. We have $, I ( x )=\frac{1000}{x^2}+\frac{125}{(600-x)^2}$
$\Rightarrow I ^{\prime}( x )=\frac{-2000}{x^3}+\frac{250}{(600-x)^3}$ and
$\Rightarrow I ^{\prime \prime}( x )=\frac{6000}{x^4}+\frac{750}{(600-x)^4}$
For maxima/minima $, I ^{\prime}( x )-0$
$\Rightarrow \frac{2000}{x^3}=\frac{250}{(600-x)^3} $
$\Rightarrow 8(600-x)^3=x^3$
Taking cube root on both sides, we get
$2(600-x)=x $
$\Rightarrow 1200=3 x $
$\Rightarrow x=400$
Thus, $I\ (x)$ is minimum when you are at $400$ feet from the strong intensity lamp post.
ii. At a distance of $200$ feet from the weaker lamp post.
Since $I(x)$ is minimum when $x = 400$ feet,
therefore the darkest spot between the two light is at a distance of $400$ feet from a stronger lamp post,
i.e., at a distance of $600-400 = 200 $ feet from the weaker lamp post.
iii. $\frac{1000}{x^2}+\frac{125}{(600-x)^2}$
Since, the distance is $x$ feet from the stronger light,
therefore the distance from the weaker light will be $600 - x.$
So, the combined light intensity from both lamp posts is given by
$\frac{1000}{x^2}+\frac{125}{(600-x)^2}$
OR
We know that $l ( x )=\frac{1000}{x^2}+\frac{125}{(600-x)^2}$
When $x=400$
$ l(x)=\frac{1000}{160000}+\frac{125}{(600-400)^2}$
$=\frac{1}{160}+\frac{125}{40000}$
$=\frac{1}{160}+\frac{1}{320}$
$=\frac{3}{320} \text { units }$

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Question 34 Marks

Read the following text carefully and answer the questions that follow :
A plane started from airport $O$ with a velocity of $120 m/s$ towards east.
Air is blowing at a velocity of $50 m/s$ towards the north As shown in the figure.
The plane travelled $1$ hr in $OA$ direction with the resultant velocity. From $A$ and $B$ travelled $1$ hr with keeping velocity of $120 m/s$ and finally landed at $B$.

i. What is the resultant velocity from $O$ to $A$?
ii. What is the direction of travel of plane $O$ to $A$ with east?
iii. What is the total displacement from $O$ to $A$?
OR
What is the resultant velocity from $A$ to $B$ ?

Answer

$(i)$. Resultant velocity from $O$ to $A$
$ =\sqrt{\left(V_{\text {Plane }}\right)^2+\left(V_{\text {wind }}\right)^2}$
$=\sqrt{(120)^2+(50)^2}$
$=\sqrt{14400+2500}$
$=\sqrt{16900}$
$=130 m / s $
$(ii) \ \tan \theta=\frac{V_{\text {wind }}}{V_{\text {aeroplane }}}$
$\tan \theta=\frac{50}{120}$
$\tan \theta=\frac{5}{12}$
$\theta=\tan ^{-1}\left(\frac{5}{12}\right) $
$(iii)$. Displacement from $O$ to $A =$ Resultant velocity $\times$ time
$ |\overrightarrow{O A}|=|\vec{V}| \times t$
$=130 \times \frac{18}{5} \times 1$
$=468 \ km $
OR
Since, from $A$ to $B$ both Aeroplane and wind have velocity in North direction.
So $ ,\vec{V}_{\text {plane }}, A $ to $B =120+50$
$=170 m / s$

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MATHS Case study (4 Marks)

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