MATHS M.C.Q (1 Marks)

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

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

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

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

2. $\frac{4}{11}$

Solution:

Total number of alphabets in probability = 11

Number of vowels = 4

$\therefore\ \text{Required probability}=\frac{4}{11}$

1.  $\frac{1}{3}$

Solution:

Sample space = 1, 2, 3, 4, 5, 6

A perfect square in samples =1, 4 = 2

Probability of getting perfect square $=\frac{1}{3}$ 

4. $\sqrt{20}$

Solution:

int of intersection of the given diameters is (8, -2) which is the centre of

the circleAlso the circle pass through the point (6, 2), so the radius is

$=\sqrt{(8-6)^2+(-2-2)^2}=\sqrt{20}$

2.  $\frac{1}{2}$

Solution:

Total number of possible cases = 36

Favourable cases of getting even number as the sum

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

Total number of favourable cases = 18

P(getting even number as the sum) $=\frac{18}{36}=\frac{1}{2}$ 

2. $\text{P(M)}+\text{P(N)}-\text{P(M}\cap\text{N)}$

Solution:

If M and N are any two events.

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

1. $\frac{1}{7}$

Solution:

There are 365 days in non-leap year and there are 7 days in a week

$\therefore\ 365\div7=52$ weeks + 1 days

So, this day may be Tuesday or Wednesday.

So, the required probability $=\frac{1}{7}$

2. $\frac{2}{5}$

Solution:

Given digits are 0, 1, 3, 5, 7

Now we have to form 4 digit numbers greater than 5000.

So leftmost digit is either 5 or 7.

When digits are repeated

Number of ways for filling left most digit = 2

Now remaining 3 digits can be filled = 5 × 5 × 5

So total number of ways of 4 digits greater than 5000 = 2 × 5 × 5 × 5 = 250

Again a number is divisible by 5 if the unit digit is either 0 or 5. So there are 2 ways to fill the unit place.

So total number of ways of 4 digits greater than 5000 and divisible by 5 = 2 × 5 × 5 × 2 = 100

Now probability of 4 digit numbers greater than 5000 and divisible by 5

$=\frac{100}{250}$

$=\frac{2}{5}$

2. $\frac{7}{15}$

Solution:

While placing 7 while balls in a row, total gaps = 8

3 black balls can be placed in 8 gaps

$=\text{C}=\frac{(8\times7\times6)}{(3\times2\times1)}=8\times7=56$

So, the total number of ways of arranging white and black balls such that no two black balls are adjacent = 56 × 3! × 7!

Actual number of arrangement possible with 7 white and 3 black balls = (7 + 3)! = 10!

So, the required Probability 

$=\frac{(56\times3!\times7!)}{10!}$

$=\frac{(56\times3!\times7!)}{(10\times9\times8\times7!)}$

$=\frac{(56\times3!)}{(10\times9\times8)}$

$=\frac{(56\times3\times2\times1)}{(10\times9\times8)}$

$=\frac{(7\times3\times2\times1)}{(10\times9)}$

$=\frac{(7\times2)}{(10\times3)}$

$=\frac{7}{(5\times3)}$

$=\frac{7}{15}$

4. 4

Solution:

y² + 4x + 4y + k = 0

y² + 2 × 2y + 4 - 4 + 4x + k = 0

$(\text{y}+2)^2=-4\text{x}-\text{k}+4$

$(\text{y}+2^2)=-4\Big(\text{x}-\frac{4\ +\ \text{k}}{4}\Big)$

$∴$ Latus rectum = 4 units

2. 10

Solution:

$\text{d}=\frac{\text{c}_1-\text{c}_2}{\sqrt{\text{a}^2+\text{b}^2}}$

where dd is the distance between lines whose equations are ax + by + C_1​ = 0 & ax + by + C_2​ = 0

$\text{d}=\frac{27-2}{\sqrt{4^2+3^2}}$

= 5 d = 5

If the distance between vertex and latus rectum = distance of vertex from directri x = a

= then d = 2a = 5

⇒ Length of latus rectum = 4a = 10

3. 3.32

Solution:

The radius of given circle is $\sqrt{11}=3.32$

Let X and Y be two events given by,

X : A fails in an examination

Y : B fails in an examination

P(A fails) = P(X) = 0.2

P(B fails) = P(Y) = 0.3

Now, P(either A or B fails) $=\text{P}(\text{X}\cup\text{Y})$

We know that,

$=\text{P}(\text{X}\cup\text{Y})\leq\text{P(X)}+\text{P()Y}=0.2+0.3=0.5$

$\Rightarrow\text{P}(\text{X}\cup\text{Y})\leq0.5$

$\therefore\text{P}\text{(either A or B fails)}\leq0.5$

Hence, the correct answer is option (c).

1. $\frac{1}{6}$

Solution:

Total number of possibilities = 36

Number of doublet = 6

Thus, probability

$=\frac{6}{36}=\frac{1}{6}$

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

Solution:

Total number of outcomes = 52

Favourable outcomes (A face card) = 12

Probability $=\frac{12}{52}=\frac{3}{13}$

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

Solution:

When a die is rolled, total number of outcomes = 6 (1, 2, 3, 4, 5, 6)

Total even number = 3 (2, 4, 6)

So, the probability that an even number is obtained 

$=\frac{3}{6}=\frac{1}{2}$

4. 21

Solution:

When two dice are thrown, then total outcome = 6 × 6 = 36

A: Getting an odd number on the first die.

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

Total outcome = 18 B: Getting a total of 7 on the two dice.

B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} Total outcome = 6

C: Getting a total of greater than or equal to 8 on the two dice.

C = {(2, 6), (3, 5), (3, 6), (4, 4),(4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} Total outcome = 15

Now $\text{n(A}\cup\text{B)}=\text{n(A)}+\text{n(B)}-\text{n(A}\cup\text{B)}$

$\Rightarrow\text{n(A}\cup\text{B)}=18+6-3$

$\Rightarrow\text{n(A}\cup\text{B)}=21$

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. $\frac{1}{4}$

Solution:

Since m and n are selected between 1 and 100,

Hence total sample space = 100 × 100

Again, 71 = 7, 72 = 49, 73 = 343, 74 = 2401, 75 = 16807, etc

Hence 1, 3, 7 and 9 will be the last digit in the power of 7.

Now, favourable number of case are

→ 1,1 1,2 1,3 …………. 1,100

2,1 2,2 2,3 …………. 2,100

3,1 3,2 3,3 …………. 3,100

100,1 100,2 100,3 …………. 100,100

Now, for m = 1, n = 3, 7, 11, ………, 97

So, favourable cases = 25

Again for m = 2, n = 4, 8, 12, ………, 100

So, favourable cases = 25

Hence for every m, favourable cases = 25

So, total favourable cases = 100 × 25 Required Probability

$=\frac{(100\times25)}{(100\times100)}$

$=\frac{25}{100}$

$=\frac{1}{4}$

4. $\frac{84}{452}$

Solution:

Total number of cards = 52 and one card is lost. Case 1: if lost card is a diamond card

Total number of cards = 51 Number of diamond cards = 12 Now two cards are drawn.

P(both cards are diamonds) $\frac{=^{12}\text{C}_2}{^{51}\text{c}_2}$

Total number of cards = 52 and one card is lost.

Case 2: If lost card is not a diamond card Total number of cards = 51

Number of diamond cards = 13Now two cards are drawn.

P(both cards are diamonds) $\frac{=^{13}\text{C}_2}{^{51}\text{c}_2}$

Now probability that both cards are diamond  $\frac{=^{12}\text{C}_2}{^{51}\text{c}_2}+\frac{=^{13}\text{C}_2}{^{51}\text{c}_2}$

$=\frac{^{12}\text{C}_2+^{13}\text{C}_2}{^{51}\text{C}_2}$

$\Big\{\frac{(12\times11)}{(2\times1)}+\frac{(13\times12)}{(2\times1)}\Big\}\Big\{\frac{(51\times50)}{(2\times1}\Big\}$

$=\frac{(12\times11\times+13\times12)}{(51\times50)}$

$=\frac{288}{2550}$

$=\frac{96}{850}$ (288 and 2550 divided by 3)

$=\frac{48}{425}$ (96 and 850 divided by 2)

So probability that both cards are diamond is $\frac{48}{425}$

4. $2\sqrt{\text{r}^2-\text{a}^2+\text{b}^2}$

Solution:

Let radius of after circle be $\text{RC}_1:(\text{x}-\text{a)}^2+(\text{y}-\text{c)}^2=\text{r}^2\text{c}_2\\:(\text{x}-\text{b}^2+(\text{y}-\text{c)}^2=\text{R}^2$

$\text{Radius axis}\rightarrow\text{x}=0\text{ OR}\text{ c}_1-\text{c}_2=0\text{ c}_1-\text{c}_2$

$= \text{x}2 \text{b}\text{x} - 2\text{a}\text{x} - \text{r}^2 + \text{R}^2 + \text{a}^2 - \text{b}^2$

$= \text{x} \text{ Put } \text{x} = \text{0 R}^2 = \text{r}^2 + \text{b}^2 - \text{a}^2\text{R}$

$=\sqrt{\text{r}^2+\text{b}^2-\text{a}^2}$ Diamete

$=2\sqrt{\text{r}^2+\text{b}^2-\text{a}^2}$ 

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

Solution:

Favourable outcomes ={H T H, T H T} = 2 outcomes

Total number of outcomes = 8

Probability $=\frac{2}{8}=\frac{1}{4}$

2. $\frac{187}{190}$

Solution:

Since, the set of three consecutive numbers from 1 to 20 are (1, 2, 3), (2, 3, 4), (3, 4, 5), ....... , (18, 19, 20), i.e., 18

P(numbers are consecutive)

$=\frac{18}{^{20}\text{C}_3}=\frac{18}{\frac{20\times19\times18}{3!}}=\frac{3}{190}$

P(three number are not consecutive)

$=1-\frac{3}{190}=\frac{187}{190}$

4. $\frac{1}{7}$

Solution:

In a leap year, total number of days = 366 days.

In 366 days, there are 52 weeks and 2 days.

Now two days may be

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

Now in total 7 possibilities, Sunday and Monday both come together is 1 time.

So probabilities of 53 Sunday and Monday in a leap year

$=\frac{1}{17}$

1. ​​​​​​$\frac{1}{2}$

Solution:

The outcome of a throw of coin can be 2.

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

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).

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

Solution:

When a die is rolled, total number of outcomes = 6 (1, 2, 3, 4, 5, 6)

Total even number = 3 (2, 4, 6)

So, the probability that an even number is obtained

$=\frac{3}{6}=\frac{1}{2}$

3.  $\frac{1}{2}$

Solution:

Sample space = {1,2,3,4,5,6}=1,2,3,4,5,6

odd nos. = 1,3,5=1,3,5

probability of getting odd nos

$=\frac{3}{6}=\frac{1}{2}$ 

3. ​​​​​​$\frac{3}{4}$

Solution:

Favourable number of outcomes, getting at least one head = 3[H H, H T, T H]

Total number of outcomes = 4[H H, H T, T H, T T]

Probability $=\frac{3}{4}$

The given digits are 0, 2, 3 and 5. 

Now, there are 3 ways to fill the thousands place (0 cannot occupy the thousands place), 3 ways to fill the hundreds place, 2 ways to fill the tens place and 1 way to fill the ones place.

Total number of four digit numbers formed = 3 × 3 × 2 × 1 = 18

We know that a number is divisible by 5 if it ends in 0 or 5.

When 0 is at the ones place,

Number of four digits numbers divisible by 5 formed = 3 × 2 × 1 = 6

When 5 is at the ones place,

Number of four digits numbers divisible by 5 formed = 2 × 2 × 1 = 4 (0 cannot occupy the thousands place)        

Total number of four digit numbers divisible by 5 = 6 + 4 = 10

$\therefore$ P(four digit number formed is divisible by 5) 

$=\frac{\text{Total Number of four digit numbers divisible by 5}}{\text{Total Number of w4 digit numbers formed}}$

$=\frac{10}{18}=\frac{5}{9}$

Hence, the correct answer is option (d).

3. $\leq0.5$

Solution:

Given, P(A fail) = 0.2

and P(B fail) = 0.3

$\therefore\ \text{P(either A or B fail)}\leq\text{P(A fail)}+\text{P(B fail)}$

$\leq0.2+0.3$

$\leq0.5$

1. $\frac{1}{462}$

Solution:

Given, 6 boys and 6 girls sit in a row at random.

Then, the total number of arrangement of 6 boys and 6 girls = arrangement of 12 persons = 12! Now, boys and girls sit alternatively.

So, the total number of arrangement = 2 × 6! × 6!

Now, P(boys and girls sit alternatively) $=\frac{(2\times6!\times6!)}{12!}$

$=\frac{(2\times6!\times6!)}{(12!\times11!)}$

$=​​\frac{(5!\times6!)}{11!}$

$=\frac{(5\times4\times3\times2\times1\times6!)}{(11\times10\times9\times8\times7\times6!)}$

$=\frac{(5\times4\times3\times2)}{(11\times10\times9\times8\times7)}$

$=\frac{(4\times3)}{(11\times9\times8\times7)}$

$=\frac{3}{(11\times9\times2\times7)}$

$=\frac{1}{(11\times9\times2\times7)}$

$=\frac{1}{462}$

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

Solution:

If two persons sit next to each other, then consider these two person as 1 geoup.

Now, we have to arrange 6 persons.

$\therefore$ Number of arrangement = 2! × 6!

Total number of arrangement of 7 person = 7!

$\therefore\ \text{Required probability}=\frac{2!6!}{7!}=\frac{2}{7}$

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

Solution:

Since, Total possibilities are = {H H, H T, T H, T T}

no. of cases with atmost one head are = {H T, T H, T T}

$=\frac{3}{4}$

1. $\frac{1}{13}$

Solution:

Favourable number of outcomes i.e., numbers of jack cards = 4

Total number of outcomes = 52

Thus, probability $=\frac{4}{52}=\frac{1}{13}$

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

Solution:

If the parabola y² = 4ax passes through (3, 2)

therefore, 4 = 4a (3)

$\Rightarrow\text{a}=\frac{1}{3}$

Therefore, length of latus rectum:

$1=4\text{a}=\frac{4}{3}$

3. $1.2$

Solution:

We have, $\text{P}(\text{A}\cup\text{B})=0.6$ and $\text{P}(\text{A}\cap\text{B})=0.2$

$\therefore\ \text{P}(\text{A}\cup\text{B})=0.6$ and $\text{P}(\text{A}\cap\text{B})=0.2$

$\Rightarrow0.6=\text{P(A)}+\text{P(B)}-0.2$

$\Rightarrow\text{P(A)}+\text{P(B)}=0.8$

$\therefore\ \text{P}(\bar{\text{A}})+\text{P}(\bar{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$

$=2-\big[\text{P(A)}+\text{P(B)}\big]=2-0.8=1.2$