M.C.Q (1 Marks)

MCQ 11 Mark

What is the total number of sample spaces when a die is thrown $2$ times?

A
$6$
B
$12$
C
$18$
✓
$36$

Answer

Correct option: D.

$36$

The possible outcomes when a die is thrown are $1, 2, 3, 4, 5,$ and $6.$
Given, a die is thrown two times.
Then, the total number of sample space $= (6 \times 6)= 36$

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MCQ 21 Mark

Choose the correct answer. In a non$-$leap year, the probability of having $53$ tuesdays or $53$ wednesdays is:

✓
$\frac{1}{7}$
B
$\frac{2}{7}$
C
$\frac{3}{7}$
D
none os these.

Answer

Correct option: A.

$\frac{1}{7}$

There are $365$ days in non$-$leap year and there are $7$ days in a week
$\therefore\ 365\div7=52$ weeks $+\ 1$ days
So, this day may be Tuesday or Wednesday.
So, the required probability $=\frac{1}{7}$

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MCQ 31 Mark

If $4-$digit numbers greater than $5000$ are randomly formed from the digits $0, 1, 3, 5$ and $7,$ then the probability of forming a number divisible by $5$ when the digits are repeated is:

A
$\frac{1}{5}$
✓
$\frac{2}{5}$
C
$\frac{3}{5}$
D
$\frac{4}{5}$

Answer

Correct option: B.

$\frac{2}{5}$

Given digits are $0, 1, 3, 5, 7$
Now we have to form $4$ digit numbers greater than $5000.$
So leftmost digit is either $5$ or $7.$
When digits are repeated
Number of ways for filling left most digit $= 2$
Now remaining $3$ digits can be filled $= 5 \times 5 \times 5$
So total number of ways of $4$ digits greater than $5000 = 2 \times 5 \times 5 \times 5 = 250$
Again a number is divisible by $5$ if the unit digit is either $0$ or $5.$
So there are $2$ ways to fill the unit place.
So total number of ways of $4$ digits greater than $5000$ and divisible by $5 = 2 \times 5 \times 5 \times 2 = 100$
Now probability of $4$ digit numbers greater than $5000$ and divisible by $5$
$=\frac{100}{250}$
$=\frac{2}{5}$

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MCQ 41 Mark

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals:

A
$\frac{1}{2}$
✓
$\frac{7}{15}$
C
$\frac{2}{15}$
D
$\frac{1}{3}$

Answer

Correct option: B.

$\frac{7}{15}$

While placing $7$ while balls in a row, total gaps $= 8$
$3$ black balls can be placed in $8$ gaps
$=\text{C}=\frac{(8\times7\times6)}{(3\times2\times1)}=8\times7=56$
So, the total number of ways of arranging white and black balls such that no two black balls are adjacent $= 56 \times 3! \times 7!$
Actual number of arrangement possible with $7$ white and $3$ black balls $= (7 + 3)! = 10!$
So, the required Probability
$=\frac{(56\times3!\times7!)}{10!}$
$=\frac{(56\times3!\times7!)}{(10\times9\times8\times7!)}$
$=\frac{(56\times3!)}{(10\times9\times8)}$
$=\frac{(56\times3\times2\times1)}{(10\times9\times8)}$
$=\frac{(7\times3\times2\times1)}{(10\times9)}$
$=\frac{(7\times2)}{(10\times3)}$
$=\frac{7}{(5\times3)}$
$=\frac{7}{15}$

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MCQ 51 Mark

The equation $y^2+ 4x + 4y + k = 0$ represents a parabola whose latus rectum is:

A
$1$
B
$2$
C
$3$
✓
$4$

Answer

Correct option: D.

$4$

$y^2+ 4x + 4y + k = 0$
$y^2+ 2 \times 2y + 4 - 4 + 4x + k = 0$
$(\text{y}+2)^2=-4\text{x}-\text{k}+4$
$(\text{y}+2^2)=-4\Big(\text{x}-\frac{4\ +\ \text{k}}{4}\Big)$
$\therefore$ Latus rectum $= 4$ units

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MCQ 61 Mark

All possible outcomes of a random experiment forms the:

A
Events
✓
Sample space
C
Both
D
None of these

Answer

Correct option: B.

Sample space

Sample Space is the set of all possible outcomes of an experiment.
It is denoted by $S.$
Examples:
When a coin is tossed, $S = \text{\{H, T\}}$ where $H =$ Head and $T =$ Tail
When a dice is thrown, $S = \{1, 2 , 3, 4, 5, 6\}$
When two coins are tossed, $S = \text{\{HH, HT, TH, TT\}}$
where $H =$ Head and $T =$ Tail

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MCQ 71 Mark

The equation of directrix and latus rectum of a parabola are $3x - 4y + 27 = 0$ and $3x - 4y + 2 = 0.$ Then the length of latus rectum is:

A
$5$
✓
$10$
C
$15$
D
$20$

Answer

Correct option: B.

$10$

$\text{d}=\frac{\text{c}_1-\text{c}_2}{\sqrt{\text{a}^2+\text{b}^2}}$
where dd is the distance between lines whose equations are $ax + by + C_1= 0$
$ax + by + C_2= 0$
$\text{d}=\frac{27-2}{\sqrt{4^2+3^2}}= 5 d = 5$
If the distance between vertex and latus rectum $=$ distance of vertex from directri $x = a=$ then $d = 2a = 5$
$\Rightarrow$ Length of latus rectum $= 4a = 10$

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MCQ 81 Mark

What is number of outcomes when tossing two coins together?

A
$1$
B
$2$
C
$3$
✓
$4$

Answer

Correct option: D.

$4$

$\text{H T, H H, T H, T T} ($Here $\text{H T}$ means Head on first coin and Tail on the second coin and so on$).$

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MCQ 91 Mark

What is the approximate radius of the circle whose equation is $\text{(x}-\sqrt{3)}^2+(\text{y}+2)^2=11?$

A
$1.71$
B
$2.33$
✓
$3.32$
D
$3.85$

Answer

Correct option: C.

$3.32$

The radius of given circle is $\sqrt{11}=3.32$

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MCQ 101 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 111 Mark

If the probability for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is:

A
$>0.5$
B
$0.5$
✓
$\leq0.5$
D
$0$

Answer

Correct option: C.

$\leq0.5$

Let X and Y be two events given by,
X : A fails in an examination
Y : B fails in an examination
P(A fails) = P(X) = 0.2
P(B fails) = P(Y) = 0.3
Now, P(either A or B fails) $=\text{P}(\text{X}\cup\text{Y})$
We know that,
$=\text{P}(\text{X}\cup\text{Y})\leq\text{P(X)}+\text{P()Y}=0.2+0.3=0.5$
$\Rightarrow\text{P}(\text{X}\cup\text{Y})\leq0.5$
$\therefore\text{P}\text{(either A or B fails)}\leq0.5$
Hence, the correct answer is option (c).

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MCQ 121 Mark

The probabilities of three mutually exclusive events $A, B$ and $C$ are given by $\frac{2}{3}, \frac{1}{4}$ and $\frac{1}{6}$respectively. The statement

A
Is true.
✓
Is false.
C
Nothing can be said.
D
Could be either.

Answer

Correct option: B.

Is false.

Since the events $A, B$ and $C$ are mutually exclusive, we have:
$\text{P}(\text{A}\cup\text{B}\cup\text{C})=\frac{2}{3}+\frac{1}{4}+\frac{1}{6}=\frac{13}{12}>1$
which is not possible.
Hence, the given statement is false.

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MCQ 131 Mark

In a simultaneous throw of two dice what is the probability of getting a doublet ?

✓
$\frac{1}{6}$
B
$\frac{1}{4}$
C
$\frac{3}{4}$
D
$\frac{2}{3}$

Answer

Correct option: A.

$\frac{1}{6}$

Total number of possibilities $= 36$
Number of doublet $= 6$
Thus, probability
$=\frac{6}{36}$
$=\frac{1}{6}$

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MCQ 141 Mark

One card is drawn from a well$-$shuffled deck of $52$ cards. Find the probability of getting a face card.

A
$\frac{1}{13}$
B
$\frac{1}{26}$
✓
$\frac{3}{13}$
D
$\text{None}\text{ of}\text{ these}$

Answer

Correct option: C.

$\frac{3}{13}$

Total number of outcomes $= 52$
Favourable outcomes $($A face card$) = 12$
Probability $=\frac{12}{52}=\frac{3}{13}$

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MCQ 151 Mark

A die is rolled. What is the probability that an even number is obtained?

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

Answer

Correct option: A.

$\frac{1}{2}$

When a die is rolled, total number of outcomes $= 6 (1, 2, 3, 4, 5, 6)$
Total even number $= 3 (2, 4, 6)$
So, the probability that an even number is obtained
$=\frac{3}{6}$
$=\frac{1}{2}$

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MCQ 161 Mark

Two dice are thrown the events $\text{A, B, C}$ are as follows
$A:$ Getting an odd number on the first die.
$B:$ Getting a total of $7$ on the two dice.
$C:$ Getting a total of greater than or equal to $8$ on the two dice. Then $\text{AUB}$ is equal to

A
$15$
B
$17$
C
$19$
✓
$21$

Answer

Correct option: D.

$21$

When two dice are thrown, then total outcome $= 6 \times 6 = 36$
$A:$ Getting an odd number on the first die.
$A = \{(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4),(3, 5), (3, 6),\\ (5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6)\}$
Total outcome $= 18$
$B:$ Getting a total of $7$ on the two dice.
$B = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$ Total outcome $= 6$
$C:$ Getting a total of greater than or equal to $8$ on the two dice.
$C = \{(2, 6), (3, 5), (3, 6), (4, 4),(4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$ Total outcome $= 15$
Now $\text{n(A}\cup\text{B)}=\text{n(A)}+\text{n(B)}-\text{n(A}\cup\text{B)}$
$\Rightarrow\text{n(A}\cup\text{B)}=18+6-3$
$\Rightarrow\text{n(A}\cup\text{B)}=21$

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MCQ 171 Mark

If $S$ is the sample space and $ \text{P(A)} = \frac{1}{3} \text{P(B)}$ and $\text{S} = \text{A}\cup\text{B}$ where $A$ and $B$ are two mutually exclusive events, then $P (A) =$

✓
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$\frac{3}{4}$
D
$\frac{3}{8}$

Answer

Correct option: A.

$\frac{1}{4}$

Let $\text{P(B)}=\text{P}$
Than $\text{P(A)}=\frac{1}{3}\text{P}$
Since $A$ and $B$ are two mutually exclusive events, we have:
$\text{A}\cup\text{B}=\text{S}$
$\Rightarrow\text{P}\text{(A}\cup\text{B)}=\text{P}\text{(S)}$
$\Rightarrow\text{P}\text{(A}\cup\text{B)}=1$
$\Rightarrow\text{P}\text{(A)}+\text{(B)}=1$
$\Rightarrow\frac{1}{2}\text{P}+\text{P}=1$
$\Rightarrow\frac{4\text{p}}{3}=1$
$\therefore\text{P(A)}=\frac{1}{3}\text{P}=\frac{1}{3}\times\frac{3}{4}=\frac{1}{4}$

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MCQ 181 Mark

If the integers $m$ and $n$ are chosen at random between $1$ and $100$, then the probability that the number of the from $7^m+ 7^n$ is divisible by $5$ equals:

✓
$\frac{1}{4}$
B
$\frac{1}{7}$
C
$\frac{1}{8}$
D
$\frac{1}{49}$

Answer

Correct option: A.

$\frac{1}{4}$

Since m and n are selected between $1$ and $100,$
Hence total sample space $= 100 \times 100$
Again, $71 = 7, 72 = 49, 73 = 343, 74 = 2401, 75 = 16807,$ etc
Hence $1, 3, 7$ and $9$ will be the last digit in the power of $7.$
Now, favourable number of case are
$\rightarrow 1,1 1,2 1,3 …………. 1,100$
$2,1 2,2 2,3 …………. 2,100$
$3,1 3,2 3,3 …………. 3,100$
$100,1 100,2 100,3 …………. 100,100$
Now, for $m = 1, n = 3, 7, 11, ………, 97$
So, favourable cases $= 25$
Again for $m = 2, n = 4, 8, 12, ………, 100$
So, favourable cases $= 25$
Hence for every $m,$ favourable cases $= 25$
So, total favourable cases $= 100 \times 25$ Required Probability
$=\frac{(100\times25)}{(100\times100)}$
$=\frac{25}{100}$
$=\frac{1}{4}$

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MCQ 191 Mark

A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn. Then the probability that they both are diamonds is:

A
$\frac{81}{452}$
B
$\frac{48}{452}$
C
$\frac{84}{452}$
✓
$\frac{48}{425}$

Answer

Correct option: D.

$\frac{48}{425}$

Total number of cards $= 52$ and one card is lost.
Case $1:$ if lost card is a diamond card
Total number of cards $= 51$ Number of diamond cards $= 12$ Now two cards are drawn.
$P($both cards are diamonds$) \frac{=\ ^{12}\text{C}_2}{{51}\text{c}_2}$
Total number of cards $= 52$ and one card is lost.
Case $2:$ If lost card is not a diamond card Total number of cards $= 51$
Number of diamond cards $= 13$ Now two cards are drawn.
P(both cards are diamonds) $\frac{=^{13}\text{C}_2}{^{51}\text{c}_2}$
Now probability that both cards are diamond $\frac{=^{12}\text{C}_2}{^{51}\text{c}_2}+\frac{=^{13}\text{C}_2}{^{51}\text{c}_2}$
$=\frac{^{12}\text{C}_2+^{13}\text{C}_2}{^{51}\text{C}_2}$
$\Big\{\frac{(12\times11)}{(2\times1)}+\frac{(13\times12)}{(2\times1)}\Big\}\Big\{\frac{(51\times50)}{(2\times1}\Big\}$
$=\frac{(12\times11\times+13\times12)}{(51\times50)}$
$=\frac{288}{2550}$
$=\frac{96}{850} (288$ and $2550$ divided by $3)$
$=\frac{48}{425} (96$ and $850$ divided by $2)$
So probability that both cards are diamond is $\frac{48}{425}$

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MCQ 201 Mark

$(a, c)$ and $(b, c)$ are the centres of two circles whose radical axis is the $y-$ axis. If the radius of first circle is $r$ then the diameter of the other circle is:

A
$2\sqrt{\text{r}^2-\text{b}^2+\text{a}^2}$
B
$\sqrt{\text{r}^2-\text{a}^2+\text{b}^2}$
C
$\Big({\text{r}^2-\text{b}^2+\text{a}^2}\Big)$
✓
$2\sqrt{\text{r}^2-\text{a}^2+\text{b}^2}$

Answer

Correct option: D.

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

Let radius of after circle be $\text{RC}_1:(\text{x}-\text{a)}^2+(\text{y}-\text{c)}^2=\text{r}^2\text{c}_2\\:(\text{x}-\text{b}^2+(\text{y}-\text{c)}^2=\text{R}^2$
Radius axis $\rightarrow\text{x}=0\text{ OR}\text{ c}_1-\text{c}_2=0\text{ c}_1-\text{c}_2$
$= x^2 \text{b}\text{x} - 2\text{a}\text{x} - \text{r}^2 + \text{R}^2 + \text{a}^2 - \text{b}^2$
$= \text{x}$ Put $\text{x} = \text{0 R}^2 = \text{r}^2 + \text{b}^2 - \text{a}^2\text{R}$
$=\sqrt{\text{r}^2+\text{b}^2-\text{a}^2}$ Diamete
$=2\sqrt{\text{r}^2+\text{b}^2-\text{a}^2}$

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MCQ 211 Mark

A circle with center $(3, 8)$ contains the point $(2, -1)$. Another point on the circle is:

A
$(1, -10)$
✓
$(4, 17)$
C
$(5, -9)$
D
$(7, 15)$

Answer

Correct option: B.

$(4, 17)$

$=r^2= (x - 3)^2+ (y - 8)^2$
Given $(2, -1)$ lies on the circle
$= r^2= (2 - 3)^2+ (-1 -8)^2$
$= r^2= 1 + 81$
$= r^2= 82$
Circle equation is: $(x - 3)^2+ (y - 8)^2= 82$
By trial and error, substitute the point
in the above equation
$(4 - 3)^2+ (17 - 8)^2= 82$
hence , $(4, 17)$ satisfy the circle euation.

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MCQ 221 Mark

In tossing a coin, the chance of throwing head and tail alternatively in $3$ successive trials is:

✓
$\frac{1}{4}$
B
$\frac{1}{6}$
C
$\frac{1}{5}$
D
$\frac{1}{48}$

Answer

Correct option: A.

$\frac{1}{4}$

Favourable outcomes $=\text{\{H T H, T H T\}} = 2$ outcomes
Total number of outcomes $= 8$
Probability $=\frac{2}{8}=\frac{1}{4}$

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MCQ 231 Mark

Choose the correct answer. Three numbers are chosen from $1$ to $20.$ Find the probability that they are not consecutive:

A
$\frac{186}{190}$
✓
$\frac{187}{190}$
C
$\frac{188}{190}$
D
$\frac{18}{^{20}\text{C}_3}$

Answer

Correct option: B.

$\frac{187}{190}$

Since, the set of three consecutive numbers from $1$ to $20$ are $(1, 2, 3), (2, 3, 4), (3, 4, 5), ....... , (18, 19, 20),$
i.e., $18$
$P($numbers are consecutive$)$
$=\frac{18}{^{20}\text{C}_3}=\frac{18}{\frac{20\times19\times18}{3!}}=\frac{3}{190}$
$P($three number are not consecutive$)$
$=1-\frac{3}{190}$
$=\frac{187}{190}$

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MCQ 241 Mark

The probability that the leap year will have $53$ sundays and $53$ monday is:

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

Answer

Correct option: D.

$\frac{1}{7}$

In a leap year, total number of days $= 366$ days.
In $366$ days, there are $52$ weeks and $2$ days.
Now two days may be
$(i)$ Sunday and Monday
$(ii)$ Monday and Tuesday
$(iii)$ Tuesday and Wednesday
$(iv)$ Wednesday and Thursday
$(v)$ Thursday and Friday
$(vi)$ Friday and Saturday
$(vii)$ Saturday and Sunday
Now in total $7$ possibilities, Sunday and Monday both come together is $1$ time.
So probabilities of $53$ Sunday and Monday in a leap year $=\frac{1}{7}$

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MCQ 251 Mark

One coin is tossed once. Find the probability of getting A head.

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

Answer

Correct option: A.

$\frac{1}{2}$

The outcome of a throw of coin can be $2.$
$\text{P}=\frac{1}{2}$

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MCQ 261 Mark

A box contains $10$ good articles and $6$ with defects. One item is drawn at random. The probability that it is either good or has a defect is:

✓
$\frac{64}{64}$
B
$\frac{49}{64}$
C
$\frac{40}{64}$
D
$\frac{24}{64}$

Answer

Correct option: A.

$\frac{64}{64}$

Let $A$ be the event of drawing one good article whereas $B$ be the event of drawing one defected article.
Here,
$\text{P(A)}=\frac{10}{10+6}=\frac{10}{16}$
$\text{P(B)}=\frac{6}{10+6}=\frac{6}{16}$
The events $A$ and $B$ are mutually exclusive.
Thus, the required probability,
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\frac{10}{16}+\frac{6}{16}=\frac{16}{16}=1$
Hence, the correct option is $(a).$

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MCQ 271 Mark

A die is rolled, then the probability that an even number is obtained is:

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

Answer

Correct option: A.

$\frac{1}{2}$

When a die is rolled, total number of outcomes $= 6 (1, 2, 3, 4, 5, 6)$
Total even number $= 3 (2, 4, 6)$
So, the probability that an even number is obtained
$=\frac{3}{6}$
$=\frac{1}{2}$

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MCQ 281 Mark

A die is thrown then find the probability of getting an odd number.

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

Answer

Correct option: C.

$\frac{1}{2}$

Sample space $= \{1,2,3,4,5,6\}=1,2,3,4,5,6$
odd nos. $= 1,3,5=1,3,5$
probability of getting odd nos
$=\frac{3}{6}$
$=\frac{1}{2}$

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MCQ 291 Mark

What is called one or more outcomes of an experiment?

A
Space
B
Experiment
C
Sample
✓
Event

Answer

Correct option: D.

Event

Event is called one or more outcomes of an experiment.
Example: rolling a dice, we get a possible outcomes as $\{1, 2, 3, 4, 5, 6\}.$

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MCQ 301 Mark

Two unbiased coins are tossed simultaneously. The probability of getting at least one head is:

A
$\frac{1}{2}$
B
$\frac{1}{4}$
✓
$\frac{3}{4}$
D
$\text{none}$

Answer

Correct option: C.

$\frac{3}{4}$

Favourable number of outcomes, getting at least one head $= 3\text{[H H, H T, T H]}$
Total number of outcomes $= 4\text{[H H, H T, T H, T T]}$
Probability $=\frac{3}{4}$

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MCQ 311 Mark

Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is:

A
$\frac{1}{5}$
B
$\frac{4}{5}$
C
$\frac{1}{30}$
✓
$\frac{5}{9}$

Answer

Correct option: D.

$\frac{5}{9}$

The given digits are 0, 2, 3 and 5.

_____	_____	_____	_____
Thousands	Hundreds	Tens	Ones

Now, there are 3 ways to fill the thousands place (0 cannot occupy the thousands place), 3 ways to fill the hundreds place, 2 ways to fill the tens place and 1 way to fill the ones place.
Total number of four digit numbers formed = 3 × 3 × 2 × 1 = 18
We know that a number is divisible by 5 if it ends in 0 or 5.
When 0 is at the ones place,
Number of four digits numbers divisible by 5 formed = 3 × 2 × 1 = 6
When 5 is at the ones place,
Number of four digits numbers divisible by 5 formed = 2 × 2 × 1 = 4 (0 cannot occupy the thousands place)
Total number of four digit numbers divisible by 5 = 6 + 4 = 10
$\therefore$ P(four digit number formed is divisible by 5)
$=\frac{\text{Total Number of four digit numbers divisible by 5}}{\text{Total Number of w4 digit numbers formed}}$
$=\frac{10}{18}=\frac{5}{9}$
Hence, the correct answer is option (d).

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MCQ 321 Mark

Choose the correct answer. If the probabilities for $A$ to fail in an examination is $0.2$ and that for $B$ is $0.3,$ then the probability that either $A$ or $B$ fails is:

A
$>0.5$
B
$0.5$
✓
$\leq0.5$
D
$0$

Answer

Correct option: C.

$\leq0.5$

Given, $\text{P(A fail) = 0.2}$
and $\text{P(B fail) = 0.3}$
$\therefore\ \text{P(either A or B fail)}\leq\text{P(A fail)}+\text{P(B fail)}$
$\leq0.2+0.3$
$\leq0.5$

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MCQ 331 Mark

Six boys and six girls sit in a row at random. The probability that the boys and girls sit alternatively is:

✓
$\frac{1}{462}$
B
$\frac{11}{462}$
C
$\frac{5}{51}$
D
$\frac{7}{123}$

Answer

Correct option: A.

$\frac{1}{462}$

Given, $6$ boys and $6$ girls sit in a row at random.
Then, the total number of arrangement of $6$ boys and $6$ girls $=$ arrangement of $12$ persons $= 12!$ Now, boys and girls sit alternatively.
So, the total number of arrangement $= 2 \times 6! \times 6!$
Now, $P($ boys and girls sit alternatively$) =\frac{(2\times6!\times6!)}{12!}$
$=\frac{(2\times6!\times6!)}{(12!\times11!)}$
$=\frac{(5!\times6!)}{11!}$
$=\frac{(5\times4\times3\times2\times1\times6!)}{(11\times10\times9\times8\times7\times6!)}$
$=\frac{(5\times4\times3\times2)}{(11\times10\times9\times8\times7)}$
$=\frac{(4\times3)}{(11\times9\times8\times7)}$
$=\frac{3}{(11\times9\times2\times7)}$
$=\frac{1}{(11\times9\times2\times7)}$
$=\frac{1}{462}$

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MCQ 341 Mark

Choose the correct answer. Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is:

A
$\frac{1}{3}$
B
$\frac{1}{6}$
✓
$\frac{2}{7}$
D
$\frac{1}{2}$

Answer

Correct option: C.

$\frac{2}{7}$

If two persons sit next to each other, then consider these two person as $1$ geoup.
Now, we have to arrange $6$ persons.
$\therefore$ Number of arrangement $= 2! \times 6!$
Total number of arrangement of $7$ person $= 7!$
$\therefore\ \text{Required probability}=\frac{2!6!}{7!}=\frac{2}{7}$

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MCQ 351 Mark

Probability is $0.45$ that a dealer will sell at least $20$ television sets during a day, and the probability is $0.74$ that he will sell less that $24$ televisions. The probability that he will sell $20, 21, 22$ or $23$ televisions during the day, is:

✓
$0.19$
B
$0.32$
C
$0.21$
D
None of these

Answer

Correct option: A.

$0.19$

Let $A$ be the event that the sale is at least $20$ televisions,
i.e. $20, 21, 22,...$ and $B$ be the event that sale is less than $24$
i.e. $0.1.2.3...23.$
Then $\text{A∩B}$ will denote the sale of $20, 21, 22$ and $23$ televisions,
We are given $P(A) = 0.45$ and $P(B) = 0.74.$
It is required to find $\text{P(A ∩ B).}$
Also $\text{P(A ∪ B)} = P($ sale of $0, 1, 2, 3,...,20, 21, 22, 23$ televisions$) = P(S) = 1.$
From addition rule, required probability is $\text{P(A ∩ B) = P(A) + P(B) - P(A ∪ B)} = 0.45 + 0.74 - 1 = 0.19.$

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MCQ 361 Mark

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

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

Answer

Correct option: C.

$\frac{3}{4}$

Since, Total possibilities are $=\text{\{H H, H T, T H, T T\}}$
no. of cases with atmost one head are $= \text{\{H T, T H, T T\}}=\frac{3}{4}$

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MCQ 371 Mark

The equation circle whose center is $(0, 0)$ and radius is $4$ is:

A
$x^2+y^2=4$
✓
$x^2+y^2=16$
C
$x^2+y^2=2$
D
None

Answer

Correct option: B.

$x^2+y^2=16$

The equation of circle is $ x^2+y^2=r^2 $
Here the radius is $4$
So the equation is
$x^2+y^2=42 $
$x^2+y^2=16 $

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MCQ 381 Mark

One card is drawn from a pack of $52$ cards. The probability of getting a jack card is:

✓
$\frac{1}{13}$
B
$\frac{2}{13}$
C
$\frac{3}{13}$
D
$\frac{4}{13}$

Answer

Correct option: A.

$\frac{1}{13}$

Favourable number of outcomes
i.e., numbers of jack cards $= 4$
Total number of outcomes $= 52$
Thus, probability $=\frac{4}{52}=\frac{1}{13}$

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MCQ 391 Mark

If the parabola $y^2= 4ax$ passes through $(3, 2)$ then the length of latus rectum is:

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

Answer

Correct option: D.

$\frac{4}{3}$

If the parabola $y^2= 4ax$ passes through $(3, 2)$
therefore, $4 = 4a (3)$
$\Rightarrow\text{a}=\frac{1}{3}$
Therefore, length of latus rectum:
$1=4\text{a}=\frac{4}{3}$

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MCQ 401 Mark

If a coin is tossed till the first head appears, then what will be the sample space?

A
$\text{\{H\}}$
B
$\text{\{TH\}}$
C
$\text{\{T, TH, HHT, HHHT, ..........\}}$
✓
$\text{\{H, TH, TTH, TTTH, .......\}}$

Answer

Correct option: D.

$\text{\{H, TH, TTH, TTTH, .......\}}$

$\text{S: \{H, TH, TTH, TTTH, ..........\}}$ infinte elements.
If for the first toss only, we would have got the head, we have stop there itself.

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MCQ 411 Mark

Choose the correct answer.
The probability that at least one of the events $A$ and $B$ occurs is $0.6.$ If $A$ and $B$ occur simultaneously with probability $0.2,$ then $\text{P}(\bar{\text{A}})+\text{P}(\bar{\text{B}})$ is:

A
$0.4$
B
$0.8$
✓
$1.2$
D
$1.6$

Answer

Correct option: C.

$1.2$

We have, $\text{P}(\text{A}\cup\text{B})=0.6$ and $\text{P}(\text{A}\cap\text{B})=0.2$
$\therefore\ \text{P}(\text{A}\cup\text{B})=0.6$ and $\text{P}(\text{A}\cap\text{B})=0.2$
$\Rightarrow0.6=\text{P(A)}+\text{P(B)}-0.2$
$\Rightarrow\text{P(A)}+\text{P(B)}=0.8$
$\therefore\ \text{P}(\bar{\text{A}})+\text{P}(\bar{\text{B}})$
$=1-\text{P(A)}+1-\text{P(B)}$
$=2-\big[\text{P(A)}+\text{P(B)}\big]$
$=2-0.8$
$=1.2$

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MCQ 421 Mark

One coin is tossed once. Find the probability of getting $A$ tail.

✓
$\frac{1}{2}$
B
$1$
C
$\text{Data} \text{ insufficient}$
D
$\text{None}\text{ of}\text{ these}$

Answer

Correct option: A.

$\frac{1}{2}$

There are $2$ possible outcomes of a throw of coin.
$\text{P(tail)}=\frac{1}{2}$

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MCQ 431 Mark

Two dice are thrown together. The probability that at least one will show its digit greater than $3$ is:

A
$\frac{1}{4}$
✓
$\frac{3}{4}$
C
$\frac{1}{2}$
D
$\frac{1}{8}$

Answer

Correct option: B.

$\frac{3}{4}$

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 at least one digit greater than $3.$
Then $E = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 1) , (4, 2), (4, 3), (4, 4), (4, 5),\\ (4, 6),(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$
$\therefore\ \text{n(E)}=27$
Hence, required probability $=\frac{27}{36}=\frac{3}{4}$

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MCQ 441 Mark

If A and B are mutually exclusive events then:

✓
$\text{P(A)}\leq\text{P}(\overline{\text{B}})$
B
$\text{P(A)}\geq\text{P}(\overline{\text{B}})$
C
$\text{P(A)}<\text{P}(\overline{\text{B}})$
D
None of these

Answer

Correct option: A.

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

It is given that A and B are mutually exclusive events.
We know that,
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{(B})-\text{P}(\text{A}\cap\text{B})$$\big[\text{From(1)}\big]$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{P}\text{(B})$ $\big[\text{P}(\text{A}\cup\text{B})\leq1\big]$
$\Rightarrow\text{P}(\text{A)}+\text{P}\text{(B})\leq1$
$\Rightarrow\text{P}(\text{A)}\leq1-\text{P}\text{(B})=\text{P}(\overline{\text{B}})$
$\therefore\text{P}(\text{A)}\leq\text{P}\text{(B})$
Hence, the correct answer is option (a).$\therefore\text{P}(\text{A }\cap\text{B})=0\ ...(1)$

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MCQ 451 Mark

The equation of the circle passing through $(3, 6)$ and whose centre is $(2, -1)$ is:

A
$x^2+ y^2- 4x + 3y = 45$
✓
$x^2+ y^2 - 4x + 2y - 45 = 0$
C
$x^2+ y^2 + 4x - 2y = 45$
D
$x^2+ y^2- 4x + 2y + 45 = 0$

Answer

Correct option: B.

$x^2+ y^2 - 4x + 2y - 45 = 0$

$(x - 2)^2+ (y + 1)^2+ r^2 (3, 6).$ lies on it
$\Rightarrow 1 + 49 = r^2$
$\Rightarrow r^2= 50$
$\Rightarrow x^2+ 4 - 4x + y^2 + 1 + 2y = 50.$
$\Rightarrow x^2+ y^2 - 4x + 2y - 45 = 0$

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MCQ 461 Mark

The length of the latus rectum of the parabola x=ay^2+by+c is:

A
$\frac{\text{a}}{4}$
B
$\frac{\text{a}}{3}$
✓
$\frac{1}{\text{a}}$
D
$\frac{1}{4\text{a}}$

Answer

Correct option: C.

$\frac{1}{\text{a}}$

$x = ay^2+ by + c$
$ay^2+ by = x - c$
$=\text{a}\Big(\text{y}+\frac{\text{b}}{\text{a}}\Big)^2$
$=\text{x}+\frac{\text{a}\text{b}^2}{4}-\text{c}$
$=\Big(\text{y}+\frac{\text{b}}{\text{a}}\Big)^2$
$=\frac{1}{\text{a}}\Big(\text{x}+\frac{\text{a}\text{b}^2}{4}-\text{c}\Big)$
Length of latus rectum
$=\frac{1}{\text{a}}$

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MCQ 471 Mark

If A, B, C are three mutually exclusive and exhaustive events of an experiment such that 3 $\text{P(A)}=2\text{P(B)}=\text{C},$ then P(A) is equal to:

A
$\frac{1}{11}$
✓
$\frac{2}{11}$
C
$\frac{5}{11}$
D
$\frac{6}{11}$

Answer

Correct option: B.

$\frac{2}{11}$

Let 3 P(A) = 2 P(B) = P(C) = p.
Then, $\text{P(A)}=\frac{\text{P}}{3},\text{P(B)}=\frac{\text{P}}{2}$ and $\text{P(C)}=\text{P}$
It is given that A, B, C are three mutually exclusive and exhaustive events.
$\therefore\text{P(A)}+\text{P(B)}+\text{P(C)}=1$
$=0 $ and $\big[\text{P}(\text{A}\cup\text{B)=}\text{P}(\text{B}\cap\text{C)}=\text{P}(\text{C}\cap\text{A)=}\text{P}(\text{A}\cup\text{B }\cup\text{C}=1\big]$
$\Rightarrow\frac{\text{P}}{3}+\frac{\text{p}}{2}+\text{P}=1$
$\Rightarrow\frac{\text{11P}}{6}=1$
$\Rightarrow\text{P}=\frac{6}{11}$
$\therefore\text{P(A)}=\frac{\text{P}}{3}=\frac{\frac{6}{11}}{3}=\frac{2}{11}$
Hence, the correct answer is option (b).

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MCQ 481 Mark

Two coins are tossed, what is the sample space?

A
$\text{(H, T), (H, T), (T, H), (H, H)}$
✓
$\text{(H, T), (T, T), (T, H), (H, H)}$
C
$\text{(T, T), (H, H), (T, T), (H, H)}$
D
$\text{(H, T), (T, T), (H, T), (T, T)}$

Answer

Correct option: B.

$\text{(H, T), (T, T), (T, H), (H, H)}$

Sample space is the collection of all possible events.
So, sample space of tossing two coins, $\text{S=(H, T), (T, T), (T, H), (H, H)}$

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MCQ 491 Mark

Three numbers are chosen from 1 to 20. The probability that they are not consecutive is:

A
$\frac{186}{190}$
✓
$\frac{187}{190}$
C
$\frac{188}{190}$
D
$\frac{18}{\ ^{20}\text{C}_3}$

Answer

Correct option: B.

$\frac{187}{190}$

Number of ways to choose three numbers from 1 to 20 $=\ ^{20}\text{C}_3=1140$
Now, the set of three consecutive numbers from 1 to 20 are (1, 2, 3), (2, 3, 4), (3, 4, 5), ...., (18, 19, 20).
So, the number of ways to choose three numbers from 1 to 20 such that they are consecutive is 18.
P(three numbers choosen are consecutive)
$=\frac{\text{Number of ways to choose three consecutive numbers from 1 to 20 }}{\text{Number of ways to choose three numbers from 1 to 20}}$
$=\frac{18}{\ ^{20}\text{c}_3}=\frac{18}{1140}=\frac{3}{190}$
$\therefore$P(three numbers choosen are not consecutive) = 1 - P(three numbers choosen are consecutive) $=1-\frac{3}{190}=\frac{187}{190}$
Hence, the correct answer is option (b).

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MCQ 501 Mark

Sample space is a set of $.....$ of an experiment.

✓
All possible outcomes
B
Selected outcomes
C
Both
D
None of these

Answer

Correct option: A.

All possible outcomes

A sample space is usually denoted using set notation, and the possible outcomes are listed as elements in the set. For example, if the
experiment is tossing a coin, the sample space is typically the set $\{$head, tail$\}$,
i.e all possible outcomes.

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