Case study (4 Marks)

Question 14 Marks

Read the text carefully and answer the questions:

Depending upon the parents, the chances of having a left handed child are as follows:
A. When both father and mother are left handed:
Chances of left handed child is $24 \%$.
B. When father is right handed and mother is left handed:
Chances of left handed child is $22 \%$.
C. When father is left handed and mother is right handed:
Chances of left handed child is $17 \%$.
D. When both father and mother are right handed:
Chances of left handed child is $9\%.$
Assuming that $P(A)=P(B)=P(C)=P(D)=\frac{1}{4}$ and $L$ denotes the event that child is left handed.
(a) Find $P\left(\frac{L}{C}\right)$
(b) Find $\mathrm{P}\left(\frac{\overline{\mathrm{L}}}{\mathrm{A}}\right)$
(c) Find $P\left(\frac{A}{L}\right)$

Answer

(i) $\mathrm{P}\left(\frac{\mathrm{L}}{\mathrm{C}}\right)=\frac{17}{100}$
(ii) $\mathrm{P}\left(\frac{\overline{\mathrm{L}}}{\mathrm{A}}\right)=1-\mathrm{P}\left(\frac{\mathrm{L}}{\mathrm{A}}\right)=1-\frac{24}{100}=\frac{76}{100}$ or $\frac{19}{25}$
${ }^{\text {(iii) }} \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{L}}\right)=\frac{\frac{1}{4} \times \frac{24}{100}}{\frac{1}{4} \times \frac{24}{100}+\frac{1}{4} \times \frac{22}{100}+\frac{1}{4} \times \frac{17}{100}+\frac{1}{4} \times \frac{9}{100}}=\frac{24}{72}=\frac{1}{3}$

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Question 24 Marks

Read the text carefully and answer the questions:
In a game a girl rolls a die, if she gets an even number she will toss a coin if she gets head in coin, she will win ₹ 10 if she gets tail in coin she will win ₹ 5 . If she gets odd number in die she has to pay ₹ 20 to organiser.
(a) Find the total number of sample points in sample space?
(b) Find the probability that girl will win ₹ 10 ?
(c) Find the probability that girl will win ₹ 5 ?

Answer

(i) Total sample points $=9$
(ii) Girl will win ₹ 10 if she get head
$\mathrm{E}=\{(2 \mathrm{H}),(4 \mathrm{H}),(6 \mathrm{H})\}$
$\mathrm{P}(\mathrm{E})=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3 \times 1}{12}=\frac{1}{4}$
(iii)Girl will win ₹ 5 if she get tail
$\mathrm{F}=\{(2 \mathrm{~T}),(4 \mathrm{~T}),(6 \mathrm{~T})\}$
$\mathrm{P}(\mathrm{F})=\frac{3 \times 1}{12}=\frac{1}{4}$

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Question 34 Marks

Read the text carefully and answer the questions:
The marks scored by five students in a test for 25 marks are: $8,13,12,15,22$.

(a) What is the Mean of the data?
(b) What is the variance of the given data?
(c) What is the Standard Deviation of the data?
OR
What is the Mean of first 20 natural numbers?

Answer

(i) Mean $=\frac{12+8+13+15+22}{5}$
$=14$
(ii) Since, mean $(\bar{x})=14$

$x_i$	$x _{ i }-\bar{X}$	$\left( x _{ i }-\bar{X}\right)^2$
8	-6	36
12	-2	4
13	-1	1
15	1	1
22	8	64

$\operatorname{Var}(\mathrm{X})=\frac{1}{n} \sum\left(x_{i}-\bar{X}\right)$
$=\frac{106}{5}=21.2$
(iii)Standard Deviation $=\sqrt{\text { Var. }}$
$=\sqrt{21.2}=4.604$
OR
Sum of first 20 natural numbers $=\mathrm{S}_{20}$
$=\frac{20}{2}[2 \times 1+(19) 1]$
$\mathrm{S}_{20}=210$
Mean $=\frac{210}{20}$
$=10.5$

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Question 44 Marks

Read the text carefully and answer the questions:
During function lots of charts and displays are made with different colours, one such display is of concentric circles.
A circle is drawn whose equation is $x^{2}+y^{2}-4 x-6 y-12=0$ and based on this other consecutive circles are drawn.
(a) Find the centre of given circle?
(b) Find the radius of given circle?
(c) Find the point which lies in the interior of circle?
OR
Find the Equation of a circle concentric with given circle, whose radius is double the radius of given circle?

Answer

(i) $x^{2}+y^{2}-4 x-6 y-12=0$
Here, $2 \mathrm{~g}=-4,2 \mathrm{f}=-6, \mathrm{c}=-12$
$\mathrm{g}=-2, \mathrm{f}=-3, \mathrm{c}=-12$
Centre ( $-\mathrm{g},-\mathrm{f}$ ) $=(2,3)$
(ii) $\mathrm{r}=\sqrt{4+9+12}=\sqrt{25}=5$ units
(iii)as $\sqrt{(2-0)^{2}+(3-3)^{2}}<5$
hence, $(0,3)$ lies inside the circle.
Radius of given circle $=5$
Radius of required circle $=10$
$\therefore$ Circle is $(x-2)^{2}+(y-3)^{2}=(10)^{2}$
$\Rightarrow x^{2}+y^{2}-4 x-6 y-87=0$

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Applied Maths Case study (4 Marks)

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