M.C.Q (1 Marks) - Page 2 - Maths STD 11 Science Questions - Vidyadip

Questions · Page 2 of 4

M.C.Q (1 Marks)

MCQ 511 Mark

Three digit numbers are formed using the digits 0, 2, 4, 6, 8. A number is chosen at random out of these numbers. What is the probability that this number has the same digits?

A
$\frac{1}{16}$
B
$\frac{16}{25}$
C
$\frac{1}{645}$
✓
$\frac{1}{25}$

Answer

Correct option: D.

$\frac{1}{25}$

The given digits are 0, 2, 4, 6, 8.

____	____	____
Hundreds	Tens	Ones

Now, there are 4 ways to fill the hundreds place (0 cannot occupy the hundreds place), 5 ways to fill the tens place and 5 ways to fill the ones place.
Total number of 3 digit numbers formed using the given digits = 4 × 5 × 5 = 100
The three digit numbers formed using given digits that have the same digits are 222, 444, 666 and 888
Number of 3 digit numbers that have the same digits = 4
$\therefore$ P(three digit number formed has the same digits)
$\frac{\text{Number of 3 digits numbers that have the same digits}}{\text{Total number of 3 digit numbers formed using the given digits}}$
$=\frac{4}{100}=\frac{1}{25}$
Hence, the correct answer is option (d).

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MCQ 521 Mark

What is the sample space for choosing an odd number from $2$ to $10$ at random?

A
$2, 4, 6, 8, 10$
B
$1, 2, 3, 5$
✓
$3, 5, 7, 9$
D
$1, 2, 3, 5, 7$

Answer

Correct option: C.

$3, 5, 7, 9$

Sample space is the collection of all possible events.
So, the sample space for choosing an odd number from $2$ to $10$ at random $= 3, 5, 7, 9.$

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MCQ 531 Mark

One card is drawn from a pack of $52$ cards. The probability that it is the card of a king or spade is:

A
$\frac{1}{26}$
B
$\frac{3}{26}$
✓
$\frac{4}{13}$
D
$\frac{3}{13}$

Answer

Correct option: C.

$\frac{4}{13}$

If $A$ and $B$ denote the events of drawing a king and a spade card, respectively, then event $A$ consists of four sample points, whereas event $B$ consists of $13$ sample points.
Thus, $\text{P(A)}=\frac{4}{52}$ and $\text{P(B)}=\frac{13}{52}$
The compound event $\text{(A n B)}$ consists of only one sample point, king of spade.
So, $\text{P}(\text{A}\cap\text{B})=\frac{1}{52}$
By addition theorem , we have:
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}$
$=\frac{16}{52}$
$=\frac{4}{13}$
Hence, the probability that the card drawn is either a king or a spade is given by $\frac{4}{13}.$

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MCQ 541 Mark

Two dice are thrown simultaneously. The probability of obtaining total score of seven is:

A
$\frac{5}{36}$
✓
$\frac{6}{36}$
C
$\frac{7}{36}$
D
$\frac{8}{36}$

Answer

Correct option: B.

$\frac{6}{36}$

When two dices are thrown, there are $(6 \times 6) = 36$ outcomes.
The set of all these outcomes is the sample space given by
$S = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$
$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$
$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$
$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$
$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$
$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$
$\therefore n(S) = 36$
Let $E$ be the event of getting a total score of $7.$
Then $E = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$
$\therefore n(E) = 6$
Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{6}{36}$

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MCQ 551 Mark

If the vertex $= (2, 0)$ and the extremities of the latus rectum are $(3, 2)$ and $(3, -2),$ then the equation of the parabola is:

A
$y^2= 2x - 4$
B
$x^2= 4y - 8$
✓
$y^2= 4x - 8$
D
None

Answer

Correct option: C.

$y^2= 4x - 8$

$y^2=4 a x$
$(y-0)^2=4 a(x-2)$
$y^2=4 a(x-2)$
$y^2=4(x-2)$
$y^2=4 x-8$

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MCQ 561 Mark

The radius of circle $x^2+ y^2 - 6x - 8y = 0:$

✓
$5$
B
$4$
C
$3$
D
$2$

Answer

Correct option: A.

$5$

The radius of circel $x^2+ y^2- 6x - 8y =0$ is
$=\sqrt{\text{g}^2+\text{f}-\text{c}}$
Here $g = -3, f = -4, c = 0$
$\Rightarrow\text{r}=\sqrt{(-3)^2+(-4)^2}$
$=\sqrt{9+16}$
$=\sqrt{25}$
$=5$

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MCQ 571 Mark

Two dice are thrown simultaneously. The probability of getting a pair of aces is

✓
$\frac{1}{36}$
B
$\frac{1}{3}$
C
$\frac{1}{6}$
D
none of these

Answer

Correct option: A.

$\frac{1}{36}$

When two dice are thrown simultaneously,
the sample space associated with the random experiment is given by:
$S = \{(1, 1), (1, 2), (1, 3), (6, 4), (6, 5), (6, 6)\}$
Clearly, total number of elementary events $= 36$
Let $A$ be the event of getting a pair of aces.
Then $A = \{(1, 1)\}$
$\therefore\text{n(A)}=1$
Hence, required probability $=\frac{\text{n(A)}}{\text{n(S)}}=\frac{1}{36}$

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MCQ 581 Mark

The line segment joining the foci of the hyperbola $x^2- y^2 + 1 = 0$ is one of the diameters of a circle. The equation of the circle is:

A
$\text{x}^2+\text{y}^2=4$
✓
$\text{x}^2+\text{y}^2=\sqrt{2}$
C
$\text{x}^2+\text{y}^2=2$
D
$\text{x}^2+\text{y}^2=2\sqrt{2}$

Answer

Correct option: B.

$\text{x}^2+\text{y}^2=\sqrt{2}$

$x^2- y^2 + 1 = 0$
$\Rightarrow (x - 0)^2+ (y - 0)^2= 0$ Foci of the given hyperbola
$=(0,+\sqrt{2)}$ Centre of the hyperbola is $(0, 0)$ Diameter of the circle is
the distance between foci of the hyperbola and which is given by
$\text{D}=\sqrt{(0-0)^2+(\sqrt{2}+\sqrt{2})^2}=2\sqrt{2}$ Centre of the circle will be same as that of the hyperbola
Centre $= (0, 0)$ and Radius $=\sqrt{2}$
Equation of the circle is $x^2+ y^2 = 2$

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MCQ 591 Mark

A pack of cards contains $4$ aces, $4$ kings, $4$ queens and $4$ jacks. Two cards are drawn at random. The probability that at least one of them is an ace is

A
$\frac{1}{5}$
B
$\frac{3}{16}$
✓
$\frac{9}{20}$
D
$\frac{1}{9}$

Answer

Correct option: C.

$\frac{9}{20}$

We have:
$P($both are aces$)=\frac{\ ^{4}\text{C}_2}{\ ^{16}\text{C}_2}$
$=\frac{4}{16}\times\frac{3}{15}=\frac{1}{20}$
$P($one are ace$)=\frac{\ ^{4}\text{C}_1\times\ ^{12}\text{C}_1}{\ ^{16}\text{C}_2}=\frac{2}{5}$
$\therefore P($ at least one are ace$)=\frac{1}{20}+\frac{2}{5}=\frac{9}{20}$

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MCQ 601 Mark

The events $A, B, C$ are mutually exclusive events such that $\text{P(A)}=\frac{(3\text{x}+1)}{3,\text{P(B)}}=\frac{(\text{x}-1)}{4\text{ and}\text{ P}}\text{(C)}=\frac{(1-2\text{x})}{4}.$The set of possible values of $x$ are in the interval:

✓
$\Big[\frac{1}{3},\frac{1}{2}\Big]$
B
$\Big[\frac{1}{3},\frac{2}{3}\Big]$
C
$\Big[\frac{1}{3},\frac{13}{3}\Big]$
D
$\Big[0,1\Big]$

Answer

Correct option: A.

$\Big[\frac{1}{3},\frac{1}{2}\Big]$

$\text{P(A)}=\frac{(3\text{x}+1)}{3}$
$\text{P(B)}=\frac{(\text{x}-1)}{4}$
$\text{P(C)}=\frac{(1-2\text{x})}{4}$
These are mutually exclusive events.
$\Rightarrow-1\leq3\text{x}\leq2,-3\leq\text{x}\leq,-\leq2\text{x}\leq1$
$\Rightarrow-\frac{1}{3}\leq\text{x}\leq\frac{2}{3},-2\leq\text{x}\leq1,-\frac{1}{2}\leq\text{x}\leq\frac{1}{2}$
Also, $0 \leq\frac{(3\text{x}+1)}{3}+\frac{\text{x}-1}{4}+\frac{(1-2\text{x)}}{4\leq1}$
$\Rightarrow\frac{1}{3}\leq\text{x}\leq\frac{13}{3}$
$\Rightarrow\text{max}\Big\{\frac{-1}{3},-3,-\frac{1}{2},\frac{1}{3}\Big\}$
$\Rightarrow\frac{1}{3}\leq\text{x}\leq\frac{1}{2}$
$\Rightarrow\text{x}\in\Big[\frac{1}{3},\frac{1}{2}\Big]$

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MCQ 611 Mark

If $\frac{(1-3\text{P})}{2},\frac{(1+4\text{P})}{3},\frac{(1+\text{P})}{6}$ are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of $p$ is:

A
$(0,1)$
✓
$\Big(\frac{-1}{4},\frac{1}{3}\Big)$
C
$\big(0,\frac{1}{3}\big)$
D
$(0,\infty)$

Answer

Correct option: B.

$\Big(\frac{-1}{4},\frac{1}{3}\Big)$

$\text{P(A)}=\frac{(1-3\text{P})}{2}$
$\text{P(B)}=\frac{(1+4\text{P})}{3}$
$\text{P()}=\frac{(1+\text{P})}{6}$
The events are mutually exclusive and exhaustive.
$\therefore\text{P}(\text{A}\cup\text{B }\cup\text{C})=\text{P(A)}+\text{P(B)}+\text{P(C)}=1$
$\Rightarrow0\leq\text{P(A)}\leq1,0\leq\text{P(B)}\leq1,0\leq\text{P(C)}\leq1$
$\Rightarrow0\leq\frac{1-3\text{P}}{2}\leq1$, $0\leq\frac{1-4\text{P}}{3}\leq1$, $0\leq\frac{1-\text{P}}{6}\leq1$
$\Rightarrow\frac{-1}{3}\leq\text{P}\leq\frac{1}{3}\ ...(1)$
$\frac{-1}{4}\leq\text{P}\leq\frac{1}{2}\ ...(2)$
and ${-1}\leq\text{P}\leq{5}\ ...(3)$
The common solution of $(1), (2),$ and $(3)$ is $\frac{-1}{4}\leq\text{P}\leq\frac{1}{3}$
$\therefore$ The set values of $P$ are $\Big(\frac{-1}{4},\frac{1}{3}\Big)$

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MCQ 621 Mark

Two unbiased coins are tossed simultaneously. Find the probability of getting at least one head.

A
$\frac{1}{2}$
B
$\frac{1}{4}$
✓
$\frac{3}{4}$
D
$\text{None}\text{ of}\text{ these}$

Answer

Correct option: C.

$\frac{3}{4}$

If two unbiased coins are tossed simultaneously, then the sample space will be $S.$
$\text{S: \{H H, H T, T H, T T\} n(S) = 4n}$
$E:$ At least one head is obtained $\text{\{H H, H T, T H\} n(E) = 3n(E) = 3}$
Hence, $P($At least one head$) =\frac{3}{4}$

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MCQ 631 Mark

A bag contains $2$ red, $3$ green and $2$ blue balls. Two balls are drawn at random. The probability that none of the balls drawn is blue is:

✓
$\frac{10}{21}$
B
$\frac{11}{21}$
C
$\frac{2}{7}$
D
$\frac{5}{7}$

Answer

Correct option: A.

$\frac{10}{21}$

Total number of balls $= 2 + 3 + 2 = 7$
Two balls are drawn.
Now, $P($none of them is blue$) =\frac{^5\text{C}_2}{^7\text{C}_2}$
$=\frac{\Big\{\frac{(5\times4)}{(2\times1)}\Big\}}{\Big\{\frac{(7\times6)}{(2\times1)}\Big\}}$
$=\frac{(5\times4)}{(7\times6)}$
$=\frac{(5\times2)}{(7\times3)}$
$=\frac{10}{21}$

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MCQ 641 Mark

Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is $9$ will be:

A
$\frac{13}{15}$
✓
$\frac{13}{18}$
C
$\frac{1}{9}$
D
$\frac{8}{9}$

Answer

Correct option: B.

$\frac{13}{18}$

When two dices are thrown, there are $(6 \times 6) = 36$ outcomes.
The set of all these outcomes is the sample space is given by
$S = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$
$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$
$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$
$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$
$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$
$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$
$\therefore\text{n}\text{(S)} = 36$
Let $E$ be the event of getting the digits which are neither equal nor give a total of $9.$
Then $E' =$ event of getting either a doublet or a total of $9$
Thus, $E' = \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (4, 5), (5, 4), (6, 3)\}$
i.e. $n(E') = 10$
$\text{P(E}')=\frac{\text{n(E}')}{\text{n(E)}}=\frac{10}{36}=\frac{5}{18}$
Hence, required probability $P(E) = 1 - P(E')$
$=1-\frac{5}{18}=\frac{13}{18}$

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MCQ 651 Mark

The lines $2x−3y=5$ and $3x−4y=7$ are the diameters of a circle of area $154$ sq.units. The equation of the circle is:

A
$x^2+ y^2 + 2x - 2y = 62$
✓
$x^2+ y^2 - 2x + 2y = 47$
C
$x^2+ y^2+ 2x - 2y = 47$
D
$x^2+ y^2 - 2x + 2y = 62$

Answer

Correct option: B.

$x^2+ y^2 - 2x + 2y = 47$

Given equation of lines is $2x - 3y = 5 .....$
$(i)$ and $3x - 4y = 7 .....$
$(ii)$ Solving above equations, we get the point of intersection as $(1, -1).$
Eqns $(i)$ and $(ii)$ are diameters of the circle.
We know that the centre of
circle $=$ point of intersection of diameters $=(1, -1).$
Now, it is given that the area of the.
circle $=154.$
$=\pi\text{r}^2=154 $
$\Rightarrow\text{r}=7$
Hence, the equation of required circle is
$(x - 1)^2+ (y + 1)^2 = 72$
$x^2+ y^2- 2x + 2y = 47$

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MCQ 661 Mark

The length of latus rectum of the parabola $(x - 2a) 2 + y^2 = x^2$ is:

A
$2a$
B
$3a$
C
$6a$
✓
$4a$

Answer

Correct option: D.

$4a$

We have, $(x-2 a)^2+y^2=x^2$
$\Rightarrow x^2-4 a x+4 a^2+y^2=x^2 $
$\Rightarrow y^2=4 a x-4 a 2=4 a(x-a)$
Comparing it with standard parabola $Y^2=4 b X$
$Y=y, X=x-a, b=a$
We know length of latus rectum of parabola $Y^2=4 b X$ is $4 b$
Hence length of latus rectum of given parabola is $= 4 \times a = 4a$

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MCQ 671 Mark

What is the probability of getting the number $6$ at least once in a regular die if it can roll it $6$ times?

✓
$1-(5 / 6)^6$
B
$1-(1 / 6)^6$
C
$(5 / 6)^6$
D
$(1 / 6)^6$

Answer

Correct option: A.

$1-(5 / 6)^6$

Let $A$ be the event that $6$ does not occur at all.
Now, the probability of at least one $6$ occurs $= 1 – PA.$
$= 1-(5 / 6)^6 $

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MCQ 681 Mark

lf the lines $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circle of radius $7$ then the equation of the circle is:

A
$x^2+ y^2+ 2x - 4y - 47 = 0$
B
$x^2+ y^2= 49$
✓
$x^2+ y^2- 2x + 2y - 47 = 0$
D
$x^2+ y^2= 17$

Answer

Correct option: C.

$x^2+ y^2- 2x + 2y - 47 = 0$

$2x - 3y = 5......... (1)$
$3x - 4y = 7......... (2)$
Intersection of this lines given centre of circle
$\therefore$ from $(1) (2),$
$n = 1, y = -1$
So, equation of circle is
$(x - 1)^2+ (y + 1)^2= (7)^2$
$\Rightarrow x^2+ y^2- 2x + 2y - 47 = 0$

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MCQ 691 Mark

One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other, then odds in favour of the other are

A
$\text{1 : 3}$
✓
$\text{3 : 1}$
C
$\text{2 : 3}$
D
$\text{3 : 2}$

Answer

Correct option: B.

$\text{3 : 1}$

Let $\text{P(B)}=\text{X}$
Than, $\text{P(A)}=\frac{2\text{x}}{3}$
$\text{P(A)}+\text{P(B)}=\text{x}+\frac{2x}{3}=\frac{5x}{3}$
$\Rightarrow\frac{5\text{x}}{3}=1$ $\big(\therefore$ They are exhaustive events $\big)$
$\Rightarrow\text{x}=\frac{3}{5}$
Now, $\text{P(A)}=\frac{2}{5}$ and $\text{P(B)}=\frac{3}{5}$
$\therefore$ odd in favour of $B=\frac{\frac{3}{5}}{\frac{1-3}{5}}=\frac{3}{2}=\text{3 : 1}$

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MCQ 701 Mark

Choose the correct answer. $6$ boys and $6$ girls sit in a row at random. The probability that all the girls sit together is:

A
$\frac{1}{432}$
B
$\frac{12}{431}$
✓
$\frac{1}{132}$
D
none of these.

Answer

Correct option: C.

$\frac{1}{132}$

If all the girls sit together, then we consider it as $1$ group

$\therefore$ Total number of arrangement of $6 + 1 = 7$ persons in a row $= 7!$ And the girls also interchanged their places with $6!$ Ways.
$\therefore\ \text{Required probability}=\frac{6!7!}{12!}$
$=\frac{6\times5\times4\times3\times2\times7!}{12\times11\times10\times9\times8\times7!}=\frac{1}{132}$

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MCQ 711 Mark

The equation of a locus is $y^2+ 2ax + 2by + c = 0$. Then:

A
it is an ellipse
B
it is a parabola
C
latus rectum $= a$
✓
latus rectum $= 2a$

Answer

Correct option: D.

latus rectum $= 2a$

Given equation of locus is $y^2+ 2ax + 2by + c = 0$, which can be written
$= as (y + b)^2$
$=-4\Big(\frac{\text{a}}{2}\Big)\Big(\text{a}+\frac{\text{c}-\text{b}^2}{2\text{a}}\Big)$
which is locus of parabola. For parabola with equation $y^2$
$= 4ax, 4a$ is length of latus rectum.
Then for given locus, length of latus rectum is
$=4\Big(\frac{\text{a}}{\text{b}}\Big)=2\text{a}.$

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MCQ 721 Mark

An urn contains $6$ balls of which two are red and four are black. Two balls are drawn at random. What is the probability that they are of different colours?

A
$\frac{2}{5}$
B
$\frac{1}{15}$
✓
$\frac{8}{15}$
D
$\frac{4}{15}$

Answer

Correct option: C.

$\frac{8}{15}$

Given that, the total number of balls $= 6$ balls
Let $A$ and $B$ be the red and black balls, respectively,
The probability that two balls are drawn are different $= P($the first ball drawn is red$)($the second ball drawn is black$)+ P($the first ball drawn is black$)P($the second ball drawn is red$)$
$=\Big(\frac{2}{6}\Big)\Big(\frac{4}{5}\Big)+\Big(\frac{4}{6}\Big)\Big(\frac{2}{5}\Big)$
$=\Big(\frac{8}{30}\Big)+\Big(\frac{8}{30}\Big)$
$=\frac{16}{30}$
$=\frac{8}{15}$

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MCQ 731 Mark

One mapping is selected at random from all mappings of the set A = {1, 2, 3, ..., n} into itself. The probability that the mapping selected is one to one is:

A
$\frac{1}{\text{n}^n}$
B
$\frac{1}{\text{n}}$
✓
$\frac{\text{n}-1}{\text{n}^n-1}$
D
None of these

Answer

Correct option: C.

$\frac{\text{n}-1}{\text{n}^n-1}$

Number of ways to map 1st element in set A = n
Number of ways to map 2nd element in set A = n and so on
$\therefore$Total number of mapping from set A to itself
$=\text{n}\times\text{n}\times\ ...\ \times\text{n}(\text{n times})=\text{n}^\text{n}$

For one to one mapping,

Number of ways to map 1st element in set A = n

Number of ways to map 2nd element in set A = n −1
Number of ways to map nth element in set A = 1

Total number of one to one mappings from set A to itself
$=\text{n}\times\text{(n -1)}\times\text{(n - 2)}\ \times\ ...\times1=\text{n}$
$\therefore$ Required probability
$=\frac{\text{Total number of one mappings from set A to inselp} }{\text{Total number of mappings form set A to itself}}$
$=\frac{\text{n}}{\text{n}^\text{n}}=\frac{(\text{n}-1)}{\text{n}^\text{n-1}}$
Hence, the correct answer is option (c).

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MCQ 741 Mark

A bag contains $5$ brown and $4$ white socks. Ram pulls out two socks. What is the probability that both the socks are of the same colour?

A
$9/20$
B
$2/9$
C
$3/20$
✓
$4/9$

Answer

Correct option: D.

$4/9$

Total number of socks $= 5 + 4 = 9$
Two socks are pulled.
Now, $P($Both are same colour$) = (5C_2 + 4C_2)/9C_2$
$= {(5\times 4)/(2\times 1) + (4\times 3)/(2\times 1)}/{(9\times 8)/(2\times 1)}$
$= (5\times 4) + (4\times 3)/(9\times 8)$
$= (5 + 3)/(9\times 2)$
$= 8/18$
$= 4/9$

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MCQ 751 Mark

The centre of a circle is $(x - 2, x + 1)$ and it passes through the points $(4, 4)$ Find the value $($or values$)$ of $x,$ if the diameter of the circle is of length $2\sqrt{5} \text{ units.}$

A
$1$ or $3$
B
$-1$ or $4$
✓
$5$ or $4$
D
$3$ or $-2$

Answer

Correct option: C.

$5$ or $4$

Radius of the circle $=$ dist. between the center and given pt. on the circle.
Distance between two points $ < br / > < br / >(x_1, y_1)$ and $(x_2, y_2)$
can be calculated using the formula.
$=\sqrt{(\text{x}_2}-<\text{br}/><\text{br}/>\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2$
Distance between the points $(x - 2, x + 1)$ and $D (4, 4)$
$=\sqrt{(4-\text{x}+2)^2+(4-\text{x}-1)^2}$
$=\sqrt{(6-\text{x}^2+(3-\text{x)}^2}$
$=\sqrt{36+\text{x}^2-12\text{x}+9+\text{x}^2-6\text{x}}$
$=\sqrt{2}\text{x}^2-18\text{x}+45$
Given, diameter $2\sqrt{5} \Rightarrow$ Radius $=\sqrt{5}$
$\Rightarrow\sqrt{2\text{x}^2-18\text{x}+45<\text{br}/><\text{br}/>=\sqrt{5}}$
Squaring both sides,
$2x^2- 18x + 45 = 5$
$2x^2-18x + 40 = 0$
$x^2- 9x + 20 = 0$
$(x - 5)(x - 4) = 0$
$x = 5$ or $4$

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MCQ 761 Mark

What is the sample space for choosing a letter from a set of vowel ?

A
$\{a, b, c........z\}$
✓
$\{a, e, i, o, u\}$
C
$\{a, e, i, o\}$
D
None of the these

Answer

Correct option: B.

$\{a, e, i, o, u\}$

A letter is choose from set of vowels i.e. $\{a, e, i, o, u\}.$

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MCQ 771 Mark

If two coins are tossed then find the probability of the events that at the most one tail turns up:

A
$\frac{1}{4}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
✓
$\frac{3}{4}$

Answer

Correct option: D.

$\frac{3}{4}$

The sample space of $2$ coins tossed $\text{= (h, h), (h, t), (t, h), (t, t)}$
for having atmost one tail we need $v=\text{(h, t), (t, h), (h, h)}$
Thus the probability is $\frac{3}{4}$

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MCQ 781 Mark

The equation of the circle which touches $x-$axis at $(0,0)$ and touches the line $3x + 4y - 5 = 0$ is:

A
$x^2+ y^2- 4y = 0$
B
$x^2+ y^2- 10y = 0$
C
$x2 + y^2+ 10x = 0$
✓
$x^2+ y^2+ 10y = 0$

Answer

Correct option: D.

$x^2+ y^2+ 10y = 0$

Equation of circle touching $x-$ axis at $(0, 0),$ means centre of circle lie on $Y-$axis
i.e. $(0, k).$
$(x - 0)^2+ (y - k)^2= k^2 $
$S : x^2+ y^2- 2ky = 0 ......(1)$ Circle $S$ touches
$= 3x + 4y - 5 = 0 $
$\therefore \text{k}=\frac{4\text{k}-5}{5}$
$= 5k = 4k - 5 $
$= k = -5 $
$\therefore $ Equation of circle is
$=x^2+ y^2- (-5) \times 2y = 0$
$\Rightarrow x^2+ y^2 + 10y = 0$

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MCQ 791 Mark

Let $A$ and $B$ are two mutually exclusive events and if $P(A) = 0.5$ and $P(B ̅) =0.6$ then $\text{P(AUB)}$ is:

A
$0$
B
$1$
C
$0.6$
✓
$0.9$

Answer

Correct option: D.

$0.9$

Given, $A$ and $B$ are two mutually exclusive events.
So, $\text{P(A ∩ B)} = 0$
Again given $P(A) = 0.5$ and $P(B ̅) = 0.6$
$\text{P(B) = 1 – P(B ̅)} = 1 – 0.6 = 0.4$
Now, $\text{P(A ∪ B) = P(A) + P(B) – P(A ∩ B)}$
$\Rightarrow \text{P(A ∪ B) = P(A) + P(B)}$
$\Rightarrow \text{P(A ∪ B)} = 0.5 + 0.4 = 0.9$

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MCQ 801 Mark

A die is thrown then find the probability of getting a number greater than $3.$

A
$\frac{1}{3}$
✓
$\frac{1}{2}$
C
$\frac{2}{3}$
D
$0$

Answer

Correct option: B.

$\frac{1}{2}$

Sample space $= 1, 2, 3, 4, 5, 6$
a no $>3$ in sample space $= 4, 5, 6 = 3$
probability of getting no greater than $3=\frac{1}{2}$

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MCQ 811 Mark

The length of the latus rectum of the parabola whose focus is $(3, 3)$ and directrix is $3x - 4y - 2 = 0$ is

A
$1$
✓
$2$
C
$4$
D
$8$

Answer

Correct option: B.

$2$

Length of the latus rectum $(1) = 2 ($perpendicular distance from focus to directix$)$
$1=2 \ \frac{3\ (3)\ -4\ (3)\ -2}{\sqrt{3^2\ +\ 4^2}}=2$

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MCQ 821 Mark

Two numbers are chosen from $\{1, 2, 3, 4, 5, 6\}$ one after another without replacement. Find the probability that the smaller of the two is less than $4.$

✓
$\frac{4}{5}$
B
$\frac{1}{15}$
C
$\frac{1}{5}$
D
$\frac{14}{15}$

Answer

Correct option: A.

$\frac{4}{5}$

Total number of ways of choosing two numbers out of six $=\ ^6\text{C}_6=\frac{(6\times2)}{2}=3\times5=15$
If smaller number is chosen as $3$ then greater has choice are $4, 5, 6$ So, total choices $= 3$
If smaller number is chosen as $2$ then greater has choice are $3, 4, 5, 6$ So, total choices $= 4$
If smaller number is chosen as $1$ then greater has choice are $2, 3, 4, 5, 6$ So, total choices $= 5$
Total favourable case $= 3 + 4 + 5 = 12$
Now, required probability
$=\frac{12}{15}=\frac{4}{5}$

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MCQ 831 Mark

Find the sample space for choosing a prime number less than $2020$ at random.

✓
$2, 3, 5, 7, 11, 13, 17, 19$
B
$2, 3, 4, 5, 7, 11, 13, 17, 19$
C
$2, 3, 5, 7, 11, 13, 17, 19, 20$
D
$2, 3, 5, 7, 11, 13, 17, 19, 15$

Answer

Correct option: A.

$2, 3, 5, 7, 11, 13, 17, 19$

Sample space is the collection of all possible events.
So, sample space for choosing a prime number less than $20 = 2, 3, 5, 7, 11, 13, 17, 19.$

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MCQ 841 Mark

On the parabola $y = x^2$, the point least distant from the straight line $y = 2x - 4$ is:

✓
$(1, 1)$
B
$(1, 0)$
C
$(1, -1)$
D
$(0, 0)$

Answer

Correct option: A.

$(1, 1)$

Given, parabola is $y = x^2....(i)$
$d$ straight line is $y = 2x - 4 ....(ii)$
From equations $(i)$ and $(ii),$ we get
$x^2- 2x - 4 = 0$
Let $f (x) = x^2- 2x - 4$
Thus $f (x) = 2x - 2$
or least distance, put $f' (x) = 0$
$\Rightarrow 2x - 2 = 0$
$\Rightarrow x = 1$
From equation $(i),$ we have $y = 1$
Hence, the point least distant from the line is $(1, 1)$

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MCQ 851 Mark

There are $30$ tickets numbered from $1$ to $30$ in a box . A ticket is drawn at random. What is the probability that the ticket drawn bears an odd number?

✓
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{2}{3}$
D
$\frac{1}{4}$

Answer

Correct option: A.

$\frac{1}{2}$

Total number of outcomes $= 30$
Favourable outcomes $($odd number on the ticket$) = 15$
Probability $=\frac{15}{30}=\frac{1}{2}$

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MCQ 861 Mark

The probability that a leap year will have $53$ Fridays or $53$ Saturdays is:

A
$\frac{2}{7}$
✓
$\frac{3}{7}$
C
$\frac{4}{7}$
D
$\frac{1}{7}$

Answer

Correct option: B.

$\frac{3}{7}$

We know that a leap year has $366$ days $($i.e. $7 \times 52 + 2) = 52$ weeks and $2$ extra days .
The sample space for these $2$ extra days is given below:
$S = \{($Sunday, Monday$), ($Monday, Tuesday$), ($Tuesday, Wednesday$), ($Wednesday, Thursday$), ($Thursday, Friday$), ($Friday, Saturday$), ($Saturday, Sunday$)\}$ There are $7$ cases.
$\therefore\text{n(S)}=7$
Let $E$ be the event that the leap year has $53$ Fridays or $53$ Saturdays.
$E = \{($Thursday, Friday$), ($Friday, Saturday$), ($Saturday, Sunday$)\}$
i.e. $n(E) = 3$
$\therefore \text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{3}{7}$
Hence, the probability that a leap year has $53$ Fridays or $53$ Saturdays is $\frac{3}{7}$.

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MCQ 871 Mark

If $(a, b)$ lies on circle with centre as origin, then its radius will be:

A
$\text{a} - \text{ b}$
B
$\text{a + b}$
✓
$\sqrt{\text{a}^2+\text{b}^2}$
D
$\text{a}^2+\text{b}^2$

Answer

Correct option: C.

$\sqrt{\text{a}^2+\text{b}^2}$

We know the formula,
The equation of a circle of radius $r$ and centre the origin is
$x^2+ y^2= r^2$
Here the center is $(a, b)$
so Radius, $\text{r} = \sqrt{\text{a}^{2} + \text{b}^{2}}$

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MCQ 881 Mark

Choose the correct answer. If $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cap\text{B})$ for any two events $A$ and $B,$ then:

✓
$\text{P(A)}=\text{P(B)}$
B
$\text{P(A)}>\text{P(B)}$
C
$\text{P}(\text{A})<\text{P(B)}$
D
none of these.

Answer

Correct option: A.

$\text{P(A)}=\text{P(B)}$

Given that, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\big[\text{P(A)}-\text{P}(\text{A}\cap\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]=0$
But $\text{P(A)}-\text{P}(\text{A}\cap\text{B})\geq0\ ....(\text{i})$
$\big[\because\ \text{P}(\text{A}\cap\text{B})\leq\text{P(A)}$ or $\text{P(B)}\big]$
And $\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})\geq0\ ...(\text{ii})$
From eq. $(i)$ and $(ii)$ we get
$\text{P(A)}=\text{P(B)}$

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MCQ 891 Mark

If three dice are throw simultaneously, then the probability of getting a score of $5$ is:

A
$\frac{5}{216}$
B
$\frac{1}{6}$
✓
$\frac{1}{36}$
D
$\text{None of these}$

Answer

Correct option: C.

$\frac{1}{36}$

When three dice are thrown together, the sample space $S$ associated with the random experiment is given by,
$S = \{(1, 1, 1), (1, 1, 2), (1, 1, 3) ...(6, 6, 5), (6, 6, 6)\}$
Clearly, total number of elementary events $n(S) = 216$
Let $A$ be the event of getting a total score of $5.$
Then $A = \{ (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1)\}$
$\therefore$ Favourable number of elementary events $= 6$
i.e. $n(A) = 6$
Hence, required probability $=\frac{6}{216}=\frac{1}{36}$

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MCQ 901 Mark

Choose the correct answer. If $A$ and $B$ are mutually exclusive events, then:

✓
$\text{P(A)}\leq\text{P}(\bar{\text{B}})$
B
$\text{P(A)}\geq\text{P}(\bar{\text{B}})$
C
$\text{P}(\text{A})<\text{P}(\bar{\text{B}})$
D
none of these.

Answer

Correct option: A.

$\text{P(A)}\leq\text{P}(\bar{\text{B}})$

For mutually exclusive events,
$\text{P}(\text{A}\cap\text{B})=0$
$\therefore\ \text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}$ $\big[\because\text{ P}(\text{A}\cap\text{B})=0\big]$
$\Rightarrow\text{P}(\text{A})+\text{P(B)}\leq1$
$\Rightarrow\text{P(A)}+1-\text{P}(\bar{\text{B}})\leq1$ $\big[\text{P(B)}=1-\text{P}(\bar{\text{B}})\big]$
$\Rightarrow\text{P(A)}-\text{P}(\bar{\text{B}})\leq0$
$\Rightarrow\text{P(A)}-\text{P}(\bar{\text{B}})$

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MCQ 911 Mark

Find the latus rectum of the parabola $x^2+ 2y - 3x + 5 = 0:$

A
$1$
✓
$2$
C
$4$
D
$8$

Answer

Correct option: B.

$2$

$\Rightarrow\text{x}^2-3\text{x}=-2\text{y}-5$
$\Rightarrow\text{x}^2-2.$
$=\frac{3}{2}\text{x}+\frac{9}{4}$
$=-2\text{y}-5+\frac{9}{4}$
$=-2\text{y}-\frac{11}{4}$
$\Rightarrow\Big(\text{x}-\frac{3}{2}\Big)^2$
$=-2\Big(\text{y}-\frac{11}{4}\Big)$
Hence latus rectum
$=4\times\frac{1}{2}=2$

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MCQ 921 Mark

If the probability of $A$ to fail in an examination is $\frac{1}{5}$ and that of $B$ is$\frac{3}{10}$Then, the probability that either $A$ or $B$ fails is

A
$\frac{1}{2}$
B
$\frac{11}{25}$
✓
$\frac{19}{50}$
D
None of these

Answer

Correct option: C.

$\frac{19}{50}$

Given:
$\text{P(A)}=\frac{1}{5}$
$\therefore\text{P(A}')=1-\frac{1}{5}=\frac{4}{5}$
$\text{P(B)}=\frac{3}{10}$
$\therefore\text{P(B}')=1-\frac{3}{10}=\frac{7}{30}$
Hence, required probability $=\text{P}(\text{A}\cap\text{B}')+\text{P}(\text{A}'\cap\text{B})$
$=\frac{1}{5}\times\frac{7}{10}+\frac{4}{5}\times\frac{3}{10}$
$=\frac{7}{50}+\frac{12}{50}$
$=\frac{19}{50}$

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MCQ 931 Mark

Coordinates of centre and radius of the circle $(x - 3)^2+ (y + 4)^2= 25$ are respectively:

A
$(3, 4), 25$
B
$(-3, 4), 5$
✓
$(3, -4), 5$
D
$(3, -4), 25$

Answer

Correct option: C.

$(3, -4), 5$

$= (x - 3)^2+ (y + 4)^2= 25$
$= (x - 3)^2+ (y - (-4))^2- (5)^2$
$= (x - 4)^2+ (y - k)^2- (r)^2$
$= r = 5 (h, k)$
$= (3, -4)$

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MCQ 941 Mark

Two dice are thrown simultaneously. The probability of obtaining a total score of $5$ is:

A
$\frac{1}{18}$
B
$\frac{1}{12}$
✓
$\frac{1}{9}$
D
None of these

Answer

Correct option: C.

$\frac{1}{9}$

When two dice are thrown, there are $(6 \times 6) = 36$ outcomes.
The set of all these outcomes is the sample space given by
$S = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$
$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$
$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$
$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$
$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$
$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$
i.e. $n(S) = 36$
Let $E$ be the event of getting a total score of $5.$
Then $E = \{(1, 4), (2, 3), (3, 2), (4, 1)\}$
$\therefore n(E) = 4$
Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{4}{36}=\frac{1}{9}$

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MCQ 951 Mark

The length of latus rectum of the parabola $y^2+ 8x - 2y + 17 = 0$ is:

A
$2$
B
$4$
✓
$8$
D
$16$

Answer

Correct option: C.

$8$

The given parabola is, $y^2+ 8x - 2y + 17 = 0$
$\Rightarrow (y^2- 2y + 1) = -8x - 17 + 1 = -8x - 16$
$\Rightarrow (y - 1)^2 = -8 (x + 2)$
Comparing with standard parabola $Y^2= -4aX$
$Y = y - 1, X = x + 2, a = 2$
Hence length of latus rectum is $= 4a = 4 \times 2 = 8$

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MCQ 961 Mark

Equation of the circle with centre on the $y-$ axis and passing through the origin and the point $(2, 3)$ is:

A
$x^2+y^2+13 y=0 $
B
$x^2+3 y^2+13 x+3=0 $
✓
$6x^2+6 y^2-13 y=0 $
D
$x^2+y^2+13+3=0 $

Answer

Correct option: C.

$6x^2+6 y^2-13 y=0 $

We have to find equation of a circle with center on the $y-$axis.
General equation of such circle is $(x - 0)^2+ (y - k)^2= k^2$
It passes through $(2, 3)$
i.e, $2^2+ (3 - k)^2= k^2$
$\Rightarrow4 + 9 + \text{k}^2 - 6\text{k} = \text{k}^2$
$\Rightarrow\text{k}=\frac{13}{6}$
$\therefore $ Equation of circle
$=\text{x}^2+\Big(\text{y}-\frac{\text{13}}{6}\Big)^2$
$=\Big(\frac{13}{6}\Big)^2$
$=6\text{x}^2+6\text{y}^2-13\text{y}=0$

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MCQ 971 Mark

Five persons entered the lift cabin on the ground floor of an $8$ floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first, then the probability of all $5$ persons leaving at different floor is:

✓
$\frac{\ ^{7}\text{P}_5}{7^5}$
B
$\frac{\ ^{7^5}}{\ ^{7}\text{P}_5}$
C
$\frac{\ ^{6}}{\ ^{6}\text{P}_5}$
D
$\frac{\ ^{5}\text{P}_5}{5^5}$

Answer

Correct option: A.

$\frac{\ ^{7}\text{P}_5}{7^5}$

Since, it is an eight $-$ storey building.
So, there are $7$ possible options for them in $7$ floors in total if ground floor is not considered.
Hence, total possible outcomes $= 7\times 7\times 7 \times 7\times 7= 7^5$
Thus, number of ways in which $5$ persons can leave from seven floors differently $=\ ^{7}\text{P}_5$
Required probability $=\frac{\ ^{7}\text{P}_5}{7^5}$

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MCQ 981 Mark

Two dice are thrown:
$P$ is the event that the sum of the scores on the uppermost faces is a multiple of $6.$
$Q$ is the event that the sum of the scores on the uppermost faces is at least $10.$
$R$ is the event that same scores on both dice.
Which of the following pairs is mutually exclusive?

A
$P, Q$
B
$P, R$
C
$Q, R$
✓
None of these

Answer

Correct option: D.

None of these

Possibilities of $P, (3, 3),(6, 6),(1, 5),(5, 1),(4, 2),(2, 4)$
Possibilities of $Q:(5, 5),(5, 6),(6, 5),(6, 6)$
Possibilities of $R:(1, 1),(2, 2),(3, 3),(4, 4),(5, 5),(6, 6)$
Thus, the possibilities are neither exhaustive, nor mutually exclusive nor these are complementary probabilities.

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MCQ 991 Mark

A person write $4$ letters and addresses $4$ envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelopes, is

A
$\frac{1}{4}$
B
$\frac{11}{24}$
C
$\frac{15}{24}$
✓
$\frac{23}{24}$

Answer

Correct option: D.

$\frac{23}{24}$

Total number of ways of placing four letters in $4$ envelops $= 4 = 24$
All the letters can be dispatched in the right envelops in only one way.
Therefore, the probability that all the letters are placed in the right envelops is $\frac{1}{24}$.
Hence, probability that all the letters are not placed in the right envelops $=1-\frac{1}{24}=\frac{23}{24}$

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MCQ 1001 Mark

The probability of getting a total of $10$ in a single throw of two dices is:

A
$\frac{1}{9}$
✓
$\frac{1}{12}$
C
$\frac{1}{6}$
D
$\frac{5}{36}$

Answer

Correct option: B.

$\frac{1}{12}$

When two dices are thrown, there are $(6 \times 6) = 36$ outcomes.
The set of all these outcomes is the sample space, given by
$S = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$
$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$
$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$
$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$
$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$
$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$
i.e. $n(S) = 36$
Let $E$ be the event of getting a total score of $10.$
Then $E = \{(4, 6), (5, 5), (6, 4)\}$
$\therefore n(E) = 3$
Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{3}{36}=\frac{1}{12}$

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