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Maths M.C.Q (1 Marks)

Probability · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 111 questions.

Questions · Page 1 of 3

M.C.Q (1 Marks)

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50 questions · auto-graded multiple-choice test.

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 $3=\frac{4}{25}$
$\text{both }2\text{ and }3=\frac{4}{25}$
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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) = 6Now 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 = 30The 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, 9To Find: Probability of getting P(xy) < 9
We will make multiplication table for x and y such tha (xy) < 9
xy = 1 × 1 = 1
= 1 × 4 = 4
= 1 × 9 = 9
= 2 × 1 = 2
= 2 × 4 = 8
= 2 × 9 = 18
= 3 × 1 = 3
= 3 × 4 = 12
= 3 × 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 × 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 yearTo 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 "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 MOBILE = 6No, 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 ⇒ x > -2
$\therefore$ x
⇒ 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 × 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 BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.
Total number of outcomes = 8
The favourable outcomes are 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. (H, H), (H, T), (T, H), (T, T)
Number of favourable outcomes for at least one head = 3, i.e. (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
    $\frac{1}{3}$
  • ✓
    None of above
Answer
Correct option: D.
None of above
A number is selected from the numbers $1$ to $25$
Probability of prime number which are $2, 3, 5, 7, 11, 13, 17, 19, 23 = 9$
$\therefore\text{P(E)}=\frac{\text{m}}{\text{n}}=\frac{9}{25}$
$\Rightarrow\ \frac{9}{25}$
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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 × 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 × 2 × 2 = 8
The results will be
(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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MCQ 411 Mark
Cards numbered 7 to 40 were put in a box. A card is selected at random from the box. The probability that the selected card has a number, which is a multiple of 7, is
  • A
    $\frac{7}{34}$
  • B
    $\frac{7}{35}$
  • C
    $\frac{6}{35}$
  • ✓
    $\frac{5}{34}$
Answer
Correct option: D.
$\frac{5}{34}$
d
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MCQ 421 Mark
The probability of getting a bad egg in a lot of 400 eggs is 0.045. The number of good eggs in the lot is
  • A
    18
  • B
    180
  • ✓
    382
  • D
    220
Answer
Correct option: C.
382
c
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MCQ 431 Mark
For an event $E$, if $P(E)+P(\bar{E})=q$, then the value of $q^2-4$ is
  • ✓
    -3
  • B
    3
  • C
    5
  • D
    -5
Answer
Correct option: A.
-3
a
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MCQ 441 Mark
Two dice are tossed simultaneuously. The probability of getting odd numbers on both the dice is
  • A
    $\frac{6}{36}$
  • B
    $\frac{3}{36}$
  • C
    $\frac{12}{36}$
  • ✓
    $\frac{9}{36}$
Answer
Correct option: D.
$\frac{9}{36}$
d
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MCQ 451 Mark
A box contains cards numbered 6 to 55. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square, is
  • A
    $\frac{7}{50}$
  • B
    $\frac{7}{55}$
  • ✓
    $\frac{1}{10}$
  • D
    $\frac{5}{49}$
Answer
Correct option: C.
$\frac{1}{10}$
c
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MCQ 461 Mark
Two dice are rolled together. The probability of getting the sum of two numbers to be more than 10, is
  • A
    $\frac{1}{9}$
  • B
    $\frac{1}{6}$
  • C
    $\frac{7}{12}$
  • ✓
    $\frac{1}{12}$
Answer
Correct option: D.
$\frac{1}{12}$
d
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MCQ 471 Mark
A card is drawn at random from a well shuffled pack of 52 playing cards. The probability of getting a face card is
  • A
    $\frac{1}{2}$
  • ✓
    $\frac{3}{13}$
  • C
    $\frac{4}{13}$
  • D
    $\frac{1}{13}$
Answer
Correct option: B.
$\frac{3}{13}$
b
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MCQ 481 Mark
A bag contains 5 pink, 8 blue and 7 yellow balls. One ball is drawn at random from the bag. What is the probability of getting neither a blue nor a pink ball?
  • A
    $\frac{1}{4}$
  • B
    $\frac{2}{5}$
  • ✓
    $\frac{7}{20}$
  • D
    $\frac{13}{20}$
Answer
Correct option: C.
$\frac{7}{20}$
c
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MCQ 491 Mark
Three coins are tossed simultaneously, what is the probability of getting at most one tail?
  • A
    $\frac{3}{8}$
  • ✓
    $\frac{4}{8}$
  • C
    $\frac{5}{8}$
  • D
    $\frac{7}{8}$
Answer
Correct option: B.
$\frac{4}{8}$
b
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MCQ 501 Mark
A bag contains 100 cards numbered 1 to 100. A card is drawn at random from the bag. What is the probability that the number on the card is a perfect cube?
  • A
    $\frac{1}{20}$
  • B
    $\frac{3}{50}$
  • ✓
    $\frac{1}{25}$
  • D
    $\frac{7}{100}$
Answer
Correct option: C.
$\frac{1}{25}$
c
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M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip