M.C.Q (1 Marks)

MCQ 11 Mark

The probability of an impossible event is:

✓
$0$
B
$1$
C
$\frac{1}{2}$
D
Non$-$existent.

Answer

Correct option: A.

$0$

Probability of an impossible event $= 0$

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MCQ 21 Mark

A bag contains cards numbered from $1$ to $25$. A card is drawn at random from the bag. The probability that the number on this card is divisible by both $2$ and $3$ is:

A
$\frac{1}{5}$
B
$\frac{3}{25}$
✓
$\frac{4}{25}$
D
$\frac{2}{25}$

Answer

Correct option: C.

$\frac{4}{25}$

Total number of outcomes $= 25$
The number which is divisible by both $2$ and $3$ are $6, 12, 18, 24.$
Number of favourable outcomes $= 4$
Probability of number which is divisible by both $2$ and $3=\frac{4}{25}$

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MCQ 31 Mark

Two dice are rolled simultaneously. The probability that they show different faces is:

A
$\frac{2}{3}$
B
$\frac{1}{6}$
C
$\frac{1}{3}$
✓
$\frac{5}{6}$

Answer

Correct option: D.

$\frac{5}{6}$

Two dice are rolled simultaneously
$\therefore$ No. of total events $= 6^2 = 36$
$\therefore$ No. of different face can be
$= 36 - ($same faces$)$
Same face are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)$ and $(6, 6) = 6$
$\therefore$ $36 - 6 = 30$
$\therefore\ \text{Probability P (E)}=\frac{\text{m}}{\text{n}}=\frac{30}{36}=\frac{5}{6}$

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MCQ 41 Mark

A bag contains three green marbles, four blue marbles and two orange marbles. If a marble is picked at random, then the probability that it is not an orange marble is:

A
$\frac{1}{4}$
B
$\frac{1}{3}$
C
$\frac{4}{9}$
✓
$\frac{7}{9}$

Answer

Correct option: D.

$\frac{7}{9}$

In a bag, there are $3$ green, $4$ blue and $2$ orange marbles
$\therefore$ Total marbles $(n) = 3 + 4 + 2 = 9$
No. of marbles which is not orange $= 3 + 4 = 7$
$\therefore m = 7$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{7}{9}$

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MCQ 51 Mark

In a single throw of a die, the probability of getting a multiple of $3$ is:

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

Answer

Correct option: B.

$\frac{1}{3}$

A die is thrown, the possible number of events $(n) = 6$
Now multiple of $3$ are $3, 6$ which are $2$
$\therefore m = 2$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{2}{6}=\frac{1}{3}$

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MCQ 61 Mark

Which of the following cannot be the probability of occurrence of an event?

A
$0.2$
B
$0.4$
C
$0.8$
✓
$1.6$

Answer

Correct option: D.

$1.6$

Probability of an event occurrence can not be $= 1.6$

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MCQ 71 Mark

A number is selected at random from the numbers $1$ to $30.$ The probability that it is a prime number is:

A
$\frac{2}{3}$
B
$\frac{1}{6}$
✓
$\frac{1}{3}$
D
$\frac{11}{30}$

Answer

Correct option: C.

$\frac{1}{3}$

Total number of outcomes $= 30$
The prime numbers from $1$ to $30$ are $2, 3, 5, 7, 11, 13, 17, 19, 23$ and $29.$
So, the favourable number of outcomes are $10.$
$\therefore P($selected number is a prime number$)$
$=\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}$
$=\frac{10}{30}$
$=\frac{1}{3}$
Hence, the correct answer is option $c.$

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MCQ 81 Mark

A card is accidently dropped from a pack of $52$ playing cards. The probability that it is an ace is:

A
$\frac{1}{4}$
✓
$\frac{1}{13}$
C
$\frac{1}{52}$
D
$\frac{12}{13}$

Answer

Correct option: B.

$\frac{1}{13}$

No. of card in a pack $(n) = 52$
A card is drawn at random
$\therefore$ No. of ace $(m) = 4$
$\therefore$ Probability of an ace $=\frac{\text{m}}{\text{n}}=\frac{4}{52}=\frac{1}{13}$

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MCQ 91 Mark

A box contains $90$ discs, numbered from $1$ to $90.$ If one disc is drawn at random from the box, the probability that it bears a prime number less than $23,$ is:

A
$\frac{7}{90}$
B
$\frac{10}{90}$
✓
$\frac{4}{45}$
D
$\frac{9}{89}$

Answer

Correct option: C.

$\frac{4}{45}$

Number of discs in a box $= 90$
Numbered on it are $1$ to $90$
Prime numbers less than $23$ are $= 2, 3, 5, 7, 11, 13, 17, 19 = 8$
Probability of a number being a prime less than $23=\frac{8}{90}=\frac{4}{45}$

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MCQ 101 Mark

If $P(E) = 0.05,$ then $P($not $E) =$

A
$-0.5$
B
$0.5$
C
$0.9$
✓
$0.95$

Answer

Correct option: D.

$0.95$

$P(E) = 0.05$
$\because P(E) + P($ not $E) = 1$
$\therefore P($ not $ E) = 1 - P(E)$
$= 1 - 0.05$
$= 0.95$

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MCQ 111 Mark

The probability of throwing a number greater than $2$ with a fair dice is:

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

Answer

Correct option: C.

$\frac{2}{3}$

Given: A dice is thrown once
To Find: Probability of getting a number greater than $2.$
Total number on a dice is $6.$
Number greater than $2$ is $3, 4, 5$ and $6$
Total number greater than $2$ is $4$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a number greater than $2$ is equal to $\frac{4}{6}=\frac{2}{3}$
Hence the correct option is $c.$

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MCQ 121 Mark

If a number $x$ is chosen from the numbers $1, 2, 3$, and a number $y$ is selected from the numbers $1, 4, 9.$ Then, $P(xy < 9)$

A
$\frac{7}{9}$
✓
$\frac{5}{9}$
C
$\frac{2}{3}$
D
$\frac{1}{9}$

Answer

Correct option: B.

$\frac{5}{9}$

Given: $x$ is chosen from the numbers $1, 2, 3$ and $y$ is chosen from the numbers $1, 4, 9$
To Find: Probability of getting $P(xy) < 9$
We will make multiplication table for x and y such tha
$(xy) < 9$
$xy = 1 \times 1 = 1$
$= 1 \times 4 = 4$
$= 1 \times 9 = 9$
$= 2 \times 1 = 2$
$= 2 \times 4 = 8$
$= 2 \times 9 = 18$
$= 3 \times 1 = 3$
$= 3 \times 4 = 12$
$= 3 \times 9 = 27$
So the numbers such that $(xy) < 9$ is $5$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence $\text{P}(\text{xy}<9)=\frac{5}{9}$
Hecne the correct option is $b.$

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MCQ 131 Mark

If two different dice are rolled together, the probability of getting an even number:

A
$\frac{1}{36}$
B
$\frac{1}{2}$
C
$\frac{1}{6}$
✓
$\frac{1}{4}$

Answer

Correct option: D.

$\frac{1}{4}$

Rolling two different dice,
Number of total events $= 6 \times 6 = 36$
Number of even number on both dice are $(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) = 9$
$\therefore\ \text{Probability}=\frac{9}{36}=\frac{1}{4}$

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MCQ 141 Mark

What is the probability that a leap year has $52$ Mondays?

A
$\frac{2}{7}$
B
$\frac{4}{7}$
✓
$\frac{5}{7}$
D
$\frac{6}{7}$

Answer

Correct option: C.

$\frac{5}{7}$

Given: A leap year
To Find: Probability that a leap year has $52$ Mondays.
Total number of days in leap year is $366$ days
Hence number of weeks in a leap year is $\frac{366}{7}=52$ weeks and $2$ day
In a leap year we have $52$ complete weeks and $2$ day which can be any pair of the day of the week i.e.
$($Sunday, Monday$)$
$($Monday, Tuesday$)$
$($Tuesday, Wednesday$)$
$($Wednesday, Thursday$)$
$($Thursday, Friday$)$
$($Friday, Saturday$)$
$($Saturday, Sunday$)$
To make $52$ Mondays the additional days should not include Monday
Hence total number of pairs of days is $7$
Favorable day
i.e. in which Mondays is not there is $5$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability that a leap year has $52$ Mondays is equal to $\frac{5}{7}$
Hence the correct option is $c.$

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MCQ 151 Mark

The probability that a non$-$leap year has $53$ Sundays, is:

A
$\frac{2}{7}$
B
$\frac{5}{7}$
C
$\frac{6}{7}$
✓
$\frac{1}{7}$

Answer

Correct option: D.

$\frac{1}{7}$

In a non leap years, number of days $= 365$
i.e. $52$ weeks $+ 1$ day
$\therefore$ Probability of being $53$ Sundays
$=\frac{\text{m}}{\text{n}}=\frac{1}{\text{No. of day in a week}}=\frac{1}{7}$

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MCQ 161 Mark

Which of the following cannot be the probability of an event?

A
$\frac{2}{3}$
✓
$-1.5$
C
$15\%$
D
$0.7$

Answer

Correct option: B.

$-1.5$

Given: $4$ options of probability of some events
To Find: Which of the given options cannot be the probability of an event?
We know that $0\leq\text{p}\leq1$.
As the probability of an event cannot be negative
In option $(b) P = -1.5$
Hence the correct answer is option $b.$

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MCQ 171 Mark

If a digit is chosen at randon from the digits $1, 2, 3, 4, 5, 6, 7, 8, 9,$ then the probability that the digit is a multiple of $3$ is:

✓
$\frac{1}{3}$
B
$\frac{2}{3}$
C
$\frac{1}{9}$
D
$\frac{2}{9}$

Answer

Correct option: A.

$\frac{1}{3}$

Given: digits are chosen from $1, 2, 3, 4, 5, 6, 7, 8, 9$ are placed in a box and mixed thoroughly.
One digit is picked at random.
To Find: Probability of getting a multiple of $3$
Total number of digits is $9$
Digits that are multiple of $3$ are $3, 6$ and $9$
Total digits that are multiple of $3$ are $3$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a multiple of $3$ is $\frac{3}{9}=\frac{1}{3}$
Hence the correct option is option $a.$

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MCQ 181 Mark

From the letters of the word $"\text{MOBILE}",$ a letter is selected. The probability that the letter is a vowel, is:

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

Answer

Correct option: D.

$\frac{1}{2}$

No. of total letters in the word $\text{MOBILE} = 6$
No, of vowels $= o, i, e = 3$
$\therefore$ Probability of being a vowel $=\frac{3}{6}=\frac{1}{2}$

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MCQ 191 Mark

If a two digit number is chosen at random, then the probability that the number chosen is a multiple of $3,$ is:

A
$\frac{3}{10}$
B
$\frac{29}{100}$
✓
$\frac{1}{3}$
D
$\frac{7}{25}$

Answer

Correct option: C.

$\frac{1}{3}$

Total number of two digit numbers are $10$ to $99$
$= 99 - 9 = 90$
Multiples of $3$ will be $12, 15, 18, 21,…. 99$
$= 33 - 3 = 30$
$\therefore\ \text{Probability}=\frac{30}{90}=\frac{1}{3}$

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MCQ 201 Mark

Two dice are thrown together. The probability of getting the same number on both dice is:

A
$\frac{1}{2}$
B
$\frac{1}{3}$
✓
$\frac{1}{6}$
D
$\frac{1}{12}$

Answer

Correct option: C.

$\frac{1}{6}$

$2$ dice are thrown together
$\therefore$ Number of total outcomes $= 6 \times 6 = 36$
Number which should come together are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$
$= 6$ pairs
$\therefore\ \text{Probability P}_\text{(E)}=\frac{6}{36}=\frac{1}{6}$

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MCQ 211 Mark

A number $x$ is chosen at random from the numbers $-3, -2, -1, 0, 1, 2, 3$ the probability that $|x| < 2$ is:

A
$\frac{5}{7}$
B
$\frac{2}{7}$
✓
$\frac{3}{7}$
D
$\frac{1}{7}$

Answer

Correct option: C.

$\frac{3}{7}$

Total possible number of events $(n) = 7$
Now $|x| < 2$
$x < 2$ or $-x < 2$
$\Rightarrow x > -2$
$\therefore x$
$\Rightarrow x = 1, 0, -1, -2, -3$ or $x = -1, 0, 1, 2, 3$
$\therefore x = -1, 0, 1$
$\therefore m = 3$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{3}{7}$

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MCQ 221 Mark

Two numbers $'a\ ’$ and $'6\ ’$ are selected successively without replacement in that order from the integers $1$ to $10.$ The probability that $\frac{\text{a}}{\text{b}}$ is an integer, is:

A
$\frac{17}{45}$
B
$\frac{1}{5}$
✓
$\frac{17}{90}$
D
$\frac{8}{45}$

Answer

Correct option: C.

$\frac{17}{90}$

a and b are two number to be selected from the integers $= 1$ to $10$ without replacement of $a$ and $b$
i.e., $1$ to $10 = 10$
And $2$ to $10 = 9$
No. of ways $= 10 \times 9 = 90$
Probability of $\frac{\text{a}}{\text{b}}$ where it is an integer
Possible event will be

$= (2, 2), (3, 3),$

$(4, 2), (4, 4), (5, 5),$

$(6, 2), (6, 3), (6, 6),(7, 7), (8, 2), (8, 4), (8, 8),$

$(9, 3), (9, 9), (10, 2), (10, 5), (10, 10), = 17$

$\text{P(E)}=\frac{\text{m}}{\text{n}}=\frac{17}{90}$

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MCQ 231 Mark

A month is selected at random in a year. The probability that it is March or October, is:

A
$\frac{1}{12}$
✓
$\frac{1}{6}$
C
$\frac{3}{4}$
D
None of these.

Answer

Correct option: B.

$\frac{1}{6}$

No. of months in a year $= 12$
Probability of being March or October $=\frac{2}{12}$
$=\frac{1}{6}$

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MCQ 241 Mark

In a family of $3$ children, the probability of having at least one boy is:

✓
$\frac{7}{8}$
B
$\frac{1}{8}$
C
$\frac{5}{8}$
D
$\frac{3}{4}$

Answer

Correct option: A.

$\frac{7}{8}$

The possible outcomes are $\text{BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.}$
Total number of outcomes $= 8$
The favourable outcomes are $\text{BBB, BBG, BGB, GBB, BGG, GBG, GGB.}$
So, the favourable number of outcomes are $7.$
$\therefore P($at least one boy$)$
$=\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}=\frac{7}{8}$
Hence, the correct answer is option $A.$

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MCQ 251 Mark

A number is selected at random from the. Numbers $3, 5, 5, 7, 7, 7, 9, 9, 9, 9.$ The probability that the selected number is their average is:

A
$\frac{1}{10}$
B
$\frac{3}{10}$
✓
$\frac{7}{10}$
D
$\frac{9}{10}$

Answer

Correct option: C.

$\frac{7}{10}$

Total numbers are $\sum\text{x}_\text{i}=10$

$x$		$f$
$3$	$=$	$1$
$5$	$=$	$2$
$7$	$=$	$3$
$9$	$=$	$4$

$\text{Average}=\frac{3\times1+5\times2+7\times3+9\times4}{10}$
$=\frac{3+10+21+36}{10}=\frac{70}{10}=7$
$\therefore\ \text{m}=7$
$\therefore$ Probability of average number $=\frac{7}{10}$

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MCQ 261 Mark

The probability that a number selected at random from the numbers $1, 2, 3, ....., 15$ is a multiple of $4,$ is:

A
$\frac{4}{15}$
B
$\frac{2}{15}$
✓
$\frac{1}{5}$
D
$\frac{1}{3}$

Answer

Correct option: C.

$\frac{1}{5}$

The total number of given numbers is $15.$
$\therefore$ Total number of outcomes $= 15$
Among the given numbers, the multiples of $4$ are $4, 8$ and $12.$
So, the favourable number of outcomes are $3.$
$\therefore P($number selected is a multiple of $4)$
$=\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}$
$=\frac{3}{15}$
$=\frac{1}{5}$
Hence, the correct answer is option $c.$

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MCQ 271 Mark

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

A
$\frac{1}{4}$
B
$\frac{1}{8}$
✓
$\frac{3}{4}$
D
$\frac{7}{8}$

Answer

Correct option: C.

$\frac{3}{4}$

When two different coins are tossed simultaneously, then total possibilities $= 4,$
i.e. $\text{(H, H), (H, T), (T, H), (T, T)}$
Number of favourable outcomes for at least one head $= 3,$
i.e. $\text{(H, T), (T, H), (T, H).}$
$\therefore$ Probability of getting at least one nead $=\frac{3}{4}$

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MCQ 281 Mark

Aarushi sold $100$ lottery tickets in which $5$ tickets carry prizes. If Priya purchased a ticket, what is the probability of Priya winning a prize?

A
$\frac{19}{20}$
B
$\frac{1}{25}$
✓
$\frac{1}{20}$
D
$\frac{17}{20}$

Answer

Correct option: C.

$\frac{1}{20}$

No. of lottery tickets $= 100$
No. of tickets carrying prizes $= 5$
$\therefore$ Probability of ticket buying a prized one
$=\frac{\text{m}}{\text{n}}=\frac{5}{100}=\frac{1}{20}$

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MCQ 291 Mark

A number is selected from first $50$ natural numbers. What is the probability that it is a multiple of $3$ or $5?$

A
$\frac{13}{25}$
B
$\frac{21}{50}$
C
$\frac{12}{25}$
✓
$\frac{23}{50}$

Answer

Correct option: D.

$\frac{23}{50}$

Given: A number is selected from $50$ natural numbers
To Find: Probability that the number selected is a multiple of $3$ or $5$
Total number is $50$
Total numbers which are multiple of $3$ or $5$ up to $50$ natural numbers are $3, 6, 5, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35, 36, 39, 40, 42, 45, 48, 50$
Total number which are multiple of $3$ or $5$ up to $50$ natural numbers are $23$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability that the number selected is a multiple of $3$ or $5$ is equal to $\frac{23}{50}$
The correct answer is option $d.$

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MCQ 301 Mark

What is the probability that a non$-$leap year has $53$ Sundays?

A
$\frac{6}{7}$
✓
$\frac{1}{7}$
C
$\frac{5}{7}$
D
None of these.

Answer

Correct option: B.

$\frac{1}{7}$

Given: A non leap year
To Find: Probability that a non leap year has $53$ Sundays.
Total number of days in a non leap year is $365$ days
Hence number of weeks in a non leap year is $\frac{365}{7}=52$ weeks and $1$ day
In a non leap year we have $52$ complete weeks and $1$ day which can be any day of the week
i.e. Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday
To make $53$ Sundays the additional day should be Sunday
Hence total number of days which can be any day is $7$
Favorable day
i.e. Sunday is $1$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability that a non leap year has $53$ Sundays is
Hence the correct option is $b.$

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MCQ 311 Mark

If a digit is chosen at randon from the digits $1, 2, 3, 4, 5, 6, 7, 8, 9,$ then the probability that it is odd, is:

A
$\frac{4}{9}$
✓
$\frac{5}{9}$
C
$\frac{1}{9}$
D
$\frac{2}{3}$

Answer

Correct option: B.

$\frac{5}{9}$

Total number of digits from $1$ to $9 (n) = 9$
Numbers which are odd $(m) = 1, 3, 5, 7, 9 = 5$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{5}{9}$

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MCQ 321 Mark

A die is thrown once. The probability of getting a prime number is:

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

Answer

Correct option: C.

$\frac{1}{2}$

In a single throw of a die, the possible outcomes are $1, 2, 3, 4, 5,$ and $6.$
$\therefore$ Total number of outcomes $= 6$
the favourable outcomes are $2, 3$ and $5.$
So, the number of favourable outcomes are $3.$
$\therefore P($getting a prime number$)$
$=\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}$
$=\frac{3}{6}$
$=\frac{1}{2}$
Hence, the correct answer is option $c.$

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MCQ 331 Mark

The probability of getting an even, number, when a die is thrown once is:

✓
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{1}{6}$
D
$\frac{5}{6}$

Answer

Correct option: A.

$\frac{1}{2}$

Even number on a die are $2, 4, 6 = 3$
$\therefore\ \text{Probability (P)}=\frac{3}{6}=\frac{1}{2}$

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MCQ 341 Mark

The probability of guessing the correct answer to a certain test questions is $\frac{\text{x}}{12}$. If the probability of not guessing the correct answer to this question is $\frac{2}{3}$, then $x =$

A
$2$
B
$3$
✓
$4$
D
$6$

Answer

Correct option: C.

$4$

Given: Probability of guessing a correct answer to a certain question is $\frac{\text{x}}{12}$
Probability of not guessing a correct answer to a same question $\frac{2}{3}$
To find: The value of $x$
Calculation: We know that sum of probability of occurrence of an event and probability of non occurrence of an event is $1.$
If $E$ is an event of occurrence and $\bar{\text{E}}$ is its complementary then
$\text{P(E)}+\text{P}(\bar{\text{E}})=1$
According to the question we have
$\frac{\text{x}}{12}+\frac{2}{3}=1$
$\frac{\text{x}+8}{12}=1$
$\text{x}+8=12$
$\text{x}=4$
Hence the correct option is $c.$

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MCQ 351 Mark

A number is selected from numbers $1$ to $25$. The probability that it is prime is:

A
$\frac{2}{3}$
B
$\frac{1}{6}$
C
$\frac{1}{3}$
✓
$\frac{9}{25}$

Answer

Correct option: D.

$\frac{9}{25}$

A number is selected from the numbers $1$ to $25$
Probability of prime number which are $2, 3, 5, 7, 11, 13, 17, 19, 23 = 9$
$\therefore\text{P(E)}=\frac{\text{m}}{\text{n}}=\frac{9}{25}$

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MCQ 361 Mark

If a digit is chosen at randon from the digits $1, 2, 3, 4, 5, 6, 7, 8, 9,$ then the probability that the digit is even, is:

✓
$\frac{4}{9}$
B
$\frac{5}{9}$
C
$\frac{1}{9}$
D
$\frac{2}{3}$

Answer

Correct option: A.

$\frac{4}{9}$

Total number of digits from $1$ to $9 (n) = 9$
Numbers which are even $(m) = 2, 4, 6, 8 = 4$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{4}{9}$

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MCQ 371 Mark

In a single throw of a pair of dice, the probability of getting the sum a perfect square is:

A
$\frac{1}{18}$
✓
$\frac{7}{36}$
C
$\frac{1}{6}$
D
$\frac{2}{9}$

Answer

Correct option: B.

$\frac{7}{36}$

A pair of dice is thrown simultaneously
$\therefore$ No. of total events $(n) = 6 \times 6 = 36$
Which are

$(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$

$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$

$\therefore$ Event whose sum is a perfect square are $(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)$
$\therefore m = 7$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{7}{36}$

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MCQ 381 Mark

If three coins are tossed simultaneously, then the probability of getting at least two heads, is:

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

Answer

Correct option: C.

$\frac{1}{2}$

Three coins are tossed simultaneously, then possible events will be $(n) = 2 \times 2 \times 2 = 8$
The results will be
$\text{(HHT), (HTH), (THH), (THT), (TTH), (HTT), (HHH), (TTT)}$
$\therefore$ Probability of getting at least two heads are
$=\frac{\text{m}}{\text{n}}=\frac{4}{8}=\frac{1}{2}$

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MCQ 391 Mark

The probability of a certain event is:

A
$0$
✓
$1$
C
$\frac{1}{2}$
D
No existent.

Answer

Correct option: B.

$1$

Given: $4$ options of probability of some events
To Find: Which of the given options is the probability of sure event?
We know that, probability of a certain event is $1.$
Hence the correct answer is option $b.$

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MCQ 401 Mark

A card is drawn at random from a pack of $52$ cards. The probability that the drawn card is not an ace is:

A
$\frac{1}{13}$
B
$\frac{9}{13}$
C
$\frac{4}{13}$
✓
$\frac{12}{13}$

Answer

Correct option: D.

$\frac{12}{13}$

Total events $= 52$ cards
Probability of card which is not in ace Number of card $= 52 - 4 = 48$
$\therefore\ \text{Probability}=\frac{48}{52}=\frac{12}{13}$

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Maths M.C.Q (1 Marks)

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