MCQ

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.

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 × 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 × 5 × 5
So total number of ways of 4 digits greater than 5000 = 2 × 5 × 5 × 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 × 5 × 5 × 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 × 3! × 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 $y^2+4 x+4 y+k=0$ represents a parabola whose latus rectum is:

A
1
B
2
C
3
✓
4

Answer

Correct option: D.

Solution:
$y^2+4 x+4 y+k=0$
$y^2+2 \times 2 y+4-4+4 x+k=0$
$(\text{y}+2)^2=-4\text{x}-\text{k}+4$
$(\text{y}+2^2)=-4\Big(\text{x}-\frac{4\ +\ \text{k}}{4}\Big)$
$∴$ 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 = {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 = {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.

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

H T, H H, T H, T T (Here 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 ofP, (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 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 AUB is equal to

A
15
B
17
C
19
✓
21

Answer

Correct option: D.

When two dice are thrown, then total outcome = 6 × 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.

$\frac{1}{4}$

$\frac{1}{4}$

Solution:
Since m and n are selected between 1 and 100,
Hence total sample space = 100 × 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
→ 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 × 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
$\frac{84}{452}$
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 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)

(4, 17)

Solution:
$=r^2=(x-3)^2+(y-8)^2$
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 211 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 ={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 221 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 231 Mark

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

A
$\frac{2}{3}$
B
$\frac{1}{2}$
C
$\frac{2}{7}$
✓
$\frac{1}{17}$

Answer

Correct option: D.

$\frac{1}{17}$

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
$=\frac{1}{17}$

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MCQ 241 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 251 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 261 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 271 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 281 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 291 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[H H, H T, T H]
Total number of outcomes = 4[H H, H T, T H, T T]
Probability $=\frac{3}{4}$

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MCQ 301 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 311 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, P(A fail) = 0.2
and 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 321 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 × 6! × 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 331 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 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 P(A ∩ B).
Also P(A ∪ B) = P( sale of 0, 1, 2, 3,...,20, 21, 22, 23 televisions) = P(S) = 1.
From addition rule, required probability is P(A ∩ B) = P(A) + P(B) - P(A ∪ B) = 0.45 + 0.74 - 1 = 0.19.

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MCQ 341 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 = {H H, H T, T H, T T}
no. of cases with atmost one head are = {H T, T H, T T}
$=\frac{3}{4}$

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MCQ 351 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.

$x^2+y^2=16$

$x^2+y^2=16$

Solution:
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 361 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 371 Mark

If the parabola $y^2=4 a x$ 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.

$\frac{4}{3}$

$\frac{4}{3}$

Solution:
If the parabola $y^2=4 a x$ 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 381 Mark

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

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

Answer

Correct option: D.

{H, TH, TTH, TTTH, .......}

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 391 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 401 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 411 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 × 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 421 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 431 Mark

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

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

Answer

Correct option: B.

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

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

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

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MCQ 441 Mark

The length of the latus rectum of the parabola $x=a y^2+b y+c$ 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.

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

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

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

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MCQ 451 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 461 Mark

Two coins are tossed, what is the sample space?

A
(H, T), (H, T), (T, H), (H, H)
✓
(H, T), (T, T), (T, H), (H, H)
C
(T, T), (H, H), (T, T), (H, H)
D
(H, T), (T, T), (H, T), (T, T)

Answer

Correct option: B.

(H, T), (T, T), (T, H), (H, H)

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

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MCQ 471 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 481 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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MCQ 491 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 501 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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