Case study (4 Marks)

Question 14 Marks

Following is a circuit diagram showing an electronic assembly consists of two sub-systems say A and B:
From previous testing procedures, the following probabilities are assumed to be known.
$\mathrm{P}(\mathrm{A}$ fails $)=0.2 ; \mathrm{P}(\mathrm{B}$ fails alone $)=0.15 ; \mathrm{P}(\mathrm{A}$ and B fail $)=0.15$.

(a) What is probability that A fails alone?
(b) What is probability that neither A nor B fails?
(c) What is conditional probability that A fails given that B has failed?

Answer

(i) $\mathrm{P}(\mathrm{A}$ fails alone $)=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$
=0.20-0.15=0.05
$
(ii) $P(\bar{A} \cap \bar{B})=P(\overline{A \cup B})=1-P(A \cup B)$ $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.20+0.30-0.15=0.35$
$\therefore P(\overline{\mathrm{~A}} \cap \overline{\mathrm{~B}})=1-0.35=0.65$
(iii) $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0.15}{0.30}=\frac{1}{2}=0.50$.

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Question 24 Marks

Rohit and Rakesh planned to play Business Board game with two dice. In which they will get the turn one by one, they roll the dice and continue the game.

(a) Rakesh got first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 8 ?
(b) Rohit got next chance. What is the probability that he got the sum of the two number appearing on the top face of the dice is 13 ?
(c) Rohit got next change. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is equal 7 ?

Answer

(i) Favourable cases (2, 6), (6, 2), (4, 4), (5, 3), (3, 5)
Probability of getting the sum as eight $=\frac{5}{36}$
(ii) As the sum of numbers on two dice is 13 will not be possible, therefore zero probability.
(iii)Favourable events of getting sum = 7
i.e, $(1,6),(6,1),(5,2),(4,3),(3,4),(2,5)$
Required probability $=\frac{6}{36}=\frac{1}{6}$

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Question 34 Marks

Read the text carefully and answer the questions:
Different organisations collect the data and analyse it quantitatively. During one such analysis some mistake crept in. The result given was that mean and variance of 100 observations as 40 and 5.1 but later on rechecking it was found that one observation was mistakenly taken as 50 instead of 40 .
(a) What is incorrect sum of variates?
(b) What is correct sum of observations?
(c) What is incorrect $\Sigma x^{2}$ ?
OR
What is corrected variance?

Answer

(i) Mean $=40, \mathrm{n}=100$, sum $=100 \times 40=4000$
(ii) Corrected sum $=4000-50+40=3990$
(iii) $\sigma^{2}=\frac{\Sigma x^{2}}{n}-\left(\frac{\Sigma x}{n}\right)^{2}$
$\Rightarrow(5.1)^{2}=\frac{\Sigma x^{2}}{100}-(40)^{2}$
$\Rightarrow(26.01+1600) 100=\Sigma x^{2}$
OR
Corrected $\Sigma x^{2}=162601-(50)^{2}+(40)^{2}$
$=162601-2500+1600=161701$
Corrected $\sigma^{2}=\frac{161701}{100}-(39.9)^{2}$
$=1617.01-1592.01=25$
38. Read the text carefully and answer the questions:

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Question 44 Marks

Children always keep of thinking something new, sometimes reaching a destination they say we will go by longer sometimes by shorter route, etc.
A child is standing at the point $A(2,3)$ and far away he can notice a straight road path represented by the equation $3 x-4 y-12=0$. He wants to reach path in the minimum possible time.
(a) Find the slope of the path followed by child?
(b) Find the Equation of the path followed by child?
(c) Find the Distance covered by child in reaching the path?
OR
If child wants to reach at point $(4,0)$ on the straight road, then find the equation of path he should follow.

Answer

(i) $\mathrm{m} \times \frac{3}{4}=-1 \Rightarrow \mathrm{~m}=\frac{-4}{3}$
(ii) Equation of line through $(2,3)$ and having shope $\frac{-4}{3}$ is
$y-3=\frac{-4}{3}(x-2)$
$\Rightarrow 3 y-9=-4 x+8$
$\Rightarrow 4 \mathrm{x}+3 \mathrm{y}-17=0$
(iii) Distance $=\left|\frac{6-12-12}{\sqrt{9+16}}\right|$
$=\left|\frac{-18}{5}\right|=\frac{18}{5}$ units
OR
Equation of path joining $(2,3)$ and $(4,0)$ is
$
\begin{aligned}
& y-0=\frac{0-3}{4-2}(x-4) \\
& \Rightarrow 2 y=-3 x+12 \\
& \Rightarrow 3 x+2 y-12=0
\end{aligned}
$

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Applied Maths Case study (4 Marks)

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