MCQ 1011 Mark
6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is
- A
$\frac{1}{432}$
- B
$\frac{12}{431}$
- ✓
$\frac{1}{132}$
- D
$\text{none of these}$
AnswerCorrect option: C. $\frac{1}{132}$
Total number of ways in which 6 boys and 6 girls can sit in a row = 12
Consider 6 girls as one group, then 6 boys and one group can arrange in 7 ways.
Now, 6 girls in the group can arrange among themselves in 6.
So, the number of ways in which all the girls sit together is 7 × 6
$\therefore$P(all girls sit together)
$=\frac{\text{Number of ways in which all girls sit together }}{\text{Total Number of ways in which 6 boys and 6 girls sit in a row}}$
$=\frac{7\times6}{12}=\frac{6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8}=\frac{1}{132}$
Hence, the correct answer is option (c).
View full question & answer→MCQ 1021 Mark
An urn contains $9$ balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is:
- ✓
$\frac{5}{84}$
- B
$\frac{3}{9}$
- C
$\frac{3}{7}$
- D
$\frac{7}{17}$
AnswerCorrect option: A. $\frac{5}{84}$
Three balls can be drawn randomly from nine balls in$ \ ^{9}\text{C}_3 = 84$ ways.
Three balls cannot be red as there are only two red balls.
Three balls of the same colour can be drawn in the following ways :
$3$ blue out of a total of $3$ blue balls.
The probability for which is $\frac{\ ^{3}\text{C}_3}{84}=\frac{1}{84}$
$3$ black out of a total of $4$ black balls.
The probability for which is $\frac{\ ^{4}\text{C}_3}{84}=\frac{4}{84}$
Hence, required probability $=\frac{1}{84}+\frac{4}{84}=\frac{5}{84}$
View full question & answer→MCQ 1031 Mark
If $\text{P}\text{(A}\cup\text{B)} = \text{P}\text{(A}\cap\text{B)}$ for any two events A and B, the
- ✓
$\text{P(A)}=\text{P(B)}$
- B
$\text{P(A)}>\text{P(B)}$
- C
$\text{P(A)}<\text{P(B)}$
- D
AnswerCorrect option: A. $\text{P(A)}=\text{P(B)}$
We know that,
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{(B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{P}\text{(B})-\text{P}(\text{A}\cap\text{B})$ $\big[\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cup\text{B})\big]$
$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})=0\ ...(1)$
But,
$\text{P(A)}-\text{P}(\text{A }\cap\text{ B})\geq0$
$\text{P(B)}-\text{P}(\text{A }\cap\text{ B})\geq0$
$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})\geq0\ ...(2)$
From (1) and (2), we have,
$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})=0$
$\Rightarrow\text{P}(\text{A)}=\text{P}(\text{A}\cap\text{B})\text{ and }\text{P}(\text{B)}=\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P(A)}=\text{P(B)}$
Hence, the correct answer is option (a).
View full question & answer→MCQ 1041 Mark
The vertex of the parabola $y^2- 4y - x + 3 = 0$ is:
- A
$(-1, 3)$
- ✓
$(-1, 2)$
- C
$(2, -1)$
- D
$(3, -1)$
AnswerCorrect option: B. $(-1, 2)$
We have,
$= y^2- 4y - x + 3 = 0$
$\Rightarrow (y - 2)^2- 4 - x + 3 = 0$
$\Rightarrow (y - 2)^2= (x + 1)$
$\therefore $ Vertex of the parabola $= (-1, 2)$
View full question & answer→MCQ 1051 Mark
Two dice are thrown simultaneously. Find the probability of getting a multiple of $2$ on first dice and a multiple of $3$ on the second dice.
- A
$\frac{4}{6}$
- B
$\frac{2}{6}$
- ✓
$\frac{1}{6}$
- D
$\frac{1}{36}$
AnswerCorrect option: C. $\frac{1}{6}$
Total cases $= 6 \times 6$
Let A be the event of getting a multiple of $2$ on first dice and a multiple of $3$ on the second dice.
Hence, $A = \{(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6)\} n(A) = 66$
$\therefore \text{P(A)}=\frac{6}{36}=\frac{1}{6}$
View full question & answer→MCQ 1061 Mark
If two coins are tossed then find the probability of the event that no head turns up.
- ✓
$\frac{1}{4}$
- B
$\frac{1}{3}$
- C
$\frac{1}{2}$
- D
$\frac{3}{4}$
AnswerCorrect option: A. $\frac{1}{4}$
$2$ coins tossed sample space $= (h, h), (h, t), (t, h), (t, t)$
have no head $= 1 ($when both tail$)$
probability is $\frac{1}{4}$
View full question & answer→MCQ 1071 Mark
The lines $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circel of area $154$ sq. units. Then the equation of circle is:
- A
$(x + 1)^2+ (y + 1)^2= 49$
- B
$(x - 1)^2+ (y - 1)^2= - 49$
- ✓
$(x - 1)^2+ (y + 1)^2= 49$
- D
$(x + 1)^2 + (y - 1)^2= 49$
AnswerCorrect option: C. $(x - 1)^2+ (y + 1)^2= 49$
Circle area $=\pi\text{r}^2=154$
$\Rightarrow r = 7$ sq.units
intersection of diameter $(2x - 3y = 5) \times 3$
$(3x - 4y = 7) \times 2- y = 1$
$\Rightarrow y = -1$
$12x = 5 - 3 = 2$
$\Rightarrow x = 1$
eqn of circle $(x - 1)^2+ (y + 1)^2= 49$
View full question & answer→MCQ 1081 Mark
A die is rolled, then the probability that a number $1$ or $6$ may appear is
- A
$\frac{2}{3}$
- B
$\frac{5}{6}$
- ✓
$\frac{1}{3}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $\frac{1}{3}$
Total number of sample space, $S = \{1, 2, 3, 4, 5, 6\}$
$\therefore\text{n}\text{(S)} = 6$
Let $A$ be the event of getting the number $1$ or $6.$
$A = \{1, 6\}$
i.e. $n(A) = 2$
Hence, required probability $=\frac{\text{n(A)}}{\text{n(S)}}=\frac{2}{6}=\frac{1}{3}$
View full question & answer→MCQ 1091 Mark
Six boys and six girls sit in a row randomly. The probability that all girls sit together is
- A
$\frac{1}{122}$
- B
$\frac{1}{112}$
- C
$\frac{1}{102}$
- ✓
$\frac{1}{132}$
AnswerCorrect option: D. $\frac{1}{132}$
Total number of ways in which six boys and six girls can be seated in a row $= (12)$
Taking all the six girls as one person, seven persons can be seated in a row in $7$ ways.
The six girls can be arranged among themselves in $6$ ways.
Then number of ways in which six boys and six girls can be seated in a row so that all
the girls sit together $= 7 \times 6$
$\therefore\text{Required Probability}=\frac{7\times6}{12}$
$\frac{720}{12\times11\times10\times9\times8}=\frac{1}{132}$
View full question & answer→MCQ 1101 Mark
If events $A$ and $B$ are independent and $P(A) = 0.15, P(A ∪ B) = 0.45,$ then $P(B)=:$
- A
$136$
- ✓
$176$
- C
$196$
- D
$236$
AnswerCorrect option: B. $176$
Given, $\text{P(A) = 0.15, P(A ∪ B)} = 0.45$
We have $\text{P(A ∪ B) = P(A) + P(B) - P(A ∩ B)}$
and $\text{P(A ∩ B) = P(A). P(B)}$
Therefore, $\text{0.45 = 0.15 + P(B) - 0.15 P(B)}$
$\Rightarrow 0.30 = 0.85 \text{P(B)}$
$\Rightarrow \text{P(B)} = 8530 = 176$
View full question & answer→MCQ 1111 Mark
The ends of the latus rectum of the conic $x^2+ 10x - 16y + 25 = 0$ are:
- A
$(3, -4), (13, 4)$
- B
$(-3, -4), (13, -4)$
- ✓
$(3, 4), (-13, 4)$
- D
$(5, -8), (-5, 8)$
AnswerCorrect option: C. $(3, 4), (-13, 4)$
$x^2+ 10x - 16y + 25 = 0$
$\Rightarrow (x + 5)^2= 16y$
$\Rightarrow X^2= 4A Y$,
where $X = x + 5, A = 4, Y = y$
The ends of the latus rectum are $(2A, A)$ and $(-2A, A)$
$\Rightarrow x + 5 = 2 (4)$
$\Rightarrow x = -8 - 5 = 3, y = 4$ and $x + 5 = -2 (4) x + 5 = -2(4)$
$\Rightarrow x = -8 - 5 = -13, y = 4$
$\Rightarrow (3, 4)$ and $(-13, 4).$
View full question & answer→MCQ 1121 Mark
A card is drawn at random from a pack of $100$ cards numbered $1$ to $100.$ The probability of drawing a number which is a square is:
- A
$\frac{1}{5}$
- B
$\frac{2}{5}$
- ✓
$\frac{1}{10}$
- D
AnswerCorrect option: C. $\frac{1}{10}$
Clearly, the sample space is given by
$S = \{1, 2, 3, 4, 5 ....97, 98, 99, 100\}$
$\therefore n(S) = 100$
Let $E =$ event of getting a square.
Then $E = \{1, 4, 9, 16, 25, 36, 49, 64, 81, 100\}$
$\therefore n(E) = 10$
Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{10}{100}=\frac{1}{10}$
View full question & answer→MCQ 1131 Mark
The point $(3, 4)$ is the focus and $2x - 3y + 5 = 0$ is the directrix of a parabola. Lenghth of latus rectum is:
- ✓
$\frac{2}{\sqrt{13}}$
- B
$\frac{4}{\sqrt{13}}$
- C
$\frac{1}{\sqrt{13}}$
- D
$\frac{3}{\sqrt{13}}$
AnswerCorrect option: A. $\frac{2}{\sqrt{13}}$
We know that, Latus rectum $= 2\times ($distance from focus to directrix$)$
$=2.\frac{6\ -\ 12\ +\ 5}{\sqrt{4\ +\ 9}}$
$=\frac{2}{\sqrt{13}}$
View full question & answer→MCQ 1141 Mark
What is the probability of selecting a vowel in the word $“\text{PROBABILITY}”\ ?$
- A
$\frac{2}{11}$
- ✓
$\frac{3}{11}$
- C
$\frac{4}{11}$
- D
$\frac{5}{11}$
AnswerCorrect option: B. $\frac{3}{11}$
View full question & answer→MCQ 1151 Mark
The equation of the circle passing through $(2, 0)$ and $(0, 4)$ and having the minimum radius is:
- A
$x^2+ y^2= 20$
- ✓
$x^2+ y^2- 2x - 4y = 0$
- C
$x^2+ y^2= 4$
- D
AnswerCorrect option: B. $x^2+ y^2- 2x - 4y = 0$
Given points are $(2,0)$ and $(0,4)$
Therefore, equation of circle is $(x - 2)(x - 0) + (y - 0)(y - 4) = 0$
By expanding, we get
$x^2- 2x + y^2- 4y = 0$
Option $B$ is correct.
View full question & answer→MCQ 1161 Mark
The length of the latus rectum of the parabola whose vertex is $(2, -3)$ and the directrix $x = 4$ is:
AnswerDistance of vertex $(2, -3)$ from directrix $x = 4$ is
$=\frac{2\ -\ 4}{\sqrt{1^2\ +\ 0^2}}$
$=2$
So length of latue rectum of above parabola is $=4\times$ distance of vertex to directrix $= 4 \times 2 = 8.$
View full question & answer→MCQ 1171 Mark
If the circle $x^2+ y^2= 9$ passesthrough $(2, c)$ then $cc$ is equal to:
- ✓
$\sqrt5$
- B
$\sqrt 6$
- C
$\sqrt 3$
- D
AnswerCorrect option: A. $\sqrt5$
The equation of circle $x^2+ y^2= 9$ The point is $(2,c)$
$\Rightarrow 2^2+ c^2= 94 + c^2= 9$
$c^2= 9 - 4$
$c^2= 5$
$\text{c}=\sqrt{5}$
View full question & answer→MCQ 1181 Mark
The equation of circle center at $(0,0)$ and Radius $8\ cm:$
- ✓
$x^2+ y^2= 64$
- B
$x^2+ y^2= 8$
- C
$x^2+ y^2= 16$
- D
$x^2+ y^2= 4$
AnswerCorrect option: A. $x^2+ y^2= 64$
The equation of circle is $x^2+ y^2= r^2$
$x^2+ y^2= 8^2$
$x^2+ y^2= 64$
View full question & answer→MCQ 1191 Mark
$A$ and $B$ are two events such that $P (A) = 0.25$ and $P (B) = 0.50.$ The probability of both happening together is $0.14.$ The probability of both $A$ and $B$ not happening is
AnswerCorrect option: A. $0.39$
$\text{p(A)}=0.25$ and $\text{P(B)=0.50}$
$\text{P}(\text{A}\cap\text{B})=0.14$
$\therefore$ Required probability $=1-\text{P}(\text{A}\cup\text{B})$
$=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$=1-\big[0.25+0.50-0.14\big]$
$=1-0.61$
$=0.39$
View full question & answer→MCQ 1201 Mark
If the centroid of an equilateral triangle is $(1, 1)$ and its one vertex is $(−1,2)$ then the equation of its circumcircle is:
- ✓
$x^2+ y^2- 2x - 2y - 3 = 0$
- B
$x^2+ y^2+ 2x - 2y - 3 = 0$
- C
$x^2+ y^2+ 2x + 2y - 3 = 0$
- D
AnswerCorrect option: A. $x^2+ y^2- 2x - 2y - 3 = 0$
Given centroid of an equilateral triangle is $G(1,1).$
We know that in an equilateral triangle, centroid, circumcenter and incenter are at the same point.
So, the circumcenter is at $G(1,1).$
Given one vertex of equilateral triangle at $A(-1,2)$
So, circumradius
$=\text{AG}=\sqrt{5}$
$=$ So, equation of circumcircle is
$= (x - 1)^2+ (y - 1)^2= 5$
$\Rightarrow x^2+ y^2- 2x - 2y - 3 = 0$
View full question & answer→MCQ 1211 Mark
Choose the correct answer.
While shuffling a pack of $52$ playing cards, $2$ are accidentally dropped. Find the probability that the missing cards to be of different colours:
- A
$\frac{29}{52}$
- B
$\frac{1}{2}$
- ✓
$\frac{26}{51}$
- D
$\frac{27}{51}$
AnswerCorrect option: C. $\frac{26}{51}$
We know that out of $52$ playing cards $26$ are of red and $26$ are of black colour.
$\therefore P($both cards of differents colour$)$
$=\frac{26}{52}\times\frac{26}{51}+\frac{26}{52}\times\frac{26}{51}$
$=2\times\frac{26}{52}\times\frac{26}{51}$
$=\frac{26}{51}$
View full question & answer→MCQ 1221 Mark
Which ordered number pair represents the center of the circle $x^2+ y^2- 6x + 4y - 12 = 0?$
- A
$(9, 4)$
- B
$(3, 2)$
- ✓
$(3, -2)$
- D
$(6, 4)$
AnswerCorrect option: C. $(3, -2)$
The equation of circle is $x^2+ y^2- 6x + 4y - 12 = 0$
$\Rightarrow x^2- 6x + 9 +y^2+ 4y + 4 = 12 + 13$
$\Rightarrow (x - 3)^2+ (y + 2)^2= 25$ Comparing above equation with $(x - h)^2+ (y - k)^2= r^2$
Therefore, we get the center of circle as $(3, -2)(3, -2).$
View full question & answer→MCQ 1231 Mark
Find the center$-$radius form of the equation of the circle with center $( 4, 0)$ and radius $7:$
- ✓
$(x-4)^2+y^2=49$
- B
$x^2+(y+4)^2=7$
- C
$x^2+(y-4)^2=7$
- D
$(x+4)^2+y^2=49$
AnswerCorrect option: A. $(x-4)^2+y^2=49$
If $(-g, -f)$ is the center and $rr$ is radius
The $(x + g)^2+ (y + f)^2= r^2 $ is the equation of the circle There
$= C = (4, 0), r = 7$
$\Rightarrow (x - 4)2 + (y - 0)2 = 72$
$= (x-4)^2+y^2=49 $
View full question & answer→MCQ 1241 Mark
The probabilities of happening of two events $A$ and $B$ are $0.25$ and $0.50$ respectively. If the probability of happening of $A$ and $B$ together is $0.14,$ then probability that neither $A$ nor $B$ happens is:
- ✓
$\text{0.39}$
- B
$0.29$
- C
$0.11$
- D
AnswerCorrect option: A. $\text{0.39}$
$\text{P(A)}=0.25, \text{P(B)=0.50}$ and $\text{P(A}\cap\text{B)}=0 .14$
$\therefore$ Required Probability $=1-\text{P}(\text{A}\cup\text{B})$
$=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B}\big]$
$=1-\big[0.25+0.50-0.14\big]$
$=1-0.61$
$=0.39$
View full question & answer→MCQ 1251 Mark
If a card is drawn from a pack of cards. The probability of getting black ace is:
- A
$\frac{1}{52}$
- ✓
$\frac{1}{26}$
- C
$\frac{1}{13}$
- D
AnswerCorrect option: B. $\frac{1}{26}$
Total number of cases $= 52$
Number of favourable cases $($getting a black ace$) = 2$
Thus, Probability $($getting a black ace$)$
$=\frac{5}{52}$
$=\frac{1}{26}$
View full question & answer→MCQ 1261 Mark
A box contains $10$ good articles and $6$ defective articles. One item is drawn at random. The probability that it is either good or has a defect, is:
- ✓
$\frac{64}{64}$
- B
$\frac{49}{64}$
- C
$\frac{40}{69}$
- D
$\frac{24}{64}$
AnswerCorrect option: A. $\frac{64}{64}$
The answer is one, because the article would be either good or defective as per the question.
Hence, the only option is $\frac{64}{64}=1$
View full question & answer→MCQ 1271 Mark
All the three face cards of spades are removed from a well$-$shuffled pack of $52$ cards. A card is then drawn at random from the remaining pack. Find the probability of getting black face card:
- A
$\frac{6}{49}$
- ✓
$\frac{3}{49}$
- C
$\frac{5}{49}$
- D
AnswerCorrect option: B. $\frac{3}{49}$
Total number of possibilities Total number of possibilities $= 49 ($Since, $3$ cards of spade are removed$)$
Number of black face cards $= 3 (3$ cards of clubs$)$
Thus, required probability
$=\frac{3}{49}$
View full question & answer→MCQ 1281 Mark
Four persons are selected at random out of $3$ men, $2$ women and $4$ children. The probability that there are exactly $2$ children in the selection is:
- A
$\frac{11}{21}$
- B
$\frac{9}{21}$
- ✓
$\frac {10}{21}$
- D
AnswerCorrect option: C. $\frac {10}{21}$
There are nine persons $($three men, two women and four children$)$ out of which four persons can be selected in $\ ^{9}\text{C}_4 = 126$ ways.
Total number of elementary events $= 126$
Exactly two children means selecting two children and two other people from three men and two women.
This can be done in $\ ^{4}\text{C}_2\times\ ^{ 5}\text{C}_2$ ways.
Favourable number of elementary events$=\ ^{4}\text{C}_2\times\ ^{ 5}\text{C}_2 = 60$
So, required probability $=\frac{60}{120} =\frac{10}{21}$
View full question & answer→MCQ 1291 Mark
Out of $30$ consecutive integers, $2$ are chosen at random. The probability that their sum is odd, is
- A
$\frac{14}{29}$
- B
$\frac{16}{29}$
- ✓
$\frac {15}{29}$
- D
$\frac{10}{29}$
AnswerCorrect option: C. $\frac {15}{29}$
The total number of ways in which two integers can be chosen from the given $30$ integers is $\ ^{30}\text{C}_2$.
The sum of the selected numbers is odd if exactly one of them is even or odd.
$\therefore$ Favourable number of outcomes $=\ ^{15}\text{C}_1\times\ ^{15}\text{C}_1$
Hence, required probability $=\frac{\ ^{15}\text{C}_1\times\ ^{15}\text{C}_1}{\ ^{30}\text{C}_2}=\frac{15}{29}$
View full question & answer→MCQ 1301 Mark
$20$ cards are numbered from $1$ to $20.$ If one card is drawn at random, what is the probability that the number on the card is a prime number?
- A
$\frac{1}{5}$
- ✓
$\frac{2}{5}$
- C
$\frac{3}{5}$
- D
$5$
AnswerCorrect option: B. $\frac{2}{5}$
Let $E$ be the event of getting a prime number.
$E = \{2, 3, 5, 7, 11, 13, 17, 19\}$
Hence, $P(E)$
$\frac{8}{20}=\frac{2}{5}.$
View full question & answer→MCQ 1311 Mark
Events $A$ and $B$ are said to be mutually exclusive if:
AnswerCorrect option: A. $\ce{P (A U B) = P A. + P B.}$
If $A$ and $B$ are mutually exclusive events,
Then $\ce{P(A ∩ B)} = 0$
Now, by the addition theorem,
$\ce{P(A U B) = PA. + PB. – P(A ∩ B)}$
$\Rightarrow \ce{P(A U B) = PA. + PB}$
View full question & answer→MCQ 1321 Mark
A glass jar contains $10$ red, $12$ green, $14$ blue and $16$ yellow marbles. If a single marble is chosen at random from the jar, find the sample space.
- ✓
$\{$red, green, blue, yellow$\}$
- B
$\{$red, blue, yellow$\}$
- C
$\{$red, yellow$\}$
- D
$\{$red, green, blue$\}$
AnswerCorrect option: A. $\{$red, green, blue, yellow$\}$
Sample space is the collection of all possible outcomes.
So, sample space $== \{$red, green, blue, yellow$\}$
View full question & answer→MCQ 1331 Mark
The area of the circle represented by the equation $(x + 3)^2+ (y + 1)^2= 25$ is:
- A
$4\pi$
- B
$5\pi$
- C
$16\pi$
- ✓
$25\pi$
AnswerCorrect option: D. $25\pi$
The general equation of circle with centre $(h, k)$ and radius $rr$ is given by
$(x - h)^2+(y - k)^2= (r)^2$.
It is given that the equation of the circle is
$(x + 3)^2+ (y + 1)^2= 25$ Therefore,
$= -3, k = -1$ and $r = 5.$
The area of the circle with radius $r$ is
$=\pi\text{r}^2$ thus, the area of the circle with radius $r = 5$ is:
$=\pi\text{r}^2$
$=\pi(5)^2$
$=25\pi$
Hence, the area of the circle is
$=25\pi.$
View full question & answer→MCQ 1341 Mark
A bag contains $5$ black balls, $4$ white balls and $3$ red balls. If a ball is selected randomwise, the probability that it is black or red ball is:
- A
$\frac{1}{3}$
- B
$\frac{1}{4}$
- C
$\frac{5}{12}$
- ✓
$\frac{2}{3}$
AnswerCorrect option: D. $\frac{2}{3}$
Out of $12$ balls, one ball can be drawn in $\ ^{12}\text{C}_1$ ways.
Total number of elementary events $= \ ^{12}\text{C}_1 = 12$
Out of fivne black balls, one black ball can be chosen in $\ ^{5}\text{C}_1 = 5$ ways.
Out of three red balls, one red ball can be chosen in $\ ^{3}\text{C}_1 = 3$ ways
Favourable number of events $= 5 + 3 = 8$
Hence, required probability $=\frac{8}{12}=\frac{2}{3}$
View full question & answer→MCQ 1351 Mark
A card is drawn at random from a pack of $52$ cards. What is the probability that the card drawn is a spade or a king ?
- ✓
$\frac{4}{13}$
- B
$\frac{3}{13}$
- C
$\frac{2}{13}$
- D
$\frac{1}{13}$
AnswerCorrect option: A. $\frac{4}{13}$
Since, Total cards $= 52$
No. of kings $= 4$
No. of cards that are spade $= 13$
There is one card of king which is of spad
$\therefore $ no. of cards which are spade or a king $= 16$
$\therefore $ Required probability $=\frac{16}{52}=\frac{4}{13}$
View full question & answer→MCQ 1361 Mark
Choose the correct answer.
Without repetition of the numbers, four digit numbers are formed with the numbers $0, 2, 3, 5$. The probability of such a number divisible by $5$ is:
- A
$\frac{1}{5}$
- B
$\frac{4}{5}$
- C
$\frac{1}{30}$
- ✓
$\frac{5}{9}$
AnswerCorrect option: D. $\frac{5}{9}$
We have digite $0, 2, 3, 5.$
Number of divisible by $5$ if unit place digit is $'0\ '$ or $'5\ '$
If unit place is $'0\ '$ then first three places can be filled in $3!$ ways.
If unit place is $'5\ '$ then first place can be filled in two ways and second and thried place can be filled in $2!$ ways.
So, number of numbers ending with digit $'5\ '$ is $2 \times 2! = 4$
$\therefore$ Total number of numbers divisible by $5 = 3! + 4 = 110 = n(E)$
Also total number of numbers $= 3 \times 3! = 18$
$\therefore$ Required probability $=\frac{10}{18}=\frac{5}{9}$
View full question & answer→MCQ 1371 Mark
Toss a fair coin $3$ times in a row, how many elements are in the sample space?
AnswerSample space is the collection of all possible events.
So, sample space, $\ce{S = (H, H, H), (H, H, T), (H, T, T), (H, T, T), (T, H, H), (T, H, T ), (T, T, H), (T, T, T)}$
Therefore, there are $8$ elements in sample space.
View full question & answer→MCQ 1381 Mark
Three identical dice are rolled. What is the probability that the same number will appear on each of them?
- A
$\frac{1}{6}$
- ✓
$\frac{1}{36}$
- C
$\frac{1}{18}$
- D
$\frac{3}{28}$
AnswerCorrect option: B. $\frac{1}{36}$
Total number of cases $= 6^3 = 216$
The same number can appear on each of the dice in the following ways:
$(1, 1, 1), (2, 2, 2), ………….(3, 3, 3)$
So, favourable number of cases $= 6$
Hence, required probability $= 6/216 = 1/36$
View full question & answer→MCQ 1391 Mark
A bag contains $3$ red, $4$ white and $5$ blue balls. All balls are different. Two balls are drawn at random. The probability that they are of different colour is:
- ✓
$\frac{47}{66}$
- B
$\frac{10}{33}$
- C
$\frac{1}{3}$
- D
$1$
AnswerCorrect option: A. $\frac{47}{66}$
Out of $12$ balls, two balls can be drawn in $^{12}\text{C}_2$ ways.
$\therefore$ Total number of elementary events, $\text{n}(\text{S})=\ ^{12}\text{C}_2=66$
We observe that at least one ball of each colour can be drawn in one of the following mutually exclusive ways:
Red and $1$ white
$1$ red and $1$ blue
$1$ white and $1$ blue
Thus, if we define three events $A, B$ and $C$ as follows:
$A =$ drawing $1$ red and $1$ white
$B =$ drawing $1$ red and $1$ blue
$C =$ drawing $1$ white and $1$ blue
then, $A, B$ and $C$ are mutually exclusive events.
$\therefore$ Required probability $=\text{P}(\text{A}\cup\text{B}\cup\text{C})$
$=\text{P}(\text{A})+\text{P}(\text{B})+\text{P}(\text{C})$
$=\frac{^3\text{C}_1\times^4\text{C}_1}{^{12}\text{C}_2}+\frac{^3\text{C}_1\times^5\text{C}_1}{^{12}\text{C}_2}+\frac{^4\text{C}_1\times^5\text{C}_1}{^{12}\text{C}_2}$
$=\frac{3\times4}{66}+\frac{3\times5}{66}+\frac{4\times5}{66}$
$=\frac{12}{66}+\frac{15}{66}+\frac{20}{56}$
$=\frac{47}{66}$
View full question & answer→MCQ 1401 Mark
Choosing a queen from a deck of cards is an example of:
AnswerChoosing a queen from a deck of cards is an example of compound event.
Because the total number of card $= 52$
Choosing a queen card $= 4= ($spade, diamond, heart, club$)$
So, more than one element of the sample space in the set representing an event, then this event is called a compound event.
View full question & answer→MCQ 1411 Mark
What is the radius of the circle passing through the point $(2, 4)$ and having centre at the intersection of the lines $4x - y = 4$ and $2x + 3y + 7\ ?$
- A
$3\text{ units}$
- B
$5\text{ units}$
- C
$3\sqrt{3}\text{ units}$
- ✓
AnswerThe line $x - y = 4$ and $2x + 3y + 7 = 0$ intersect at the point $(1, -3)$
So, the centre of circle lies at $(1, -3)$
Point $(2, 4)$ lies on the circle.
So, radius $=$ distance of $(2, 4)$ from $(1, -3)$
$=\sqrt{(2-1)^2+(4-(-3)^2}$
$=\sqrt{1^2+7^2}$
$=\sqrt{50}$
$=5\sqrt{2}$
View full question & answer→MCQ 1421 Mark
The name of the conic represent by the equation $x^2+ y^2- 2y + 20x + 10 = 0$ is:
AnswerFor a standard second degree equation $ax^2+ 2hxy + by^2+ 2gx +2fy + c = 0$ to be a circle $a = b$
$h = 0$ Here $a = b = 1$
$h =0$
So The given equation is Circle.
View full question & answer→MCQ 1431 Mark
Three integers are chosen at random from the first $20$ integers. The probability that their product is even is:
- A
$\frac{2}{19}$
- B
$\frac{3}{29}$
- ✓
$\frac{17}{19}$
- D
$\frac{4}{19}$
AnswerCorrect option: C. $\frac{17}{19}$
Number of ways in which we can choose three distinct integers from $20$ integers $\ ^{20}\text{C}_3=1140$
We know that, if we take three odd numbers, there product will always be an odd number.
Out of $20$ consecutive integers, $10$ are even and $10$ are odd integers.
Number of ways in which we can choose three distinct odd integers from $10$ odd integers $=\ ^{10}\text{C} _3=120$
$P($product is even$) = 1 - P($product is odd$),$
$=1-\frac{120}{1140}$
$=\frac{1140-120}{1140}$
$=\frac{1020}{1140}$
$=\frac{17}{19}$
View full question & answer→MCQ 1441 Mark
A box contains $3$ red, $3$ white and $3$ green balls. A ball is selected at random. Find the probability that the ball picked up is neither a white nor a red ball:
- A
$\frac{1}{4}$
- ✓
$\frac{1}{3}$
- C
$\frac{1}{2}$
- D
$\frac{3}{4}$
AnswerCorrect option: B. $\frac{1}{3}$
Total number of outcomes $= 9$
Favourable outcomes $($the ball is neither white nor red$) = 3$
Probability $=\frac{3}{9}=\frac{1}{3}$
View full question & answer→MCQ 1451 Mark
The equation of the tangent to the curve $y = 2\sin x + \sin2x$ at $\text{x}=\frac{\pi}{3}$ on it is:
- ✓
$(2,3)$
- B
$\text{y}+\sqrt{3}=0$
- C
$2\text{t}-3=0$
- D
$2\text{y}-3\sqrt{3}=0$
AnswerCorrect option: A. $(2,3)$
$(x - 2)^2+ (y - 3)^2= 0$
$\Rightarrow x^2+ 4 - 4x + y^2+ 9 - 6y = 0$
$x^2+ y^2- 4x + 6y + 13 = 0$
radius $= \sqrt{\text{g}^2+\text{f}^2-\text{cg}}=-2$
$\text{f}=-3 \text{ c}=13\text{r}$
$=\sqrt{(-2)^2+(-3)^2-13}$
$=\sqrt{13-13}$
$=$ 0
radius of the circle is zero .
Hence diameter $=0$
i.e., it is a point circle with centre at $(2 ,3).$
View full question & answer→MCQ 1461 Mark
Poonam buys a fish from a shop for her aquarium. The shopkeeper takes out one fish at random from a tank containing $5$ male fish and $8$ female fish. What is the probability that the fish taken out is a male fish?
- ✓
$\frac{5}{13}$
- B
$\frac{1}{4}$
- C
$\frac{1}{5}$
- D
$\frac{5}{14}$
AnswerCorrect option: A. $\frac{5}{13}$
Favourable outcome $($Getting a male fish$) = 5$
Total number of outcomes $($Male and female fish$) =13$
Probabilit $=\frac{5}{13}$
View full question & answer→MCQ 1471 Mark
In how many different ways can the letter of the word $\text{TOTAL}$ be arranged?
AnswerThe word $\text{′TOTAL′}$ It can be arranged in $\frac {5!}{2!}$ ways
i.e, Total no. of words $= 5$
$\therefore$ It can be arranged in $5!$
ways. No. of repeating words can be arranged in $2!$ ways.
$\therefore$ Probability $=\frac {5!}{2!}=60.$
View full question & answer→MCQ 1481 Mark
A box contains $6$ nails and $10$ nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, the probability that it is rusted or is a nail is:
- A
$\frac{3}{16}$
- B
$\frac{5}{16}$
- ✓
$\frac{11}{16}$
- D
$\frac{14}{16}$
AnswerCorrect option: C. $\frac{11}{16}$
If the numbers of nails and nuts are $6$ and $10,$ respectively, then the numbers of rusted nails and rusted nuts are $3$ and $5,$ respectively.
Total number of items $= 6 + 10 = 16$
Total number of rusted items $= 3 + 5 = 8$
Total number of ways of drawing one item $=\ ^{16}\text{C}_1$
Let $R$ and $N$ be the events where both the items drawn are rusted items and nails, respectively.
$R$ and $N$ are not mutually exclusive events, because there are $3$ rusted nails.
$P($either a rusted item or a nail$) = \text{P} \text{(R}\cup\text{N})$
$= \text{P} \text{(R})+\text{P}\text{(N})-\text{P}(\text{R}\cap\text{N})$
$=\frac{\ ^{6}\text{C}_1}{{\ ^{16}\text{C}_1}}+\frac{\ ^{8}\text{C}_1}{{\ ^{16}\text{C}_1}}-\frac{\ ^{3}\text{C}_1}{{\ ^{16}\text{C}_1}}$
$=\frac{6}{16}+\frac{8}{16}-\frac{3}{16}$
$=\frac{11}{16}$
View full question & answer→MCQ 1491 Mark
When a die is thrown, list the outcomes of an event of getting:
$I.$ A prime number,
$II.$ Not a prime number.
- ✓
Prime number are $2, 3$ and $5$ Not a prime number are $1, 4$ and $6$
- B
Prime number are $2, 7$ and $9$ Not a prime number are $1, 4$ and $6$
- C
Prime number are 2, 3 and 5 Not a prime number are $1, 5$ and $7$
- D
AnswerCorrect option: A. Prime number are $2, 3$ and $5$ Not a prime number are $1, 4$ and $6$
In a throw of die, the total outcomes are $\{1, 2, 3, 4, 5, 6\}$
A prime number are $2, 3$ and $5.$
Not a prime number are $1, 4$ and $6.$
View full question & answer→MCQ 1501 Mark
Choose the correct answer.
A single letter is selected at random from the word $\ce{‘PROBABILITY’.}$ The probability that it is a vowel is:
- A
$\frac{1}{3}$
- ✓
$\frac{4}{11}$
- C
$\frac{2}{11}$
- D
$\frac{3}{11}$
AnswerCorrect option: B. $\frac{4}{11}$
Total number of alphabets in probability $= 11$
Number of vowels $= 4$
$\therefore$ Required probability $=\frac{4}{11}$
View full question & answer→