MCQ 11 Mark
The probability of a certain event is
- A$0$
- ✓1
- C$1/2$
- D-1
Answer
View full question & answer→Correct option: B.
1
(b):The probability of a certain event is always 1 .
Questions · Page 1 of 2
M.C.Q (1 Marks)
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MCQ 21 Mark
Which of the following cannot be the probability of an event?
- A$\frac{1}{2}$
- B1
- C$0$
- ✓1.5
Answer
View full question & answer→Correct option: D.
1.5
(d):For probability of an event $E, 0 \leq P(E) \leq 1$.
MCQ 31 Mark
If probability of success is $0.9 \%$, then probability of failure is
- A0.91
- B0.091
- C99.1
- ✓0.991
Answer
View full question & answer→Correct option: D.
0.991
(d):Probability of failure $=1$ - Probability of success
$=1-\frac{0.9}{100}=1-\frac{9}{1000}=\frac{1000-9}{1000}=\frac{991}{1000}=0.991$
$=1-\frac{0.9}{100}=1-\frac{9}{1000}=\frac{1000-9}{1000}=\frac{991}{1000}=0.991$
MCQ 41 Mark
If $P(E)=0.01$, then $P(\operatorname{not} E)$ is equal to
- A0.09
- B0.1
- C0.9
- ✓0.99
Answer
View full question & answer→Correct option: D.
0.99
(d):$P(\operatorname{not} E)=1-P(E)=1-0.01=0.99$
MCQ 51 Mark
A single letter is selected at random from the word 'PROBABILITY'. The probability that it is a vowel is
- A$\frac{3}{11}$
- ✓$\frac{4}{11}$
- C$\frac{2}{11}$
- D$\frac{5}{11}$
Answer
View full question & answer→Correct option: B.
$\frac{4}{11}$
(b):Total number of letters in the word
'PROBABILITY' = 11
$\therefore \quad$ Total number of possible outcomes $=11$
Vowels in the word 'PROBABILITY' are $\{ O , A , I , I \}$
So, favourable number of outcomes $=4$
$\therefore \quad$ Required probability $=\frac{4}{11}$
'PROBABILITY' = 11
$\therefore \quad$ Total number of possible outcomes $=11$
Vowels in the word 'PROBABILITY' are $\{ O , A , I , I \}$
So, favourable number of outcomes $=4$
$\therefore \quad$ Required probability $=\frac{4}{11}$
MCQ 61 Mark
A letter is chosen at random from the English alphabets. Find the probability that the letter chosen succeeds $V$.
- ✓$\frac{2}{13}$
- B$\frac{5}{26}$
- C$\frac{1}{26}$
- D$\frac{1}{2}$
Answer
View full question & answer→Correct option: A.
$\frac{2}{13}$
(a):Total number of outcomes $=26$
Letters succeeding $V$ are $\{W, X, Y, Z\}$ i.e.,
favourable number of outcomes is 4 .
$\therefore \quad$ Required probability $=\frac{4}{26}=\frac{2}{13}$
Letters succeeding $V$ are $\{W, X, Y, Z\}$ i.e.,
favourable number of outcomes is 4 .
$\therefore \quad$ Required probability $=\frac{4}{26}=\frac{2}{13}$
MCQ 71 Mark
Two coins are tossed simultaneously. The probability of getting no head is
- ✓$\frac{1}{4}$
- B$\frac{1}{2}$
- C$\frac{3}{4}$
- D1
Answer
View full question & answer→Correct option: A.
$\frac{1}{4}$
(a):Total possible outcomes are $\{H H, H T, T H, T T\}$ i.e., 4 in number.
Favourable outcome is $\{T T\}$ i.e., 1 in number.
$\therefore \quad P($ getting no head $)=\frac{1}{4}$
Favourable outcome is $\{T T\}$ i.e., 1 in number.
$\therefore \quad P($ getting no head $)=\frac{1}{4}$
MCQ 81 Mark
In a single throw of a die, the probability of getting a multiple of 2 is
- ✓$\frac{1}{2}$
- B$\frac{1}{3}$
- C$\frac{1}{6}$
- D$\frac{2}{3}$
Answer
View full question & answer→Correct option: A.
$\frac{1}{2}$
(a):Total number of possible outcomes $=6$
Favourable outcomes of getting multiple of 2 are $\{2,4,6\}$ i.e., 3 in number.
$\therefore \quad$ Required probability $=\frac{3}{6}=\frac{1}{2}$
Favourable outcomes of getting multiple of 2 are $\{2,4,6\}$ i.e., 3 in number.
$\therefore \quad$ Required probability $=\frac{3}{6}=\frac{1}{2}$
MCQ 91 Mark
A card is drawn from a pack of 52 cards. The probability of drawing a black face card is
- A$\frac{2}{13}$
- ✓$\frac{3}{26}$
- C$\frac{1}{13}$
- D$\frac{3}{52}$
Answer
View full question & answer→Correct option: B.
$\frac{3}{26}$
(b):Total number of cards $=52$
Number of black face cards $=6$
$ \therefore \quad P(\text { getting a black face card })=\frac{6}{52}=\frac{3}{26} $
Number of black face cards $=6$
$ \therefore \quad P(\text { getting a black face card })=\frac{6}{52}=\frac{3}{26} $
MCQ 101 Mark
A number $x$ is chosen at random from the numbers $-2,-1,0,1,2$. What is the probability that $x^2<2$ ?
- A$\frac{1}{5}$
- B$\frac{2}{5}$
- ✓$\frac{3}{5}$
- D$\frac{4}{5}$
Answer
View full question & answer→Correct option: C.
$\frac{3}{5}$
(c):Clearly, number $x$ can take any one of the five given values.
So, total number of possible outcomes $=5$
We observe that $x^2<2$, when $x$ take any one of the following three values $-1,0$ and 1 .
So, favourable number of outcomes $=3$
$ \therefore P\left(x^2<2\right)=\frac{3}{5} $
So, total number of possible outcomes $=5$
We observe that $x^2<2$, when $x$ take any one of the following three values $-1,0$ and 1 .
So, favourable number of outcomes $=3$
$ \therefore P\left(x^2<2\right)=\frac{3}{5} $
MCQ 111 Mark
A box contains 100 discs, numbered from 1 to 100. If one disc is drawn at random from the box, then the probability that it bears a prime number less than 30 , is
- A$\frac{7}{100}$
- ✓$\frac{1}{10}$
- C$\frac{4}{50}$
- D$\frac{9}{50}$
Answer
View full question & answer→Correct option: B.
$\frac{1}{10}$
(b):Total number of possible outcomes $=100$
Let $E$ be the event 'drawing a prime number less than $30^{\circ}$.
Outcomes favourable to $E$ are $\{2,3,5,7,11,13,17,19$, $23,29\}$ i.e., 10 in number.
$ \therefore \quad P(E)=\frac{10}{100}=\frac{1}{10} $
Let $E$ be the event 'drawing a prime number less than $30^{\circ}$.
Outcomes favourable to $E$ are $\{2,3,5,7,11,13,17,19$, $23,29\}$ i.e., 10 in number.
$ \therefore \quad P(E)=\frac{10}{100}=\frac{1}{10} $
MCQ 121 Mark
If a die is thrown, then what is the probability of getting a number less than 4 and greater than 3 ?
- ✓$0$
- B1
- C$\frac{1}{3}$
- D$\frac{2}{3}$
Answer
View full question & answer→Correct option: A.
$0$
(a):There is no such number lie on a die, which is less than 4 and greater than 3 .
$\therefore \quad$ Required probability $=0$
$\therefore \quad$ Required probability $=0$
MCQ 131 Mark
In a single throw of a pair of dice, the probability of getting the sum as a perfect square is
- ✓$\frac{7}{36}$
- B$\frac{5}{36}$
- C$\frac{8}{36}$
- D$\frac{11}{36}$
Answer
View full question & answer→Correct option: A.
$\frac{7}{36}$
(a):Total number of possible outcomes $=6 \times 6=36$
Let $E$ be the event 'getting the sum as a perfect square'.
$\therefore$ Outcomes favorable to $E$ are $\{(1,3),(2,2),(3,1)$, $(3,6),(4,5),(5,4),(6,3) \mid$ i.e., 7 in number.
$ \therefore \quad P(E)=\frac{7}{36} $
Let $E$ be the event 'getting the sum as a perfect square'.
$\therefore$ Outcomes favorable to $E$ are $\{(1,3),(2,2),(3,1)$, $(3,6),(4,5),(5,4),(6,3) \mid$ i.e., 7 in number.
$ \therefore \quad P(E)=\frac{7}{36} $
MCQ 141 Mark
A card is accidently dropped from a pack of 52 playing cards. The probability that it is a red card is
- ✓$\frac{1}{2}$
- B$\frac{1}{13}$
- C$\frac{1}{52}$
- D$\frac{12}{13}$
Answer
View full question & answer→Correct option: A.
$\frac{1}{2}$
(a):Total number of possible outcomes $=52$
$\because \quad$ Number of red cards $=26$
So, number of favourable outcomes $=26$
$\therefore \quad$ Required probability $=\frac{26}{52}=\frac{1}{2}$
$\because \quad$ Number of red cards $=26$
So, number of favourable outcomes $=26$
$\therefore \quad$ Required probability $=\frac{26}{52}=\frac{1}{2}$
MCQ 151 Mark
A letter is chosen at random from the English alphabet. Probability that it is a letter of the word "SIMULTANEOUSLY" is
- A$\frac{14}{26}$
- ✓$\frac{11}{26}$
- C$\frac{10}{26}$
- D$\frac{15}{26}$
Answer
View full question & answer→Correct option: B.
$\frac{11}{26}$
(b):Total number of possible outcomes $=26$
Distinct letters in the word "SIMULTANEOUSLY" are {S, I, M, U, L, T, A, N, E, O, Y} i.e., 11 in number.
So, favourable number of outcomes $=11$
$\therefore \quad$ Required probability $=\frac{11}{26}$
Distinct letters in the word "SIMULTANEOUSLY" are {S, I, M, U, L, T, A, N, E, O, Y} i.e., 11 in number.
So, favourable number of outcomes $=11$
$\therefore \quad$ Required probability $=\frac{11}{26}$
MCQ 161 Mark
A number $x$ is chosen at random from $-5,-4,-3,-2,-1,0,1,2,3,4,5$. The probability that $|x| \leq 3$ is
- A$\frac{3}{11}$
- B$\frac{6}{11}$
- ✓$\frac{7}{11}$
- D$\frac{9}{11}$
Answer
View full question & answer→Correct option: C.
$\frac{7}{11}$
(c):Total number of possible outcomes $=11$
Let $E$ be the event ' $|x| \leq 3$ '.
So, outcomes favourable to $E$ are $\{-3,-2,-1,0,1,2,3\}$ i.e., 7 in number.
$ \therefore \quad P(E)=\frac{7}{11} $
Let $E$ be the event ' $|x| \leq 3$ '.
So, outcomes favourable to $E$ are $\{-3,-2,-1,0,1,2,3\}$ i.e., 7 in number.
$ \therefore \quad P(E)=\frac{7}{11} $
MCQ 171 Mark
A child has a block in the shape of a cube with one letter written on each face as follows:
The cube is thrown once. What is the probability of getting $A$ ?
The cube is thrown once. What is the probability of getting $A$ ?
- ✓$\frac{1}{3}$
- B$\frac{1}{6}$
- C$\frac{1}{2}$
- D$\frac{1}{4}$
Answer
View full question & answer→Correct option: A.
$\frac{1}{3}$
(a):Total number of possible outcomes $=6$
As there are two $A$ 's.
$\therefore \quad$ Favourable number of outcomes $=2$
$\therefore \quad$ Required probability $=\frac{2}{6}=\frac{1}{3}$
As there are two $A$ 's.
$\therefore \quad$ Favourable number of outcomes $=2$
$\therefore \quad$ Required probability $=\frac{2}{6}=\frac{1}{3}$
MCQ 181 Mark
A bag contains 4 red, 5 black and 3 yellow balls. A ball is taken out of the bag at random. Find the probability that the ball taken out is not of yellow colour.
- A$\frac{2}{3}$
- B$\frac{1}{3}$
- ✓$\frac{3}{4}$
- D$\frac{1}{2}$
Answer
View full question & answer→Correct option: C.
$\frac{3}{4}$
(c):Total number of possible outcomes $=4+5+3=12$
Favourable number of outcomes $=5+4=9$
$\therefore \quad$ Required probability $=\frac{9}{12}=\frac{3}{4}$
Favourable number of outcomes $=5+4=9$
$\therefore \quad$ Required probability $=\frac{9}{12}=\frac{3}{4}$
MCQ 191 Mark
A bag contains 12 red roses only. Shalini takes out one rose without looking into the bag. The probability that she takes out an orange rose is
- A$\frac{1}{2}$
- ✓$0$
- C1
- D$\frac{2}{3}$
Answer
View full question & answer→Correct option: B.
$0$
(b):Since, there is no orange rose in the bag.
$\therefore \quad$ Required probability $=0$
$\therefore \quad$ Required probability $=0$
MCQ 201 Mark
A month is selected at random from a year. The probability that it is May or July is
- A$\frac{1}{12}$
- ✓$\frac{1}{6}$
- C$\frac{3}{4}$
- D$\frac{1}{3}$
Answer
View full question & answer→Correct option: B.
$\frac{1}{6}$
(b):Number of months in a year $=12$
So, total number of possible outcomes $=12$
Favourable outcomes are May or July i.e., 2
$\therefore \quad$ Required probability $=\frac{2}{12}=\frac{1}{6}$
So, total number of possible outcomes $=12$
Favourable outcomes are May or July i.e., 2
$\therefore \quad$ Required probability $=\frac{2}{12}=\frac{1}{6}$
MCQ 211 Mark
A card is drawn at random from a pack of 52 cards. The probability that the drawn card is not a king is
- A$\frac{1}{13}$
- B$\frac{9}{13}$
- C$\frac{4}{13}$
- ✓$\frac{12}{13}$
Answer
View full question & answer→Correct option: D.
$\frac{12}{13}$
(d):Total number of possible outcomes $=52$
Number of king cards in the pack $=4$
$\therefore \quad$ Number of cards that are not king $=52-4=48$
So, favourable number of outcomes $=48$
$\therefore \quad$ Required probability $=\frac{48}{52}=\frac{12}{13}$
Number of king cards in the pack $=4$
$\therefore \quad$ Number of cards that are not king $=52-4=48$
So, favourable number of outcomes $=48$
$\therefore \quad$ Required probability $=\frac{48}{52}=\frac{12}{13}$
MCQ 221 Mark
A die is thrown once. Find the probability of getting an odd prime number.
- A$\frac{1}{2}$
- ✓$\frac{1}{3}$
- C$\frac{1}{6}$
- D$\frac{1}{4}$
Answer
View full question & answer→Correct option: B.
$\frac{1}{3}$
(b):Total possible outcomes are $(1,2,3,4,5,6\}$ i.e., 6 in number.
Favourable outcomes are $\{3,5\}$ i.e., 2 in number.
$\therefore \quad$ Required probability $=\frac{2}{6}=\frac{1}{3}$
Favourable outcomes are $\{3,5\}$ i.e., 2 in number.
$\therefore \quad$ Required probability $=\frac{2}{6}=\frac{1}{3}$
MCQ 231 Mark
One ticket is selected at random from 100 tickets numbered $00,01,02, \ldots, 99$. Suppose $x$ is sum of the digits and $y$ is product of the digits. Then, probability of getting $x=8$ and $y=0$ is
- A$\frac{2}{17}$
- B$\frac{3}{27}$
- ✓$\frac{1}{50}$
- D$\frac{1}{25}$
Answer
View full question & answer→Correct option: C.
$\frac{1}{50}$
(c):Total number of possible outcomes $=100$
Sum of digits $=8$ and product $=0$
$\therefore \quad$ Favourable outcomes are 08 and 80 i.e., 2
$\therefore \quad$ Required probability $=\frac{2}{100}=\frac{1}{50}$
Sum of digits $=8$ and product $=0$
$\therefore \quad$ Favourable outcomes are 08 and 80 i.e., 2
$\therefore \quad$ Required probability $=\frac{2}{100}=\frac{1}{50}$
MCQ 241 Mark
A card is drawn from a pack of well-shuffled deck of 52 playing cards. The probability that the number on the card is a perfect square is
- ✓$\frac{2}{13}$
- B$\frac{3}{13}$
- C$\frac{5}{13}$
- D$\frac{7}{52}$
Answer
View full question & answer→Correct option: A.
$\frac{2}{13}$
(a):Perfect square numbers on the cards are $[4,9\}$ i.e., 2 in number.
$\therefore$ Total number of favourable outcomes $=4 \times 2=8$
Total number of possible outcomes $=52$
$\therefore \quad$ Required probability $=\frac{8}{52}=\frac{2}{13}$
$\therefore$ Total number of favourable outcomes $=4 \times 2=8$
Total number of possible outcomes $=52$
$\therefore \quad$ Required probability $=\frac{8}{52}=\frac{2}{13}$
MCQ 251 Mark
An integer is chosen between 0 to 50 . What is the probability that it is divisible by 4 ?
- ✓$\frac{12}{49}$
- B$\frac{13}{49}$
- C$\frac{1}{7}$
- D$\frac{4}{49}$
Answer
View full question & answer→Correct option: A.
$\frac{12}{49}$
(a):Total number of possible outcomes $=49$
Favourable outcomes are $\{4,8,12, \ldots, 48\}$ i.e., 12 in number.
$\therefore \quad$ Required probability $=\frac{12}{49}$
Favourable outcomes are $\{4,8,12, \ldots, 48\}$ i.e., 12 in number.
$\therefore \quad$ Required probability $=\frac{12}{49}$
MCQ 261 Mark
A die is thrown once. The probability of getting a non-prime number is
- A$\frac{2}{3}$
- B$\frac{1}{3}$
- ✓$\frac{1}{2}$
- D$\frac{1}{6}$
Answer
View full question & answer→Correct option: C.
$\frac{1}{2}$
(c):Total number of possible outcomes $=6$
Favourable outcomes are $\{1,4,6\}$ i.e., 3 in number.
$\therefore \quad P($ getting a non-prime number $)=\frac{3}{6}=\frac{1}{2}$
Favourable outcomes are $\{1,4,6\}$ i.e., 3 in number.
$\therefore \quad P($ getting a non-prime number $)=\frac{3}{6}=\frac{1}{2}$
MCQ 271 Mark
The probability expressed as a percentage of a particular occurrence can never be
- ALess than 100
- ✓less than 0
- Cgreater than 1
- Danything but a whole number
Answer
View full question & answer→Correct option: B.
less than 0
(b):We know that, the probability expressed as a percentage always lies from 0 to 100 . So, it cannot be less than 0 .
MCQ 281 Mark
If $P(A)$ denotes the probability of an event $A$, then
- A$P\left(A^{\prime}\right)<0$
- B$P\left(A^{\prime}\right)>1$
- ✓$0 \leq P\left(A^{\prime}\right) \leq 1$
- D$-1 \leq P\left(A^{\prime}\right) \leq 1$
Answer
View full question & answer→Correct option: C.
$0 \leq P\left(A^{\prime}\right) \leq 1$
(c):Since, probability of any event always lies from 0 to 1 .
MCQ 291 Mark
When a die is thrown, the probability of getting an odd number less than 3 is
- ✓$\frac{1}{6}$
- B$\frac{1}{3}$
- C$\frac{1}{2}$
- D$0$
Answer
View full question & answer→Correct option: A.
$\frac{1}{6}$
(a):When a die is thrown, then total number of possible outcomes $=6$
Odd number less than 3 is 1 only.
$\therefore \quad$ Number of favourable outcomes $=1$
$\therefore \quad$ Required probability $=\frac{1}{6}$
Odd number less than 3 is 1 only.
$\therefore \quad$ Number of favourable outcomes $=1$
$\therefore \quad$ Required probability $=\frac{1}{6}$
MCQ 301 Mark
A card is drawn from a deck of 52 cards. The event $E$ is that card is not an ace of hearts. The number of outcomes favourable to $E$ is
- A4
- B13
- C48
- ✓51
Answer
View full question & answer→Correct option: D.
51
(d):In a deck of 52 cards, there are 13 cards of hearts and 1 of them is ace of heart.
Hence, the number of outcomes favourable to $E$ $=52-1=51$
Hence, the number of outcomes favourable to $E$ $=52-1=51$
MCQ 311 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 5 is
- ✓$\frac{1}{5}$
- B$\frac{3}{5}$
- C$\frac{4}{5}$
- D$\frac{1}{3}$
Answer
View full question & answer→Correct option: A.
$\frac{1}{5}$
(a):Total number of outcomes $=40$
Multiples of 5 from 1 to 40 are $\{5,10,15,20,25,30,35,40\}$
So, number of favourable outcomes $=8$
$\therefore \quad$ Required probability $=\frac{8}{40}=\frac{1}{5}$
Multiples of 5 from 1 to 40 are $\{5,10,15,20,25,30,35,40\}$
So, number of favourable outcomes $=8$
$\therefore \quad$ Required probability $=\frac{8}{40}=\frac{1}{5}$
MCQ 321 Mark
A school has five houses $A, B, C, D$ and $E$. A class has 23 students, 4 from house $A, 8$ from house $B, 5$ from house $C, 2$ from house $D$ and rest from house $E$. A single student is selected at random to be the class monitor. The probability that the selected student is not from $A, B$ and $C$ is
- A$\frac{4}{23}$
- ✓$\frac{6}{23}$
- C$\frac{8}{23}$
- D$\frac{17}{23}$
Answer
View full question & answer→Correct option: B.
$\frac{6}{23}$
(b):Total number of students $=23$
Number of students in houses $A, B$ and $C=4+8+5=17$
$\therefore \quad$ Remaining students $=23-17=6$
So, probability that the selected student is not from $A, B$ and $C=\frac{6}{23}$
Number of students in houses $A, B$ and $C=4+8+5=17$
$\therefore \quad$ Remaining students $=23-17=6$
So, probability that the selected student is not from $A, B$ and $C=\frac{6}{23}$
MCQ 331 Mark
A letter is chosen at random from the letters of the word 'PRONUNCIATION'. Find the probability that the letter chosen is a vowel.
- ✓$\frac{6}{13}$
- B$\frac{2}{3}$
- C$\frac{1}{8}$
- D$\frac{7}{13}$
Answer
View full question & answer→Correct option: A.
$\frac{6}{13}$
(a):$\because$ There are 13 letters in the given word.
$\therefore \quad$Total number of possible outcomes $=13$
Vowels are O, U, I, A, I, O i.e., 6 in number.
$\therefore \quad$ Favourable number of outcomes $=6$
$\therefore \quad$ Required probability $=6 / 13$
$\therefore \quad$Total number of possible outcomes $=13$
Vowels are O, U, I, A, I, O i.e., 6 in number.
$\therefore \quad$ Favourable number of outcomes $=6$
$\therefore \quad$ Required probability $=6 / 13$
MCQ 341 Mark
A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. Find the probability of getting a black ball.
- A$\frac{1}{5}$
- B$\frac{2}{5}$
- C$\frac{4}{5}$
- ✓$\frac{3}{5}$
Answer
View full question & answer→Correct option: D.
$\frac{3}{5}$
(d):Number of red balls $=4$
Number of black balls $=6$
Total number of balls in the bag $=4+6=10$
$\therefore \quad P($ getting a black ball $)=\frac{6}{10}=\frac{3}{5}$
Number of black balls $=6$
Total number of balls in the bag $=4+6=10$
$\therefore \quad P($ getting a black ball $)=\frac{6}{10}=\frac{3}{5}$
MCQ 351 Mark
A die is thrown once. Find the probability of getting a number which is not a factor of 36 .
- A$\frac{1}{3}$
- B$\frac{1}{2}$
- ✓$\frac{1}{6}$
- D$\frac{5}{6}$
Answer
View full question & answer→Correct option: C.
$\frac{1}{6}$
(c):Let $E$ be the event of getting a number on the die which is not a factor of 36 and this number is 5 .
Total number of possible outcomes $=6$ Number of outcomes favourable to event $E=1$
$\therefore \quad$ Required probability $=\frac{1}{6}$
Total number of possible outcomes $=6$ Number of outcomes favourable to event $E=1$
$\therefore \quad$ Required probability $=\frac{1}{6}$
MCQ 361 Mark
The blood group of 16 students of class $X$ are recorded as follows :
$A , B , O , O , AB , O , A , O , B , A , O , B , A , O , O , AB$
Find the probability of selecting a student with blood group $O$.
$A , B , O , O , AB , O , A , O , B , A , O , B , A , O , O , AB$
Find the probability of selecting a student with blood group $O$.
- A$\frac{5}{16}$
- ✓$\frac{7}{16}$
- C$\frac{3}{16}$
- D$\frac{1}{4}$
Answer
View full question & answer→Correct option: B.
$\frac{7}{16}$
(b):Total number of possible outcomes $=16$
Let $E$ be an event of selecting a student with blood group $O$.
So, favourable number of outcomes $=7$
$ \therefore \quad P(E)=\frac{7}{16} $
Let $E$ be an event of selecting a student with blood group $O$.
So, favourable number of outcomes $=7$
$ \therefore \quad P(E)=\frac{7}{16} $
MCQ 371 Mark
A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be neither yellow nor a blue card.
- A$\frac{1}{7}$
- ✓$\frac{2}{7}$
- C$\frac{3}{7}$
- D$\frac{5}{7}$
Answer
View full question & answer→Correct option: B.
$\frac{2}{7}$
(b):Total number of cards $=100+200+50=350$
Total number of possible outcomes $=350$
$ \begin{aligned} \text { Number of favourable outcomes } & =\text { Number of red cards } \\ & =100 \end{aligned} $
$\therefore \quad P($ neither getting yellow nor a blue card) $=\frac{100}{350}=\frac{2}{7}$
Total number of possible outcomes $=350$
$ \begin{aligned} \text { Number of favourable outcomes } & =\text { Number of red cards } \\ & =100 \end{aligned} $
$\therefore \quad P($ neither getting yellow nor a blue card) $=\frac{100}{350}=\frac{2}{7}$
MCQ 381 Mark
Sunita picked a prime number from the integers 1 to 20 . The probability that it will be the number 13 is
- A$\frac{1}{20}$
- ✓$\frac{1}{8}$
- C$\frac{2}{7}$
- D$\frac{13}{20}$
Answer
View full question & answer→Correct option: B.
$\frac{1}{8}$
(b):$\because$ Prime number between 1 to 20 are $\{2,3,5,7$, $11,13,17,19\}$
So, total number of outcomes $=8$
Favourable outcome is $\{13\}$ i.e., 1
$\therefore \quad$ Required probability $=\frac{1}{8}$
So, total number of outcomes $=8$
Favourable outcome is $\{13\}$ i.e., 1
$\therefore \quad$ Required probability $=\frac{1}{8}$
MCQ 391 Mark
A box contains 100 memory cards out of which 25 are good and 75 are defective. A memory card is selected at random. The probability that selected memory card is defective is
- A$\frac{1}{4}$
- B$\frac{1}{2}$
- ✓$\frac{3}{4}$
- D1
Answer
View full question & answer→Correct option: C.
$\frac{3}{4}$
(c):Total number of possible outcomes $=100$
Favourable number of outcomes $=75$
$P($ selecting a defective memory card $)=\frac{75}{100}=\frac{3}{4}$
Favourable number of outcomes $=75$
$P($ selecting a defective memory card $)=\frac{75}{100}=\frac{3}{4}$
MCQ 401 Mark
A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers $1,2,3, \ldots ., 8$ as shown in the given figure which are equally likely outcomes. What is the probability that the arrow will point at a number which is less than 9 .
- A$\frac{1}{2}$
- B$\frac{1}{3}$
- C$\frac{1}{8}$
- ✓1
Answer
View full question & answer→Correct option: D.
1
(d):Total number of outcomes $=8$
Favourable outcomes are $\{1,2,3,4,5,6,7,8\}$ i.e., 8 in number.
$\therefore \quad P($ getting a number less than 9$)=\frac{8}{8}=1$
Favourable outcomes are $\{1,2,3,4,5,6,7,8\}$ i.e., 8 in number.
$\therefore \quad P($ getting a number less than 9$)=\frac{8}{8}=1$
MCQ 411 Mark
₹ 5 coins, twenty eight ₹ 10 coins and eight ₹ 20 coins. Now, they said to Nisha, their another friends, to choose a coin randomly.
Find the probability that the coin chosen is ₹ 5 coin.
Find the probability that the coin chosen is ₹ 5 coin.
- A$\frac{17}{55}$
- B$\frac{36}{85}$
- ✓$\frac{18}{85}$
- D$\frac{1}{15}$
Answer
View full question & answer→Correct option: C.
$\frac{18}{85}$
(c):Total number of coins $=50+48+36+28+8=170$
Number of ₹ 5 coins $=36$
$\therefore \quad$ Required probability $=\frac{36}{170}=\frac{18}{85}$
Number of ₹ 5 coins $=36$
$\therefore \quad$ Required probability $=\frac{36}{170}=\frac{18}{85}$
MCQ 421 Mark
₹ 5 coins, twenty eight ₹ 10 coins and eight ₹ 20 coins. Now, they said to Nisha, their another friends, to choose a coin randomly.
Find the probability that the coin chosen is ₹ 20 coin.
Find the probability that the coin chosen is ₹ 20 coin.
- A$\frac{13}{85}$
- ✓$\frac{4}{85}$
- C$\frac{3}{85}$
- D$\frac{4}{15}$
Answer
View full question & answer→Correct option: B.
$\frac{4}{85}$
(b):Number of ₹ 20 coins $=8$
$\therefore \quad$ Required probability $=\frac{8}{170}=\frac{4}{85}$
$\therefore \quad$ Required probability $=\frac{8}{170}=\frac{4}{85}$
MCQ 431 Mark
$₹\ 5$ coins, twenty eight $₹\ 10$ coins and eight $₹\ 20$ coins. Now, they said to Nisha, their another friends, to choose a coin randomly.
Find the probability that the coin chosen is not a $₹\ 10$ coin.
Find the probability that the coin chosen is not a $₹\ 10$ coin.
- A$\frac{15}{31}$
- B$\frac{36}{85}$
- C$\frac{1}{5}$
- ✓$\frac{71}{85}$
Answer
View full question & answer→Correct option: D.
$\frac{71}{85}$
Number of $₹\ 10$ coins $=28$
Probability $($coin is of $₹\ 10) =\frac{28}{170}$
$\therefore$ Required probability $=1-P\ ($coin is of $\text ₹ 10)$
$\quad=1-\frac{28}{170}=\frac{142}{170}=\frac{71}{85}$
Probability $($coin is of $₹\ 10) =\frac{28}{170}$
$\therefore$ Required probability $=1-P\ ($coin is of $\text ₹ 10)$
$\quad=1-\frac{28}{170}=\frac{142}{170}=\frac{71}{85}$
MCQ 441 Mark
₹ 5 coins, twenty eight ₹ 10 coins and eight ₹ 20 coins. Now, they said to Nisha, their another friends, to choose a coin randomly.
Find the probability that the coin chosen is of denomination of atleast ₹ 10 .
Find the probability that the coin chosen is of denomination of atleast ₹ 10 .
- ✓$\frac{18}{85}$
- B$\frac{36}{85}$
- C$\frac{1}{17}$
- D$\frac{16}{85}$
Answer
View full question & answer→Correct option: A.
$\frac{18}{85}$
(a):Total number of coins of ₹ 10 and ₹ 20
$ =28+8=36 $
$\therefore \quad$ Required probability $=\frac{36}{170}=\frac{18}{85}$
$ =28+8=36 $
$\therefore \quad$ Required probability $=\frac{36}{170}=\frac{18}{85}$
MCQ 451 Mark
₹ 5 coins, twenty eight ₹ 10 coins and eight ₹ 20 coins. Now, they said to Nisha, their another friends, to choose a coin randomly.
Find the probability that the coin chosen is of denomination of atmost ₹ 5.
Find the probability that the coin chosen is of denomination of atmost ₹ 5.
- ✓$\frac{67}{85}$
- B$\frac{36}{85}$
- C$\frac{4}{85}$
- D$\frac{18}{85}$
Answer
View full question & answer→Correct option: A.
$\frac{67}{85}$
(a):Total number of coins of ₹ 1, ₹ 2 and ₹ 5
$ =50+48+36=134 $
$\therefore \quad$ Required probability $=\frac{134}{170}=\frac{67}{85}$
$ =50+48+36=134 $
$\therefore \quad$ Required probability $=\frac{134}{170}=\frac{67}{85}$
MCQ 461 Mark
Which of the following is not the probability of an event ?
- A$\frac{2}{3}$
- ✓-1.5
- C$15 \%$
- D0.7
Answer
View full question & answer→Correct option: B.
-1.5
-1.5
MCQ 471 Mark
India reached in Semifinal match in world cup cricket match. Other than India three more team Australia, New zealand and England also plays the game then ………... is the probability of an event that India win the world cup.
- ✓$\frac{1}{4}$
- B4
- C8
- D$\frac{1}{8}$
Answer
View full question & answer→Correct option: A.
$\frac{1}{4}$
$\frac{1}{4}$
MCQ 481 Mark
$\mathrm{P}\left(\mathrm{A}^{\prime}\right)=0.57$ then $\mathrm{P}(\mathrm{A})$……………..
- A$\frac{27}{100}$
- B0.013
- ✓0.43
- D$0$
Answer
View full question & answer→Correct option: C.
0.43
0.43
MCQ 491 Mark
………. is the probability of the event that there are 5 Wednesdays in the month of April.
- A$\frac{1}{7}$
- ✓$\frac{2}{7}$
- C$\frac{5}{7}$
- D$\frac{5}{30}$
Answer
View full question & answer→Correct option: B.
$\frac{2}{7}$
$\frac{2}{7}$
MCQ 501 Mark
………… is the probability of an event that on first day of the month has Sunday.
- A$\frac{1}{14}$
- ✓$\frac{1}{7}$
- C$\frac{1}{28}$
- D$\frac{1}{30}$
Answer
View full question & answer→Correct option: B.
$\frac{1}{7}$
$\frac{1}{7}$