MCQ 1011 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
$\frac{12}{25}$
- ✓
$\frac{23}{50}$
AnswerCorrect 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 $d.$
View full question & answer→MCQ 1021 Mark
A ticket is drawn from a bag containing $100$ tickets numbered from $1$ to $100.$ The probability of getting a ticket with a number divisible by $10$ is :
- A
$\frac{1}{5}$
- B
$\frac{4}{10}$
- C
$\frac{3}{10}$
- ✓
$\frac{1}{10}$
AnswerCorrect option: D. $\frac{1}{10}$
Number of possible outcomes $= \{10, 20, 30, 40, 50, 60, 70, 80, 90, 100\} = 10$
Number of Total outcomes $= 100$
$\therefore$ Required Probability $=\frac{10}{100}=\frac{1}{10}$
View full question & answer→MCQ 1031 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
AnswerWe know that, if $E$ is an event, then $P(E) + P(E') = 1.$
Let $E$ be the event where the game is won.
So $, 0.4 + P(E') = 1$
$\Rightarrow P(E') = 1 - 0.4$
$\Rightarrow P(E') = 0.6$
So, the probability of losing the game is $0.6.$
View full question & answer→MCQ 1041 Mark
Choose the correct answer from the given four options : Which of the following cannot be the probability of an event?
- A
$\frac{1}{3}$
- ✓
$0.1$
- C
$3\%$
- D
$\frac{17}{16}$
AnswerSince, probaility of an event always lies between $0$ and $1.$
View full question & answer→MCQ 1051 Mark
Choose the correct answer from the given four options : If the probability of an event is $p,$ the probability of its complementary event will be :
- A
$p - 1$
- B
$p$
- ✓
$1 - p$
- D
$1-\frac{1}{\text{p}}$
AnswerCorrect option: C. $1 - p$
Since, probability of an event $+$ probability of its complementry event $= 1$
So, probaility of its complementry event $= 1 -$ Probability of an event $= 1 - p$
View full question & answer→MCQ 1061 Mark
Choose the correct answer from the given four options : If an event cannot occur, then its probability is :
- A
$1$
- B
$\frac{3}{4}$
- C
$\frac{1}{2}$
- ✓
$0$
AnswerThe event which cannot occur is said to be impossible event and probaillity of impossible event is zero.
View full question & answer→MCQ 1071 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
AnswerCorrect 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 $b.$
View full question & answer→MCQ 1081 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
$\frac{3}{10}$
AnswerCorrect 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 $5)$
$=\frac{4}{20}$
$=\frac{1}{5}$
View full question & answer→MCQ 1091 Mark
A bag contains $5$ red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, then the number of blue balls is :
AnswerLet the number of blue balls be $x.$
$\therefore$ Number of total outcomes $= 5 + x$
Now, $P ($getting the red ball$) =\frac{5}{5+\text{x}}$
$\therefore P ($getting the blue ball$) =2\Big(\frac{5}{5+\text{x}}\Big)$
Also $P ($getting the blue ball$) =\frac{\text{x}}{\text{x+5}}$
$\therefore 2 \Big(\frac{5}{\text{x+5}}\Big) =\frac{\text{x}}{\text{x+5}}$
$\Rightarrow\text{x }=10$
View full question & answer→MCQ 1101 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
$\frac{2}{3}$
AnswerCorrect option: B. $\frac{5}{9}$
Total number of digits from $1$ to $9 (n) = 9$
Numbers which are odd $(m) = \{1, 3, 5, 7, 9\} = 5$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{5}{9}$
View full question & answer→MCQ 1111 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
$\frac{8}{89}$
AnswerCorrect 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.
View full question & answer→MCQ 1121 Mark
A letter of English alphabets is chosen at random. The probability that the letter chosen is a consonant is :
- A
$\frac{5}{26}$
- B
$\frac{2}{26}$
- C
$\frac{1}{26}$
- ✓
$\frac{21}{26}$
AnswerCorrect option: D. $\frac{21}{26}$
We have,
Number of vowels $= 5\ \text{( a, e, i, o, u)}$
Number of consonants $= 21 (26 - 5 = 21)$
Number of possible outcomes $= 21$
Number of total outcomes $= 26$
$\therefore$ Required Probability $=\frac{21}{26}$
View full question & answer→MCQ 1131 Mark
If the probability of occurrence of an event is $p$ then the probability of non $-$ happening of this event is :
AnswerCorrect 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 $, P(E) + p(E') = 1$
$\Rightarrow p + P(E') = 1$
$\Rightarrow P(E') = 1 - p$
View full question & answer→MCQ 1141 Mark
A die is thrown once. The probability of getting an even number and a multiple of $3$ is :
- A
$\frac{1}{3}$
- B
$\frac{1}{2}$
- C
$\frac{1}{5}$
- ✓
$\frac{1}{6}$
AnswerCorrect option: D. $\frac{1}{6}$
Even outcomes $= 2, 4, 6 = 3$
Multiple of $3$ also $= 6 = 1$ only
Number of possible outcomes $= {6} = 1$
Number of Total outcomes $= 6$
$\therefore$ Required Probability $=\frac{1}{6}$
View full question & answer→MCQ 1151 Mark
A card is drawn from a pack of $52$ cards at random. The probability of getting neither an ace nor a king card is :
- A
$\frac{4}{13}$
- ✓
$\frac{11}{12}$
- C
$\frac{2}{13}$
- D
$\frac{8}{13}$
AnswerCorrect option: B. $\frac{11}{12}$
Number of Total outcomes $= 52$
Number of aces and Number of kings $= 4 + 4 = 8$
Number of cards except ace and king $= 52 - 8 = 44$
Required Probability $=\frac{44}{52}=\frac{11}{12}$
View full question & answer→MCQ 1161 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
$\frac{5}{6}$
AnswerCorrect option: A. $\frac{1}{2}$
Even number on a die are $2, 4, 6 = 3$
$\therefore\ \text{Probability (P)}=\frac{3}{6}=\frac{1}{2}$
View full question & answer→MCQ 1171 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
$\frac{3}{14}$
AnswerCorrect option: B. $\frac{3}{26}$
The total number of cards $ = 52$
The number of black face cards $= 6$
$P($getting a black face card$)$
$=\frac{6}{52}$
$=\frac{3}{26}$
View full question & answer→MCQ 1181 Mark
Which of the following cannot be the probability of an event?
- A
$\frac{1}{3}$
- B
$0.3$
- C
$33\%$
- ✓
$\frac{7}{6}$
AnswerCorrect 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.
View full question & answer→MCQ 1191 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 $x =$
AnswerGiven : 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 $c.$
View full question & answer→MCQ 1201 Mark
A number is selected from numbers 1 to 25. The probability that it is prime is:
- A
$\frac{2}{3}$
- B
$\frac{1}{6}$
- C
$\frac{1}{3}$
- ✓
$\frac{9}{25}$
AnswerCorrect option: D. $\frac{9}{25}$
$\frac{9}{25}$
Solution:
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
$\therefore\text{P(E)}=\frac{\text{m}}{\text{n}}=\frac{9}{25}$
$\Rightarrow\ \frac{9}{25}$
View full question & answer→MCQ 1211 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
$\text{None of these}$
AnswerCorrect option: A. $\frac{1}{13}$
The total number of cards $= 52$
The number of queens $= 4$
$P($getting a queen$)$
$=\frac{4}{52}$
$=\frac{1}{13}$
View full question & answer→MCQ 1221 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
$\frac{13}{25}$
AnswerCorrect 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 $5)$
$=\frac{12}{25}$
View full question & answer→MCQ 1231 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
$\frac{2}{3}$
AnswerCorrect option: A. $\frac{4}{9}$
Total number of digits from $1$ to $9 (n) = 9$
Numbers which are even $(m) = 2, 4, 6, 8 = 4$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{4}{9}$
View full question & answer→MCQ 1241 Mark
From a well shuffled pack of $52$ cards, one card is drawn at random. The probability of getting a jack of hearts is :
- ✓
$\frac{1}{52}$
- B
$\frac{4}{52}$
- C
$\frac{6}{52}$
- D
$\frac{2}{52}$
AnswerCorrect option: A. $\frac{1}{52}$
Number of jacks of Heart in a pack of $52$ cards $= 1$
Number of possible outcomes $= 1$
Number of Total outcomes $= 52$
$\therefore$ Required Probability $=\frac{1}{52}$
View full question & answer→MCQ 1251 Mark
Choose the correct answer from the given four options : The probability that a non leap year selected at random will contain $53$ sundays is :
- ✓
$\frac{1}{7}$
- B
$\frac{2}{7}$
- C
$\frac{3}{7}$
- D
$\frac{5}{7}$
AnswerCorrect option: A. $\frac{1}{7}$
A non $-$ leap year has $365$ days and therefore $52$ weeks and $1$ day.
The $1$ day may be Sunday or Manday or Tuesday or Wednesday or Thursday ot Friday or Saturday.
Thus, out of $7$ possibilities $,1$ favourable enent is the event that the one day is Sunday.
$\therefore\ \text{Required probability}=\frac{1}{7}$
View full question & answer→MCQ 1261 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
$\frac{2}{9}$
AnswerCorrect option: B. $\frac{7}{36}$
A pair of dice is thrown simultaneously
$\therefore$ No. of total events $(n) = 6 \times 6 = 36$
Which are
$\{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\}$
$\{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)\}$
$\{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}$
$\{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$
$\{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$
$\{(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
$\therefore$ Event whose sum is a perfect square are
$\{(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)\}$
$\therefore m = 7$
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{7}{36}$
View full question & answer→MCQ 1271 Mark
Choose the correct answer from the given four options : Someone is asked to take a number from $1$ to $100$. The probability that it is a prime is :
- A
$\frac{1}{5}$
- B
$\frac{6}{25}$
- ✓
$\frac{1}{4}$
- D
$\frac{13}{50}$
AnswerCorrect option: C. $\frac{1}{4}$
Total Number of outcomes $= 100$
So, the prime number between $1$ to $100$ are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 56, 61, 67, 71, 73, 79, 83, 89,\}$ and $97.$
$\therefore$ Total number of possible outcomes $= 25$
$\therefore\ \ \text{Reqired probability}=\frac{25}{100}=\frac{1}{4}$
View full question & answer→MCQ 1281 Mark
Choose the correct answer from the given four options : If $P(A)$ denotes the probability of an event $A,$ then:
- A
$\text{P(A)}<0$
- B
$\text{P(A)}>1$
- ✓
$0\leq\text{P(A)}\leq1$
- D
$-1\leq\text{P(A)}\leq1$
AnswerCorrect option: C. $0\leq\text{P(A)}\leq1$
Since, probability of an event always lies between $0$ and $1.$
View full question & answer→MCQ 1291 Mark
Choose the correct answer from the given four options : A card is selected from a deck of $52$ cards. The probability of its being a red face card is
- ✓
$\frac{3}{26}$
- B
$\frac{3}{13}$
- C
$\frac{2}{13}$
- D
$\frac{1}{2}$
AnswerCorrect option: A. $\frac{3}{26}$
In a deck of $52$ cards, there are $12$ face cards
i.e., $6$ red and $6$ black cards.
So, probability of getting a red face card $=\frac{6}{52}=\frac{3}{26}$
View full question & answer→MCQ 1301 Mark
Choose the correct answer from the given four options : The probability of getting a bad egg in a lot of $400$ is $0.035.$ The number of bad eggs in the lot is :
AnswerHere, total number of eggs $= 400$
Probability of getting a bad egg $= 0.035$
$\Rightarrow\ \ \frac{\text{Number of bad eggs}}{\text{Total number of eggs}}=0.035$
$\Rightarrow\ \ \frac{\text{Number of bad eggs}}{400}=0.035$
$\therefore\ \ \text{Number of bad eggs}=0.035\times400=14$
View full question & answer→MCQ 1311 Mark
The probability of a sure event is :
AnswerThe probability of a sure event is always $1.$
View full question & answer→MCQ 1321 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
$\frac{11}{30}$
AnswerCorrect 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$)$
$=\frac{\text{Number of primes between 1 and 30}}{\text{Total}}$
$=\frac{10}{30}$
$=\frac{1}{3}$
View full question & answer→MCQ 1331 Mark
There are $8$ defective items in a sample of $24$ items. One item is drawn at random. The probability that it is a non $-$ defective item is:
- A
$1$
- ✓
$\frac{2}{4}$
- C
$\frac{1}{2}$
- D
$\frac{1}{3}$
AnswerCorrect option: B. $\frac{2}{4}$
Number of Non $-$ defective items $= 24 - 8 = 16$
Number of possible outcomes $= 16$
Number of Total outcomes $= 24$
Probability of getting a Non $-$ defective item $=\frac{16}{24}=\frac{2}{4}$
View full question & answer→MCQ 1341 Mark
Raju bought a fish from a shop for his aquarium. The shop keeper takes out one fish from a tank containing $15$ male fish and $18$ female fish. The probability that the fish taken out is a male fish is :
- A
$\frac{7}{11}$
- B
$\frac{6}{11}$
- ✓
$\frac{5}{11}$
- D
$\frac{5}{12}$
AnswerCorrect option: C. $\frac{5}{11}$
Total number of fish $= 15 + 18 = 33$
Male fish $= 15$
Number of possible outcomes $= 15$
Number of total outcomes $= 15 + 18 = 33$
Required Probability $=\frac{15}{33}=\frac{5}{11}$
View full question & answer→MCQ 1351 Mark
A bag contains $5$ black balls $,4$ white balls and $3$ red balls. If a ball is selected at random, the probability that it is not red is :
- A
$\frac{1}{2}$
- B
$\frac{5}{12}$
- C
$\frac{3}{12}$
- ✓
$\frac{3}{4}$
AnswerCorrect option: D. $\frac{3}{4}$
Numbers of balls which are not red $= 12 - 3 = 9$
Number of possible outcomes $= 9$
Number of Total outcomes $= 12$
Required Probability$=\frac{9}{12}=\frac{3}{4}$
View full question & answer→MCQ 1361 Mark
The probability that a student will score centum in Mathematics is $\frac{3}{10}$. The probability that the student will not score centum is :
- ✓
$\frac{7}{10}$
- B
$\frac{6}{10}$
- C
$\frac{1}{2}$
- D
$\frac{1}{3}$
AnswerCorrect option: A. $\frac{7}{10}$
given : $\text{P(E)}=\frac{3}{10}$
$\text{p}\text{(notE)=1-p (E)}$
$=1-\frac{3}{10}\frac{7}{10}$
View full question & answer→MCQ 1371 Mark
A piggy bank contains $100$ fifty paise coins, $50$ one rupee coins, $20$ two rupee coins and $10$ five rupee coins. One coin is drawn at random. The probability that the coin drawn will not be a five rupee coin is :
- A
$\frac{5}{9}$
- ✓
$\frac{17}{18}$
- C
$\frac{7}{18}$
- D
$\frac{8}{9}$
AnswerCorrect option: B. $\frac{17}{18}$
Number of total coins $= 100 + 50 + 20 + 10 = 180$
Number of coins except five rupee coins $= 180 - 10 = 170$
$\therefore$ Required Probability $=\frac{170}{180}=\frac{17}{18}$
View full question & answer→MCQ 1381 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
$\frac{1}{4}$
AnswerCorrect option: C. $\frac{1}{2}$
Three coins are tossed simultaneously, then possible events will be $(n) = 2 × 2 × 2 = 8$
The results will be
$\text{\{(HHT), (HTH), (THH), (THT), (TTH), (HTT), (HHH), (TTT)\}}$
$\therefore$ Probability of getting at least two heads are
$=\frac{\text{m}}{\text{n}}=\frac{4}{8}=\frac{1}{2}$
View full question & answer→MCQ 1391 Mark
One number is chosen randomly from the integers $1$ to $50$. The probability that it is divisible by $4$ or $6$ is :
- A
$\frac{6}{25}$
- B
$\frac{2}{25}$
- ✓
$\frac{8}{25}$
- D
$\frac{4}{25}$
AnswerCorrect option: C. $\frac{8}{25}$
numbers divisible by $4 = \{4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48\}$
Numbers divisible by $6 = \{6, 12, 18, 24, 30, 36, 42, 48\}$
Number of possible outcomes $= \{4, 6, 8, 12, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 44, 48\} = 16$
Number of Total outcomes $= 50$
$\therefore$ Required Probability $=\frac{16}{50}=\frac{8}{25}$
View full question & answer→MCQ 1401 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
$\frac{1}{3}$
AnswerCorrect 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$)$
$=\frac{13}{15}$
View full question & answer→MCQ 1411 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
$\frac{5}{9}$
AnswerCorrect 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 $P(A)$
$=\frac{\text{Number of outcomes in favour of A }}{\text{Total number of outcomes i}}$
$=\frac{5}{10}$
$=\frac{1}{2}$
View full question & answer→MCQ 1421 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
$\text{None of these}$
AnswerCorrect option: C. $\frac{1}{13}$
The total number of cards $= 52$
The number of $6$ in the deck of cards $= 4$
$P($getting a $6)$
$=\frac{4}{52}$
$=\frac{1}{13}$
View full question & answer→MCQ 1431 Mark
The probability of an impossible event is :
AnswerAn event which has no chance of occurrence is called an impossible event.
for example : The probability of getting more than $6$ when a die is thrown is an impossible event because the highest number in a die is $6$
The probability of an impossible event is always zero.
View full question & answer→MCQ 1441 Mark
A number is selected at random from the numbers $\{7, 3, 9, 7, 9, 5, 7, 9, 9, 5\}$. The probability that the selected number is their average is :
- ✓
$\frac{3}{10}$
- B
$\frac{7}{10}$
- C
$\frac{5}{10}$
- D
$\frac{1}{10}$
AnswerCorrect option: A. $\frac{3}{10}$
Average of given numbers $=\frac{7+3+9+7+9+5+7+9+9+5}{10}=\frac{70}{10}=7$
$($There are $3$ times $7$ in the given numbers$)$
Therefore Number of outcomes $= 3$
Number of total outcomes $= 10$
$\therefore$ Required Probability $=\frac{3}{10}$
View full question & answer→MCQ 1451 Mark
Two dice are thrown simultaneously. The probability of getting a doublet is :
- A
$\frac{2}{3}$
- B
$\frac{1}{3}$
- C
$\frac{1}{4}$
- ✓
$\frac{1}{6}$
AnswerCorrect option: D. $\frac{1}{6}$
Doublet means getting same number on both dice simultaneously
Doublets $= \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}$
Number of possible outcomes $= 6$
Total number of ways to throw a dice $= 36$
Probability of getting a doublet $=\frac{6}{36}=\frac{1}{6}$
View full question & answer→MCQ 1461 Mark
A card is drawn from a pack of $52$ cards at random. The probability of getting either an ace or a king card is :
- ✓
$\frac{2}{13}$
- B
$\frac{4}{13}$
- C
$\frac{8}{13}$
- D
$\frac{3}{13}$
AnswerCorrect option: A. $\frac{2}{13}$
Number of Total outcomes $= 52$
Number of aces and Number of kings $= 4 + 4 = 8$
Required Probability $=\frac{8}{52}=\frac{2}{13}$
View full question & answer→MCQ 1471 Mark
The probability that a non leap year will have $53 $ Fridays and $53$ Saturdays is :
- A
$\frac{3}{7}$
- ✓
$0$
- C
$\frac{2}{7}$
- D
$\frac{1}{7}$
AnswerNon $-$ leap year contains $366$ days $= 52$ weeks $+\ 1$ day
$52$ weeks contain $52$ Fridays and $52$ weeks contain $52$ Saturdays
We will get $53$ Fridays or $53$ Saturdays if remaining one day is a Friday and Saturday
Therefore, Total possibility is zero.
Number of Total possible outcomes $= 7$
Number of possible outcomes Friday and Saturday $= 0$
Required Probability $=\frac{0}{7}=0$
View full question & answer→MCQ 1481 Mark
A box contains $3$ blue balls, $2$ white balls and $4$ red balls. If a ball is drawn at random from the box, the probability of getting a white ball is :
- A
$1$
- ✓
$ \frac{2}{9}$
- C
$ \frac{4}{9}$
- D
$ \frac{3}{9}$
AnswerCorrect option: B. $ \frac{2}{9}$
Number of outcomes $= 2$
Number of total outcomes $= 3 + 2 + 4 = 9$
$\therefore$ Required Probability$=\frac{2}{9}$
View full question & answer→MCQ 1491 Mark
The probability of a certain event is :
AnswerGiven : $4$ 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 $b.$
View full question & answer→MCQ 1501 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
$\text{None of these}$
AnswerCorrect 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$)$
$=\frac{8}{24}$
$=\frac{1}{3}$
View full question & answer→