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Maths M.C.Q (1 Marks)

Probability · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 37 questions.

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M.C.Q (1 Marks)

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37 questions · auto-graded multiple-choice test.

MCQ 11 Mark
Two coins are tossed simultaneously. What is the probability of getting at most one head?
  • A
    $\frac{1}{4}$
  • B
    $\frac{1}{2}$
  • C
    $\frac{2}{3}$
  • ✓
    $\frac{3}{4}$
Answer
Correct option: D.
$\frac{3}{4}$
When two coins are tossed the simultanecusly the
outcomes are:
{HH, HT, TH, TT}
So, there are 4 outcomes.
Getting atmost one head means the possible outcomes are:
{HT, TH, TT}
So, there are 3 possible outcomes.
P(getting atmost one head)
$=\frac{3}{4}$
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MCQ 21 Mark
Cards, each marked with one of the numbers 6, 7, 8, ..., 15 are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with number less than 10?
  • A
    $\frac{3}{5}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{2}$
  • ✓
    $\frac{2}{5}$
Answer
Correct option: D.
$\frac{2}{5}$
The total number of tickets = 10
The numbers less than 10 are 6, 7, 8 and 9.
So, there are 4 numbers.
P(getting a number less than 10)
$=\frac{4}{10}$
$=\frac{2}{5}$
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MCQ 31 Mark
A die is thrown once. The probability of getting an odd number greater than 3 is:
  • A
    $\frac{1}{3}$
  • ✓
    $\frac{1}{6}$
  • C
    $\frac{1}{2}$
  • D
    $0$
Answer
Correct option: B.
$\frac{1}{6}$
The number on a die are 1, 2, 3, 4, 5 and 6.
So, there are 6 numbers in total.
The odd number on a die greater than 3 is 5.
So, there is only 1 number.
P(getting an odd number greater than 3)
$=\frac{1}{6}$
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MCQ 41 Mark
The probability of throwing a number greater than 2 with a fair die is:
  • A
    $\frac{2}{5}$
  • B
    $\frac{5}{6}$
  • C
    $\frac{1}{3}$
  • ✓
    $\frac{2}{3}$
Answer
Correct option: D.
$\frac{2}{3}$
The numbers on a fair die are 1, 2, 3, 4, 5 and 6.
So, there are 6 numbers in total.
The number greater than 2 are 3, 4, 5 and 6.
So, there are 4 numbers.
P(getting a number greater than 2)
$=\frac{4}{6}$
$=\frac{2}{3}$
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MCQ 51 Mark
One card is drawn at random from a well-shuffled deck of 52 cards. What is th probability of getting a face card?
  • A
    $\frac{1}{26}$
  • B
    $\frac{3}{26}$
  • ✓
    $\frac{3}{13}$
  • D
    $\frac{4}{13}$
Answer
Correct option: C.
$\frac{3}{13}$
The total number of cards = 52
The number of queens = 12
P(getting a face card)
$=\frac{12}{52}$
$=\frac{3}{13}$
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MCQ 61 Mark
A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black king?
  • A
    $\frac{1}{13}$
  • ✓
    $\frac{1}{26}$
  • C
    $\frac{2}{39}$
  • D
    $\text{None of these}$
Answer
Correct option: B.
$\frac{1}{26}$
The total number of cards = 52
The number of black kings = 2
P(getting a black king)
$=\frac{2}{52}$
$=\frac{1}{26}$
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MCQ 71 Mark
Which of the following cennot be the probability of an event?
  • ✓
    1.5
  • B
    $\frac{3}{5}$
  • C
    25%
  • D
    0.3
Answer
Correct option: A.
1.5
We know that, the probability of an event E will always lie betweent 0 and 1.
Since 1.5 > 1, it cannot be the probability of an event.
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MCQ 81 Mark
If an event cannot occur then its probability is:
  • A
    $1$
  • B
    $\frac{1}{2}$
  • C
    $\frac{3}{4}$
  • ✓
    $0$
Answer
Correct option: D.
$0$
An event that cannot ocour is called an impossibel event.
the probability of an impossible event is 0.
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MCQ 91 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.
$\frac{1}{2}$
The number on a die are 1, 2, 3, 4, 5 and 6.
So, there are 6 numbers in total.
The prime numbers on the die are 2, 3, and 5.
So, there are 3 numbers.
P(getting a prime number on the die)
$=\frac{3}{6}$
$=\frac{1}{2}$
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MCQ 101 Mark
The probability of getting 2 heads, when two coins are tossed, is:
  • A
    $1$
  • B
    $\frac{3}{4}$
  • C
    $\frac{2}{2}$
  • ✓
    $\frac{1}{4}$
Answer
Correct option: D.
$\frac{1}{4}$
When two coins are tossed the outcomes are:
{HH, HT, TH, TT}
So, there are 4 numbers in total
P(getting one head)
$=\frac{1}{4}$
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MCQ 111 Mark
One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number, which is a multiple of 7, is:
  • A
    $\frac{1}{7}$
  • ✓
    $\frac{1}{8}$
  • C
    $\frac{1}{5}$
  • D
    $\frac{7}{40}$
Answer
Correct option: B.
$\frac{1}{8}$
The total number of tickets = 40
The multiples of 7 between 1 and 40 are 7, 14, 21, 28 and 35.
So. there are 5 numbers.
P(getting a multiple of 7)
$=\frac{5}{40}$
$=\frac{1}{8}$
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MCQ 121 Mark
In a lottery, there are 6 prizes and 24 blanks. What is the probability of not getting a prize?
  • A
    $\frac{3}{4}$
  • B
    $\frac{3}{5}$
  • ✓
    $\frac{4}{5}$
  • D
    $\text{None of these}$
Answer
Correct option: C.
$\frac{4}{5}$
The number of prizes = 6
The number of blanks = 24
So, the total number of tickets = 6 + 24 = 30
P(not getting a prize)
$=\frac{24}{30}$
$=\frac{4}{5}$
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MCQ 131 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 selected numbers would be 4, 8, and 12.
So, there are 3 number.
P(number of multiples of 4)
$=\frac{\text{Number of multipes of 4}}{\text{Total}}$
$=\frac{3}{15}$
$=\frac{1}{5}$
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MCQ 141 Mark
A die is thrown once. The probability of getting an even number is:
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{6}$
  • D
    $\frac{5}{6}$
Answer
Correct option: A.
$\frac{1}{2}$
The numbers on a die are 1, 2, 3, 4, 5 and 6.
So, there are 6 numbers in total.
The even number on the die are 2, 4 and 6.
So, there are 3 even number.
P(getting an even number)
$=\frac{3}{6}$
$=\frac{1}{2}$
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MCQ 151 Mark
A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball?
  • A
    $\frac{2}{5}$
  • ✓
    $\frac{3}{5}$
  • C
    $\frac{1}{10}$
  • D
    $\text{None of these}$
Answer
Correct option: B.
$\frac{3}{5}$
The bag contains 4 red and 6 black balls.
So, the total number of balls = 4 + 6 = 10
The number of black balls = 6
P(getting a black ball)
$=\frac{6}{10}$
$=\frac{3}{5}$
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MCQ 161 Mark
What is the probability of a sure event?
  • A
    0
  • B
    $\frac{1}{2}$
  • ✓
    1
  • D
    Less than 1
Answer
Correct option: C.
1
The probability of a sure event is always 1.
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MCQ 171 Mark
What is the probability of an impossible event?
  • A
    $\frac{1}{2}$
  • ✓
    0
  • C
    1
  • D
    More than 1
Answer
Correct option: B.
0
The probability of an impossible event is always 0.
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MCQ 181 Mark
A bag contains 3 white, 4 red and 5 black balls. One ball is drawn at random. What is the probability that the ball drawn is neither black nor white?
  • A
    $\frac{1}{4}$
  • B
    $\frac{1}{2}$
  • ✓
    $\frac{1}{3}$
  • D
    $\frac{3}{4}$
Answer
Correct option: C.
$\frac{1}{3}$
The bag contains 3 white, 4 red and 5 black and balls.
So, the total number of balls = 3 + 4 + 5 = 12
For the ball that is drawn to be neither black not white, it should be red.
The number of red balls = 4
P(getting a red ball)
$=\frac{4}{12}$
$=\frac{1}{3}$
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MCQ 191 Mark
Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{4}$
  • ✓
    $\frac{3}{8}$
  • D
    $\frac{3}{4}$
Answer
Correct option: C.
$\frac{3}{8}$
When three coins are tossed the simultaneously the
outcomes are:
{HHH, HHT, HTH, THT, HTT, TTH and TTT}
So, there are 8 possible outcones.
P(getting exactly two heads)
$=\frac{3}{8}$
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MCQ 201 Mark
Two dice are thrown together. The probability of getting a doulet is:
  • A
    $\frac{1}{3}$
  • ✓
    $\frac{1}{6}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{2}{3}$
Answer
Correct option: B.
$\frac{1}{6}$
The number on each die are 1, 2, 3, 4, 5 and 6.
So, the total possibilities 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)
So, there are 36 number in toral.
There are 6 possibilities when we obtain a doublet,
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
P(getting a doublet)
$=\frac{6}{36}$
$=\frac{1}{6}$
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MCQ 211 Mark
If P(E) denotes the probability of an E then:
  • A
    $\text{P(E)}<0$
  • B
    $\text{P(E)}<1$
  • ✓
    $0\leq\text{P(E)}\leq1$
  • D
    $-1\leq\text{P(E)}\leq1$
Answer
Correct option: C.
$0\leq\text{P(E)}\leq1$
We know that, the probability of an event E will always lie between 0 and 1,
Where 0 is the probability of an impossible event and 1 is the probability of a sure event.
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MCQ 221 Mark
Two dice are thrown together. The probabililty 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.
$\frac{1}{6}$
The number on each die are 1, 2, 3, 4, 5 and 6.
So, the total possibilities 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)
So, there are 36 number in toral.
There are 6 possibilities when the two die
have the same number (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
P(getting the same number on both the die)
$=\frac{6}{36}$
$=\frac{1}{6}$
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MCQ 231 Mark
If the probability of winning a game 0.4 then the probability of losing it, is:
  • A
    0.96
  • B
    $\frac{1}{0.4}$
  • ✓
    0.6
  • D
    None of these
Answer
Correct option: C.
0.6
We know that, if E is an event, then P(E) + P(E') = 1.
Let E be the event where the game is won.
So, 0.4 + P(E') = 1
⇒ P(E') = 1 - 0.4
⇒ P(E') = 0.6
So, the probability of losing the game is 0.6.
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MCQ 241 Mark
There are 20 tickets numbered as 1, 2, 3, ..., 20 respectively. One ticket is drawn at random. what is the probability that the number on the ticket drawn is a multiple of 5?
  • A
    $\frac{1}{4}$
  • ✓
    $\frac{1}{5}$
  • C
    $\frac{2}{5}$
  • D
    $\frac{3}{10}$
Answer
Correct option: B.
$\frac{1}{5}$
The total number of tickets = 20
The multiples of 5 between 1 and 20 are 5, 10, 15 and 20.
So, there are 4 numbers.
P(getting a multiple of 5)
$=\frac{4}{20}$
$=\frac{1}{5}$
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MCQ 251 Mark
A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the bex, the probability that it bears prime number less than 23 is:
  • A
    $\frac{7}{90}$
  • B
    $\frac{1}{9}$
  • ✓
    $\frac{4}{45}$
  • D
    $\frac{8}{89}$
Answer
Correct option: C.
$\frac{4}{45}$
The total number of discs = 90
The primes less than 23 are 2, 3, 5, 7, 11, 13, 17, 19,
So. there are 8 numbers.
P(getting a prime number less than 23)
$=\frac{8}{90}$
$=\frac{4}{45}$
Note: in the text, the option (c) is incorrect. it should be $\frac{4}{45}$ to go with the question asked.
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MCQ 261 Mark
If the probability of occurrence of an event is p then the probability of non-happening of this event is:
  • A
    (p - 1)
  • ✓
    (1 - p)
  • C
    p
  • D
    $\Big(1-\frac{1}{\text{p}}\Big)$
Answer
Correct option: B.
(1 - p)
Let E be the event.
So, the probability of the event happening will be P(E).
Thus, the probability of the event not happening will be P(E').
Given that, P(E) = p
We know that, P(E) + p(E') = 1
⇒ p + P(E') = 1
⇒ P(E') = 1 - p
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MCQ 271 Mark
One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black face card?
  • A
    $\frac{1}{26}$
  • ✓
    $\frac{3}{26}$
  • C
    $\frac{3}{13}$
  • D
    $\frac{3}{14}$
Answer
Correct option: B.
$\frac{3}{26}$
The total number of cards = 52
The number of black face cards = 6
P(getting a black face card)
$=\frac{6}{52}$
$=\frac{3}{26}$
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MCQ 281 Mark
Which of the following cannot be the probability of an event?
  • A
    $\frac{1}{3}$
  • B
    $0.3$
  • C
    $33\%$
  • ✓
    $\frac{7}{6}$
Answer
Correct option: D.
$\frac{7}{6}$
We know that, the probability of an event E will always lie between 0 and 1.
Since $\frac{7}{6}>1,$ it cannot be the probability of an event.
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MCQ 291 Mark
From a well-shuffled deck of 52 cards, one card is drawn at random. What is the probability of getting a queen?
  • ✓
    $\frac{1}{13}$
  • B
    $\frac{1}{26}$
  • C
    $\frac{4}{39}$
  • D
    $\text{None of these}$
Answer
Correct option: A.
$\frac{1}{13}$
The total number of cards = 52
The number of queens = 4
P(getting a queen)
$=\frac{4}{52}$
$=\frac{1}{13}$
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MCQ 301 Mark
There are 25 tickets numbered as 1, 2, 3, 4, ..., 25 respectively. One ticket is drawn at random. what is the probability that the number on the ticket is a multiple of 3 or 5?
  • A
    $\frac{2}{5}$
  • B
    $\frac{11}{25}$
  • ✓
    $\frac{12}{25}$
  • D
    $\frac{13}{25}$
Answer
Correct option: C.
$\frac{12}{25}$
The total number of tickets = 25
The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24.
The multiples of 5 are 5, 10, 15, 20 and 25.
Since 15 is a multiple of 3 as 5, it is to be caculated only once.
So, there are 12 numbers
P(getting a multiple of 3 or 5)
$=\frac{12}{25}$
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MCQ 311 Mark
A number is selected at random from the nubers 1 to 30. What is the probability that the selected number is a prime number?
  • A
    $\frac{2}{3}$
  • B
    $\frac{1}{6}$
  • ✓
    $\frac{1}{3}$
  • D
    $\frac{11}{30}$
Answer
Correct option: C.
$\frac{1}{3}$
The prime numbers from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
So, there are 10 prime numbers between 1 and 30.
P(getting a prime number)
$=\frac{\text{Number of primes between 1 and 30}}{\text{Total}}$
$=\frac{10}{30}$
$=\frac{1}{3}$
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MCQ 321 Mark
A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black?
  • A
    $\frac{8}{15}$
  • B
    $\frac{2}{15}$
  • ✓
    $\frac{13}{15}$
  • D
    $\frac{1}{3}$
Answer
Correct option: C.
$\frac{13}{15}$
The bag contains 8 red, 2 black and 5 white balls.
So, the total number of balls = 8 + 2 + 5 = 15
Since the ball should not be black, it can be red or white.
The number of red and white balls = 13
P(getting a red and white ball)
$=\frac{13}{15}$
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MCQ 331 Mark
Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is:
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{2}{5}$
  • C
    $\frac{3}{10}$
  • D
    $\frac{5}{9}$
Answer
Correct option: A.
$\frac{1}{2}$
Number on the cards are 2, 3, 4, ..., 11.
Sample space associated with the experiment, S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
$\therefore\ $Total number of outcomes = 10
Let A be the event of drawing a card with a prime number.
The cards with prime number are 2, 3, 5, 7 and 11.
Number of outcomes in favour of event A = 5
$\therefore\ $Requried probability P(A)
$=\frac{\text{Number of outcomes in favour of A }}{\text{Total number of outcomes i}}$
$=\frac{5}{10}$
$=\frac{1}{2}$
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MCQ 341 Mark
One card is drawn at random from a well$-$shuffled deck of $52$ cards. What is the probability of getting a $6?$
  • A
    $\frac{3}{26}$
  • B
    $\frac{1}{52}$
  • ✓
    $\frac{1}{13}$
  • D
    $\text{None of these}$
Answer
Correct option: C.
$\frac{1}{13}$
The total number of cards $= 52$
The number of $6$ in the deck of cards $= 4$
$P($ getting a $6)$
$=\frac{4}{52}$
$=\frac{1}{13}$
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MCQ 351 Mark
In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize?
  • A
    $\frac{1}{2}$
  • ✓
    $\frac{1}{3}$
  • C
    $\frac{2}{3}$
  • D
    $\text{None of these}$
Answer
Correct option: B.
$\frac{1}{3}$
The number of prizes = 8
The number of blanks = 16
So, the total number of tickets = 8 + 16 = 24
P(getting a prize)
$=\frac{8}{24}$
$=\frac{1}{3}$
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MCQ 361 Mark
A box contains 3 blue, 2 white and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will not be a white marble?
  • A
    $\frac{1}{3}$
  • B
    $\frac{4}{9}$
  • ✓
    $\frac{7}{9}$
  • D
    $\frac{2}{9}$
Answer
Correct option: C.
$\frac{7}{9}$
The bag contains 3 blue, 2 white and 4 red marbles.
So, the total number of marbles = 3 + 2 + 4 = 9
Since the marbles cannot be white, it can blue or red.
The number of blue or red marbles = 3 + 4 = 7
P(getting a blue or red marble)
$=\frac{7}{9}$
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MCQ 371 Mark
A box contains cards numbered 6 to 50. 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
    $\frac{1}{45}$
  • B
    $\frac{2}{15}$
  • C
    $\frac{4}{45}$
  • ✓
    $\frac{1}{9}$
Answer
Correct option: D.
$\frac{1}{9}$
The numbers on the card have to be perfect sqaures.
So, the numbers would be 9, 16, 25, 36, 49.
So, there are 5 numbers
Total number of cards = (50 - 6) + 1
= 44 + 1
= 45
P(getting a perfect square)
$=\frac{\text{Number of perfect squares}}{\text{Total}}$
$=\frac{5}{45}$
$=\frac{1}{9}$
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