MCQ 511 Mark
Which of the following cennot be the probability of an event?
- ✓
$1.5$
- B
$\frac{3}{5}$
- C
$25\%$
- D
$0.3$
AnswerWe 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.
View full question & answer→MCQ 521 Mark
The probability that a non $-$ leap year has $53$ Sundays, is :
- A
$\frac{2}{7}$
- B
$\frac{5}{7}$
- C
$\frac{6}{7}$
- ✓
$\frac{1}{7}$
AnswerCorrect 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
$=\frac{\text{m}}{\text{n}}=\frac{1}{\text{No. of day in a week}}=\frac{1}{7}$
View full question & answer→MCQ 531 Mark
A bag contains $50$ balls of which $2x$ are red, $3x$ are white and $5x$ are blue. A ball is selected at random. The probability that it is not white is :
- A
$\frac{3}{5}$
- ✓
$\frac{7}{10}$
- C
$\frac{2}{5}$
- D
$\frac{7}{45}$
AnswerCorrect option: B. $\frac{7}{10}$
Here,
$2x + 3x + 5x = 50$
$\Rightarrow 10x = 50$
$\Rightarrow x = 5$
Number of red balls $= 2 \times 5 = 10$
Number of white balls $= 3 \times 5 = 15$
Number of blue balls $= 5 \times 5 = 25$
Now, Number of possible outcomes $= 25 + 10 = 35$
And Number of total outcomes $= 50$
$\therefore$ Required Probability $=\frac{35}{50}=\frac{7}{10}$
View full question & answer→MCQ 541 Mark
If an event cannot occur then its probability is :
- A
$1$
- B
$\frac{1}{2}$
- C
$\frac{3}{4}$
- ✓
$0$
AnswerAn event that cannot ocour is called an impossibel event.
the probability of an impossible event is $0.$
View full question & answer→MCQ 551 Mark
Which of the following cannot be the probability of an event?
- A
$\frac{2}{3}$
- ✓
$-1.5$
- C
$15\%$
- D
$0.7$
AnswerCorrect 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 $b.$
View full question & answer→MCQ 561 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
$\frac{1}{6}$
AnswerCorrect 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$)$
$=\frac{3}{6}$
$=\frac{1}{2}$
View full question & answer→MCQ 571 Mark
The probability of getting $2$ heads, when two coins are tossed, is :
- A
$1$
- B
$\frac{3}{4}$
- C
$\frac{2}{2}$
- ✓
$\frac{1}{4}$
AnswerCorrect option: D. $\frac{1}{4}$
When two coins are tossed the outcomes are :
$\text{\{HH, HT, TH, TT}\}$
So, there are $4$ numbers in total
$P($getting one head$)$
$=\frac{1}{4}$
View full question & answer→MCQ 581 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
$\frac{7}{40}$
AnswerCorrect 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 $7)$
$=\frac{5}{40}$
$=\frac{1}{8}$
View full question & answer→MCQ 591 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
$\frac{2}{9}$
AnswerCorrect 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.
View full question & answer→MCQ 601 Mark
A child’s game has $8$ triangles of which $5$ are blue and rest are red and $10$ squares of which $6$ are blue and the rest are red. One piece is lost at random. The probability that it is a square of blue colour is :
- ✓
$\frac{1}{3}$
- B
$\frac{1}{3}$
- C
$\frac{1}{3}$
- D
$\frac{1}{3}$
AnswerCorrect option: A. $\frac{1}{3}$
Total number of pieces $= 8$ triangles $+\ 10$ squares $= 18$
Number of blue squares $= 6$
Number of possible outcomes $= 6$
Number of total outcomes $= 8 + 10 = 18$
$\therefore$ Required Probability $=\frac{6}{18}=\frac{1}{3}$
View full question & answer→MCQ 611 Mark
A bag contains $6$ red $, 8$ white $, 4$ green and $7$ black balls. One ball is drawn at random. The probability that the ball is drawn is neither green nor white is :
- ✓
$\frac{13}{25}$
- B
$\frac{13}{25}$
- C
$\frac{13}{25}$
- D
$\frac{13}{25}$
AnswerCorrect option: A. $\frac{13}{25}$
Total number of balls $= 25$
Number of Green and White balls $= 4 + 8 = 12$
Number of balls neither green nor white $= 25 - 12 = 13$
Number of possible outcomes $= 13$
Number of total outcomes $= 25$
$\therefore$ Required Probability $=\frac{13}{25}$
View full question & answer→MCQ 621 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
$\text{None of these}$
AnswerCorrect 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$)$
$=\frac{24}{30}$
$=\frac{4}{5}$
View full question & answer→MCQ 631 Mark
A box contains $10$ white $,6$ red, and $10$ black balls. A ball is drawn at random. The probability that the ball drawn is white or red is :
- A
$\frac{6}{13}$
- B
$\frac{10}{13}$
- ✓
$\frac{8}{13}$
- D
$\frac{4}{13}$
AnswerCorrect option: C. $\frac{8}{13}$
Number of possible outcomes $($No. of white and red balls$) = 10 + 6 =16$
Number of Total outcomes $= 10 + 6 + 10 = 26$
Required Probability $=\frac{16}{26}=\frac{8}{13}$
View full question & answer→MCQ 641 Mark
Ram and Shyam are friends. The probability that both will have the birthday on the same day is :
- A
$\frac{2}{365}$
- B
$\frac{4}{365}$
- ✓
$\frac{1}{365}$
- D
$\frac{364}{365}$
AnswerCorrect option: C. $\frac{1}{365}$
Assuming a non $-$ leap year
Ram can have the birthday on any day of the $365$ days of the year
Shyam has a different birthday if his birthday is on any of the remaining $364$ days of the year
Therefore $P ($Ram and Shyam have different birthdays$) =\frac{364}{365}$
and so, $P ($Ram and Shyam have birthdays on the same day$) = 1 - P($Ram and Shyam have different birthdays$)$
$=1-\frac{364}{365}$
$=\frac{1}{365}$
View full question & answer→MCQ 651 Mark
Choose the correct answer from the given four options : 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
$\frac{17}{23}$
AnswerCorrect option: B. $\frac{6}{23}$
Total Number of students $= 23$
Number of students in house $A, B$ and $C = 4 + 8 + 5 = 17$
$\therefore$ Remains students $= 23 - 17 = 6$
So, probability that the selected student is not from $A, B$ and $C =\frac{6}{23}$
View full question & answer→MCQ 661 Mark
A die is thrown twice. The probability that $5$ will come up at least once is :
- ✓
$\frac{11}{36}$
- B
$\frac{25}{36}$
- C
$0$
- D
$1$
AnswerCorrect option: A. $\frac{11}{36}$
Elementary events are
$\{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\}$
$\{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)\}$
$\{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}$
$\{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$
$\{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$
$\{(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
$\therefore$ Number of Total outcomes $= 36$
And Number of possible outcomes $= 11$
$\therefore$ Required Probability $=\frac{11}{36}$
View full question & answer→MCQ 671 Mark
Choose the correct answer from the given four options : 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
$\frac{1}{3}$
AnswerCorrect option: A. $\frac{1}{5}$
Number of total outcomes $= 40$
Multiples of $5$ between $1$ to $40 = \{5, 10, 15, 20, 25, 30, 35, 40\}$
$\therefore$ Total number of possible outcomes $= 8$
$\therefore\ \ \text{Reqired probability}=\frac{8}{40}=\frac{1}{5}$
View full question & answer→MCQ 681 Mark
When a die is thrown, the probability of getting an odd number less than $3$ is :
- A
$\frac{1}{4}$
- B
$\frac{1}{3}$
- ✓
$\frac{1}{6}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $\frac{1}{6}$
Odd outcomes are $1, 3, 5$ but $3$ and $5$ are not less than $3$
Number of possible outcomes which are odd and less than $3 = 1$
Number of possible outcomes $= 1$
Number of Total outcomes $= 6$
$\therefore$ Required Probability$=\frac{1}{6}$
View full question & answer→MCQ 691 Mark
The probability of selecting a queen of diamonds when a card is drawn from well shuffled pack of $52 $ cards is :
- A
$\frac{1}{13}$
- B
$\frac{16}{52}$
- C
$\frac{1}{26}$
- ✓
$\frac{1}{52}$
AnswerCorrect option: D. $\frac{1}{52}$
Number of possible outcomes $($queen of diamonds$) = 1$
Number of Total outcomes $= 52$
Required Probability $=\frac{1}{52}$
View full question & answer→MCQ 701 Mark
A letter is chosen at random from the word $\text{ACCOMMODATION}$. The probability that it is $A$ or $O$ is :
- ✓
$\frac{5}{13}$
- B
$\frac{8}{13}$
- C
$\frac{6}{13}$
- D
$\frac{10}{13}$
AnswerCorrect option: A. $\frac{5}{13}$
Number of possible outcomes $= \text{A, O, O, A O} = 5$
Number of Total outcomes $= 13$
Required Probability $=\frac{5}{13}$
View full question & answer→MCQ 711 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
$\frac{1}{6}$
- ✓
$\frac{1}{2}$
AnswerCorrect option: D. $\frac{1}{2}$
No. of total letters in the word $\text{MOBILE} = 6$
No, of vowels $\text{= o, i, e } = 3$
$\therefore$ Probability of being a vowel $=\frac{3}{6}=\frac{1}{2}$
View full question & answer→MCQ 721 Mark
A number is selected at random from $1$ to $75$. The probability that it is a perfect square is :
- ✓
$\frac{8}{75}$
- B
$\frac{6}{75}$
- C
$\frac{4}{75}$
- D
$\frac{10}{75}$
AnswerCorrect option: A. $\frac{8}{75}$
Number of possible outcomes $= \{1, 4, 9, 16, 25, 36, 49, 64\} = 8$
Number of Total outcomes $= 75$
$\therefore$ Probability (of getting a perfect square) $=\frac{8}{75}$
View full question & answer→MCQ 731 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
$\frac{7}{25}$
AnswerCorrect option: C. $\frac{1}{3}$
Total number of two digit numbers are $10$ to $99$
$= 99 - 9 = 90$
Multiples of $3$ will be $\{12, 15, 18, 21,…. 99\}$
$= 33 - 3 = 30$
$\therefore\ \text{Probability}=\frac{30}{90}=\frac{1}{3}$
View full question & answer→MCQ 741 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
$\frac{1}{3}$
AnswerCorrect option: C. $\frac{1}{5}$
The selected numbers would be $4, 8,$ and $12.$
So, there are $3$ number.
$P($number of multiples of $4)$
$=\frac{\text{Number of multipes of 4}}{\text{Total}}$
$=\frac{3}{15}$
$=\frac{1}{5}$
View full question & answer→MCQ 751 Mark
Cards marked with numbers $\{1, 2, 3, ..., 25\}$ are placed in a box and mixed thoroughly and one card is drawn at random from the box. The probability that the number on the card is a multiple of $3$ or $5$ is :
- A
$\frac{4}{25}$
- ✓
$\frac{12}{25}$
- C
$\frac{1}{5}$
- D
$\frac{8}{25}$
AnswerCorrect option: B. $\frac{12}{25}$
Number of multiples of $3 = 8 \text{\{( 3 ,6, 9 ,12, 15, 18 ,21 ,24)\}}$
Number of multiples of $5 = 5 \{\text{(5 ,10, 15, 20, 25)}\}$
Number of possible outcomes $($multiples of $3$ or $5) = 12 \{(3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25)\}$
Number of Total outcomes $= 25$
$\therefore$ Required Probability $=\frac{12}{25}$
View full question & answer→MCQ 761 Mark
The king, queen and jack of clubs are removed from a deck of $52$ cards and the remaining cards are shuffled. $A$ card is drawn from the remaining cards. The probability of getting a king is :
- ✓
$\frac{3}{49}$
- B
$\frac{3}{52}$
- C
$\frac{4}{49}$
- D
$\frac{4}{52}$
AnswerCorrect option: A. $\frac{3}{49}$
$K, Q, J$ of clubs i.e $3$ cards are removed,
therefore remaining cards $= 52 - 3 = 49$
$3$ kings are left in the pack
Number of possible outcomes $= 3$
Number of total outcomes $= 52 - 3 = 49$
$\therefore$ Required Probability $=\frac{3}{49}$
View full question & answer→MCQ 771 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
$\frac{1}{7}$
AnswerCorrect 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
$x = -1, 0, 1, 2, 3$
$\therefore x = -1, 0, 1$
$\therefore m = 3$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{3}{7}$
View full question & answer→MCQ 781 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
$\frac{8}{45}$
AnswerCorrect 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 $= \{(2, 2), (3, 3)\},$
$\{(4, 2), (4, 4), (5, 5)\},$
$\{(6, 2), (6, 3), (6, 6),(7, 7), (8, 2), (8, 4), (8, 8),\}$
$\{(9, 3), (9, 9), (10, 2), (10, 5), (10, 10)\}, = 17$
$\text{P(E)}=\frac{\text{m}}{\text{n}}=\frac{17}{90}$
View full question & answer→MCQ 791 Mark
If $\phi$ is an impossible event, then $\text{p}(\phi)=$
- A
$\frac{1}{2}$
- ✓
$0$
- C
$\frac{1}{4}$
- D
$1$
AnswerAn event which has no chance of occurring is called an impossible event.
If $\phi$ is an impossible event, then $\text{p}(\phi)=0$
View full question & answer→MCQ 801 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
$\frac{5}{6}$
AnswerCorrect 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$)$
$=\frac{3}{6}$
$=\frac{1}{2}$
View full question & answer→MCQ 811 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
$\text{None of these}$
AnswerCorrect 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$)$
$=\frac{6}{10}$
$=\frac{3}{5}$
View full question & answer→MCQ 821 Mark
A lot consists of $40$ mobile phones of which $32$ are good, $3$ have only minor defects and $5$ have major defects. Ram will buy a phone if it is good or have minor defects. One phone is selected at random. The probability that it is acceptable to Ram is :
- A
$\frac{3}{5}$
- B
$\frac{4}{5}$
- C
$\frac{3}{40}$
- ✓
$\frac{7}{8}$
AnswerCorrect option: D. $\frac{7}{8}$
Number of phones which are good and minor defects $= 32 + 3 = 35$
Number of possible outcomes $= 35$
Number of Total outcomes $= 40$
$\therefore$ Required Probability $=\frac{35}{40}=\frac{7}{8}$
View full question & answer→MCQ 831 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
AnswerCorrect option: B. $\frac{1}{6}$
No. of months in a year $= 12$
Probability of being March or October $=\frac{2}{12}$
$=\frac{1}{6}$
View full question & answer→MCQ 841 Mark
There are $20$ boys and $15$ girls in a class. A student is chosen as leader at random. The probability that the leader is a boy is :
- A
$\frac{3}{7}$
- ✓
$\frac{4}{7}$
- C
$20$
- D
$15$
AnswerCorrect option: B. $\frac{4}{7}$
Number of boys $= 20$
Number of possible outcomes $= 20$
Number of Total outcomes $= 20 + 15 = 35$
$\therefore$ Required Probability $=\frac{20}{35}=\frac{4}{7}$
View full question & answer→MCQ 851 Mark
What is the probability of an impossible event?
- A
$\frac{1}{2}$
- ✓
$0$
- C
$1$
- D
More than $1$
AnswerThe probability of an impossible event is always $ 0.$
View full question & answer→MCQ 861 Mark
From a well shuffled pack of $52$ cards, one card is drawn at random. The probability of getting a black king is :
- A
$\frac{2}{39}$
- ✓
$\frac{1}{26}$
- C
$\frac{1}{13}$
- D
AnswerCorrect option: B. $\frac{1}{26}$
black kings $=$ club king $+$ spade king $= 2$
Number of possible outcomes $ = 2$
Number of Total outcomes $= 52$
$\therefore$ Required Probability $=\frac{2}{52}=\frac{1}{26}$
View full question & answer→MCQ 871 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
$\frac{3}{4}$
AnswerCorrect 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$)$
$=\frac{4}{12}$
$=\frac{1}{3}$
View full question & answer→MCQ 881 Mark
Choose the correct answer from the given four options : The probability expressed as a percentage of a particular occurrence can never be :
- A
Less than $100.$
- ✓
Less than $0.$
- C
Greater than $1.$
- D
Anything but a whole number.
AnswerCorrect option: B. Less than $0.$
We know that, the probability expressed as a percentage always lie between $0$ and $100.$
So, it cannot be less than $0.$
View full question & answer→MCQ 891 Mark
Choose the correct answer from the given four options : 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:
AnswerIn a deck of $52$ cards, there are $13$ cards of heart and $1$ is ace of heart.
Hence, the number of outcomes favourable to $E = 51$
View full question & answer→MCQ 901 Mark
From a well shuffled pack of $52$ cards, one card is drawn at random. The probability of getting a red queen is :
- ✓
$\frac{1}{26}$
- B
$\frac{3}{26}$
- C
$\frac{1}{2}$
- D
$\frac{1}{13}$
AnswerCorrect option: A. $\frac{1}{26}$
Red Queens $=$ Diamond Queen $+$ Heart Queen $= 2$
Number of possible outcomes $= 2$
Number of Total outcomes $= 52$
$\therefore$ Required Probability $=\frac{2}{52}=\frac{1}{26}$
View full question & answer→MCQ 911 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
$\frac{9}{10}$
AnswerCorrect 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 $=\frac{7}{10}$ View full question & answer→MCQ 921 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
$\frac{3}{4}$
AnswerCorrect option: C. $\frac{3}{8}$
When three coins are tossed the simultaneously the
outcomes are :
$\text{\{HHH, HHT, HTH, THT, HTT, TTH\}}$ and $\text{\{TTT\}}$
So, there are $8$ possible outcones.
$P($getting exactly two heads$)$
$=\frac{3}{8}$
View full question & answer→MCQ 931 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
$\frac{1}{3}$
AnswerCorrect 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 $c$.
View full question & answer→MCQ 941 Mark
A letter of English alphabets is chosen at random. The probability that the letter chosen is a vowel is :
- A
$\frac{1}{26}$
- B
$\frac{2}{26}$
- C
$\frac{4}{26}$
- ✓
$\frac{5}{26}$
AnswerCorrect option: D. $\frac{5}{26}$
We know that $"\text{A, E, I, O, U}" $ are vowels
Number of vowels $= 5$
Number of possible outcomes $= 5$
Number of total outcomes $= 26$
$\therefore$ Required Probability $=\frac{5}{26}$
View full question & answer→MCQ 951 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
$\frac{2}{3}$
AnswerCorrect 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$)$
$=\frac{6}{36}$
$=\frac{1}{6}$
View full question & answer→MCQ 961 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
$\frac{7}{8}$
AnswerCorrect 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 $=\frac{3}{4}$
View full question & answer→MCQ 971 Mark
The probability of getting $3$ head or $3$ tails in tossing a coin $3$ times is :
- A
$\frac{1}{2}$
- ✓
$\frac{1}{4}$
- C
$\frac{1}{8}$
- D
$\frac{3}{8}$
AnswerCorrect option: B. $\frac{1}{4}$
Total outcomes $= \text{\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}\} = 8$
Number of possible outcomes $ = \text{(HHH)}$ or $\text{(TTT)} = 2$
Required Probability $=\frac{2}{8}=\frac{1}{4}$
View full question & answer→MCQ 981 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
$\frac{17}{20}$
AnswerCorrect option: C. $\frac{1}{20}$
No. of lottery tickets $= 100$
No. of tickets carrying prizes $= 5$
$\therefore$ Probability of ticket buying a prized one
$=\frac{\text{m}}{\text{n}}=\frac{5}{100}=\frac{1}{20}$
View full question & answer→MCQ 991 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
$-1\leq\text{P(E)}\leq1$
AnswerCorrect 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.
View full question & answer→MCQ 1001 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
$\frac{1}{12}$
AnswerCorrect 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$)$
$=\frac{6}{36}$
$=\frac{1}{6}$
View full question & answer→