MATHS M.C.Q (1 Marks)

1. $\frac{5}{84}$​

Solution:

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 $=\frac{1}{84}+\frac{4}{84}=\frac{5}{84}$

4. $\frac{1}{132}$

Solution:

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 × 6

$\therefore\text{Required Probability}=\frac{7\times6}{12}$$\frac{720}{12\times11\times10\times9\times8}=\frac{1}{132}$

3. $\frac{1}{10}$

Solution:

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 $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{10}{100}=\frac{1}{10}$

1. ​​​​​0.39

Solution:

$\text{p(A)}=0.25\text{ and }\text{P(B)=0.50}$

$\text{P}(\text{A}\cap\text{B})=0.14$

$\therefore\text{Required probability}=1-\text{P}(\text{A}\cup\text{B})$

$=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$

$=1-\big[0.25+0.50-0.14\big]$

$=1-0.61=0.39$

1. $0.39$

Solution:

$\text{P(A)}=0.25, \text{P(B)=0.50}\text{ and }\text{P(A}\cap\text{B)}=0 .14$

$\therefore\text{Required Probability}=1-\text{P}(\text{A}\cup\text{B})$

$=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B}\big]$

$=1-\big[0.25+0.50-0.14\big]$

$=1-0.61=0.39$

1. ​​​​​​$\frac{64}{64}$

Solution:

The answer is one, because the article would be either good or defective as per the question.
Hence, the only option is $\frac{64}{64}=1$

3. $\frac{10}{21}$

Solution:

There are nine persons (three men, two women and four children) out of which four persons can be selected in $\ ^{9}\text{C}_4 = 126\ \text{ways}.$

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 $\ ^{4}\text{C}_2\times\ ^{ 5}\text{C}_2 \text{ways}.$

Favourable number of elementary events$=\ ^{4}\text{C}_2\times\ ^{ 5}\text{C}_2 = 60$

So, required probability $=\frac{60}{120} =\frac{10}{21}$

3. $\frac{15}{29}$​

Solution:

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 $=\frac{\ ^{15}\text{C}_1\times\ ^{15}\text{C}_1}{\ ^{30}\text{C}_2}=\frac{15}{29}$

2. Is false.

Solution:

Since the events A, B and C are mutually exclusive, we have:

$\text{P}(\text{A}\cup\text{B}\cup\text{C})=\frac{2}{3}+\frac{1}{4}+\frac{1}{6}=\frac{13}{12}>1$

which is not possible.

Hence, the given statement is false.

1. ​​​​​​$\frac{1}{4}$​

Solution:

Let $\text{P(B)}=\text{P}$

Than $\text{P(A)}=\frac{1}{3}\text{P}$

Since A and B are two mutually exclusive events, we have:

$\text{A}\cup\text{B}=\text{S}$

$\Rightarrow\text{P}\text{(A}\cup\text{B)}=\text{P}\text{(S)}$

$\Rightarrow\text{P}\text{(A}\cup\text{B)}=1$

$\Rightarrow\text{P}\text{(A)}+\text{(B)}=1$

$\Rightarrow\frac{1}{2}\text{P}+\text{P}=1$

$\Rightarrow\frac{4\text{p}}{3}=1$

$\therefore\text{P(A)}=\frac{1}{3}\text{P}=\frac{1}{3}\times\frac{3}{4}=\frac{1}{4}$

1. $\frac{64}{64}$​

Solution:

Let A be the event of drawing one good article whereas B be the event of drawing one defected article.

Here,

$\text{P(A)}=\frac{10}{10+6}=\frac{10}{16}$

$\text{P(B)}=\frac{6}{10+6}=\frac{6}{16}$

The events A and B are mutually exclusive. Thus, the required probability,

$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}$

$\Rightarrow\text{P}(\text{A}\cup\text{B})=\frac{10}{16}+\frac{6}{16}=\frac{16}{16}=1$

Hence, the correct option is (a).

2. $\frac{3}{4}$

Solution: 

When two dice are thrown, there are (6 × 6) = 36 outcomes. The set of all these outcomes is the sample space, given by

S = (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)

i.e. n(S) = 36

Let E be the event of getting at least one digit greater than 3.

Then E = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (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) }

$\therefore\ \text{n(E)}=27$

Hence, required probability $=\frac{27}{36}=\frac{3}{4}$

2. $\frac{6}{36}$

Solution:

When two dices are thrown, there are (6 × 6) = 36 outcomes.

The set of all these outcomes is the sample space given by

S = (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)

$\therefore$ n(S) = 36

Let E be the event of getting a total score of 7.

Then E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

$\therefore$ n(E) = 6

Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{6}{36}$

1. $\frac{1}{36}$​

Solution:

When two dice are thrown simultaneously, the sample space associated with the random experiment is given by:

S = {(1, 1), (1, 2), (1, 3), (6, 4), (6, 5), (6, 6)}

Clearly, total number of elementary events = 36

Let A be the event of getting a pair of aces.

Then A = {(1, 1)}

$\therefore\text{n(A)}=1$

Hence, required probability $=\frac{\text{n(A)}}{\text{n(S)}}=\frac{1}{36}$ 

3. $$$\frac{9}{20}$

Solution:

We have:

$\text{P(both are aces)}=\frac{\ ^{4}\text{C}_2}{\ ^{16}\text{C}_2}$

$=\frac{4}{16}\times\frac{3}{15}=\frac{1}{20}$

$\text{P(one are ace)}=\frac{\ ^{4}\text{C}_1\times\ ^{12}\text{C}_1}{\ ^{16}\text{C}_2}=\frac{2}{5}$

$\therefore\text{P(at least one are ace)}=\frac{1}{20}+\frac{2}{5}=\frac{9}{20}$

2. $$$\Big(\frac{-1}{4},\frac{1}{3}\Big)$

Solution:

$\text{P(A)}=\frac{(1-3\text{P})}{2}$

$\text{P(B)}=\frac{(1+4\text{P})}{3}$

$\text{P()}=\frac{(1+\text{P})}{6}$

The events are mutually exclusive and exhaustive.

$\therefore\text{P}(\text{A}\cup\text{B }\cup\text{C})=\text{P(A)}+\text{P(B)}+\text{P(C)}=1$

$\Rightarrow0\leq\text{P(A)}\leq1,0\leq\text{P(B)}\leq1,0\leq\text{P(C)}\leq1$

$\Rightarrow0\leq\frac{1-3\text{P}}{2}\leq1$, $0\leq\frac{1-4\text{P}}{3}\leq1$, $0\leq\frac{1-\text{P}}{6}\leq1$

$\Rightarrow\frac{-1}{3}\leq\text{P}\leq\frac{1}{3}\ ...(1)$

$\frac{-1}{4}\leq\text{P}\leq\frac{1}{2}\ ...(2)$

$\text{and }{-1}\leq\text{P}\leq{5}\ ...(3)$

The common solution of (1), (2), and (3) is $\frac{-1}{4}\leq\text{P}\leq\frac{1}{3}$

$\therefore\text{The set values of P are}\Big(\frac{-1}{4},\frac{1}{3}\Big)$

3. $\frac{4}{13}$

Solution:

If A and B denote the events of drawing a king and a spade card, respectively, then event A consists of four sample points, whereas event B consists of 13 sample points.

Thus, $\text{P(A)}=\frac{4}{52}$ and $\text{P(B)}=\frac{13}{52}$

The compound event (A n B) consists of only one sample point, king of spade.

So, $\text{P}(\text{A}\cap\text{B})=\frac{1}{52}$

By addition theorem , we have:

$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$

$=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$

Hence, the probability that the card drawn is either a king or a spade is given by $\frac{4}{13}.$

2. $\frac{13}{18}$

Solution:

When two dices are thrown, there are (6 × 6) = 36 outcomes.

The set of all these outcomes is the sample space is given by

S = (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)

Let E be the event of getting the digits which are neither equal nor give a total of 9.

Then E' = event of getting either a doublet or a total of 9

Thus, E' = {{1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (4, 5), (5, 4), (6, 3)}

i.e. n(E') = 10

$\text{P(E}')=\frac{\text{n(E}')}{\text{n(E)}}=\frac{10}{36}=\frac{5}{18}$

Hence, required probability P(E) = 1 - P(E')

$=1-\frac{5}{18}=\frac{13}{18}$

4. ​​​​$\text{3 : 2}$

Solution:

 Let $\text{P(B)}=\text{X}$

 Than, $\text{P(A)}=\frac{2\text{x}}{3}$

$\text{P(A)}+\text{P(B)}=\text{x}+\frac{2x}{3}=\frac{5x}{3}$

$\Rightarrow\frac{5\text{x}}{3}=1$ $\big(\therefore$ They are exhaustive events $\big)$

$\Rightarrow\text{x}=\frac{3}{5}$

Now, $\text{P(A)}=\frac{2}{5}\text{ and }\text{P(B)}=\frac{3}{5}$

$\therefore\text{odd in favour of B}=\frac{\frac{3}{5}}{\frac{1-3}{5}}=\frac{3}{2}=\text{3 : 1}$

3. $$$\frac{1}{36}$

Solution:

When three dice are thrown together, the sample space S associated with the random experiment is given by,

S = {(1, 1, 1), (1, 1, 2), (1, 1, 3) ...(6, 6, 5), (6, 6, 6)}

Clearly, total number of elementary events n(S) = 216

Let A be the event of getting a total score of 5.

Then A = { (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1)}

$\therefore$ Favourable number of elementary events = 6

i.e. n(A) = 6

Hence, required probability $=\frac{6}{216}=\frac{1}{36}$

3. ​​​​​​$\frac{19}{50}$

Solution:

Given:

$\text{P(A)}=\frac{1}{5}$

$\therefore\text{P(A}')=1-\frac{1}{5}=\frac{4}{5}$

$\text{P(B)}=\frac{3}{10}$

$\therefore\text{P(B}')=1-\frac{3}{10}=\frac{7}{30}$

Hence, required probability $=\text{P}(\text{A}\cap\text{B}')+\text{P}(\text{A}'\cap\text{B})$

$=\frac{1}{5}\times\frac{7}{10}+\frac{4}{5}\times\frac{3}{10}$

$=\frac{7}{50}+\frac{12}{50}$

$=\frac{19}{50}$

3. $\frac{1}{9}$

Solution:

When two dice are thrown, there are (6 × 6) = 36 outcomes.

The set of all these outcomes is the sample space given by

S = (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)

i.e. n(S) = 36

Let E be the event of getting a total score of 5.

Then E = {(1, 4), (2, 3), (3, 2), (4, 1)

$\therefore$ n(E) = 4

Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{4}{36}=\frac{1}{9}$

1. $\frac{\ ^{7}\text{P}_5}{7^5}$​

Solution:

Since, it is an eight - storey building.

So, there are 7 possible options for them in 7 floors in total if ground floor is not considered.

Hence, total possible outcomes = 7× 7× 7 × 7× 7= 7⁵

Thus, number of ways in which 5 persons can leave from seven floors differently $=\ ^{7}\text{P}_5$

Required probability $=\frac{\ ^{7}\text{P}_5}{7^5}$

2. $\frac{3}{7}$

Solution:

We know that a leap year has 366 days (i.e. 7 × 52 + 2) = 52 weeks and 2 extra days .

The sample space for these 2 extra days is given below:

S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)} There are 7 cases.

$\therefore\text{n(S)}=7$

Let E be the event that the leap year has 53 Fridays or 53 Saturdays.

E = { (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}

i.e. n(E) = 3

$\therefore\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{3}{7}$

Hence, the probability that a leap year has 53 Fridays or 53 Saturdays is $\frac{3}{7}$.

4. $\frac{23}{24}$

Solution:

Total number of ways of placing four letters in 4 envelops = 4 = 24

All the letters can be dispatched in the right envelops in only one way. Therefore, the probability that all the letters are placed in the right envelops is $\frac{1}{24}$.

Hence, probability that all the letters are not placed in the right envelops $=1-\frac{1}{24}=\frac{23}{24}$

4. $\frac{2}{3}$​

Solution:

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 $=\frac{8}{12}=\frac{2}{3}$

1. $\frac{47}{66}$

Solution:

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 $=\text{P}(\text{A}\cup\text{B}\cup\text{C})$

$=\text{P}(\text{A})+\text{P}(\text{B})+\text{P}(\text{C})$

$=\frac{^3\text{C}_1\times^4\text{C}_1}{^{12}\text{C}_2}+\frac{^3\text{C}_1\times^5\text{C}_1}{^{12}\text{C}_2}+\frac{^4\text{C}_1\times^5\text{C}_1}{^{12}\text{C}_2}$

$=\frac{3\times4}{66}+\frac{3\times5}{66}+\frac{4\times5}{66}$

$=\frac{12}{66}+\frac{15}{66}+\frac{20}{56}=\frac{47}{66}$

3. $\frac{17}{19}$​

Solution:

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),

$=1-\frac{120}{1140}=\frac{1140-120}{1140}=\frac{1020}{1140}=\frac{17}{19}$

3. $\frac{1}{3}$

Solution:

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 $=\frac{\text{n(A)}}{\text{n(S)}}=\frac{2}{6}=\frac{1}{3}$

3. $\frac{11}{16}$​

Solution:

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) $= \text{P} \text{(R}\cup\text{N})$

$= \text{P} \text{(R})+\text{P}\text{(N})-\text{P}(\text{R}\cap\text{N})$

$=\frac{\ ^{6}\text{C}_1}{{\ ^{16}\text{C}_1}}+\frac{\ ^{8}\text{C}_1}{{\ ^{16}\text{C}_1}}-\frac{\ ^{3}\text{C}_1}{{\ ^{16}\text{C}_1}}$

$=\frac{6}{16}+\frac{8}{16}-\frac{3}{16}=\frac{11}{16}$

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{(B})-\text{P}(\text{A}\cap\text{B})$

$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{P}\text{(B})-\text{P}(\text{A}\cap\text{B})$ $\big[\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cup\text{B})\big]$

$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})=0\ ...(1)$

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).

2. $\frac{1}{12}$

Solution:

When two dices are thrown, there are (6 × 6) = 36 outcomes.

The set of all these outcomes is the sample space, given by

S = (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)

i.e. n(S) = 36

Let E be the event of getting a total score of 10.

Then E = {(4, 6), (5, 5), (6, 4)}

$\therefore$ n(E) = 3

Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{3}{36}=\frac{1}{12}$

We know that,