Quantitative Aptitude PYQs

Calculation

Total number of arrangements of 52 cards is 52!.

Now, Lets consider the 14^th position.

The 14^th position can be filled with ace in 4 ways.

After the 14^th position is filled, remaining 51 positions can be filled with the remaining 51 cards in 51! ways.

So total favourable arrangements is 4 × 51!

Required probability \(= {\rm{}}\frac{{4{\rm{;}} \times {\rm{;}}51!}}{{52!}} = \frac{1}{{13}}\)

Correct option is (4).

Concept

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

\({\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right) + {\rm{P}}\left( {\overline {{\rm{A}} \cup {\rm{B}}} } \right) = 1\)

Calculation

P(A) = 0.5, P(B) = 0.6, P(A ∩ B) = 0.4

⇒ P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(A ∪ B) = 0.5 + 0.6 – 0.4

⇒ P(A ∪ B) = 0.7

\(⇒ {\rm{P}}\left( {\overline {{\rm{A}} \cup {\rm{B}}} } \right) = 1 - {\rm{P}}\left( {\ {{\rm{A}} \cup {\rm{B}}} } \right)\)

⇒ 1 - P( A ∪ B) = 1 – 0.7

\(⇒ {\rm{P}}\left( {\overline {{\rm{A}} \cup {\rm{B}}} } \right) = 0.3\)

Correct option is (4).

Calculation

\({\rm{P}}\left( {\rm{A}} \right) = \frac{1}{2},{\rm{;;P}}\left( {\rm{B}} \right) = \frac{3}{4},{\rm{;;P}}\left( {\rm{C}} \right) = \frac{1}{4}{\rm{;}}\)

 All students solves the problem independently, so if at least one of them solves the problem then the problem is considered solved.

Probability that the problem is unsolved

= [1 – P(A)][1 – P(B)][1 – P(C)]

= \(\frac{3}{{32}}\)

The probability that the problem will be solved

= 1 – Probability that the problem is unsolved

\( = \frac{{29}}{{32}}\)

Correct option is (1).

Concept

The sample space when a pair of dice is rolled is as follows:

(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)

Total outcomes = 36

Calculation

The pairs in which the second number is greater than the first number are as follows:

(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)

Total favourable outcomes = 15

Required probability

\( = \frac{{15}}{{36}} = \frac{5}{{12}}\)

Correct option is (3).

Concept

The sample space when a fair coin is tossed and an unbiased dice is rolled together is as follows:

(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)

(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)

Total outcomes = 12

Calculation

Pairs in which we get a 2 or 4 or 6 along with head are as follows:

(H, 2), (H, 4), (H, 6)

Total favourable outcomes = 3

Required probability

\(= \frac{3}{{12}}\)

\(= \frac{1}{{4}}\)

Correct option is (3).

Concept

Calculation

Consider the following Venn diagram:

Atleast two of the events means exactly two events will occur or all the three events will occur.

Region b, d and f represents exactly two events.

b = (A ∩ B) - (A ∩ B ∩ C)

d = (A ∩ C) - (A ∩ B ∩ C)

f = (B ∩ C) - (A ∩ B ∩ C)

Region e represents all the three events.

e = (A ∩ B ∩ C)

The probability that at least two of these events occur together

= P(b) + P(d) + P(f) + P(e)

= P(A ∩ B) + P(A ∩ C) + P(B ∩ C) – 2P(A ∩ B ∩ C)

Correct option is (3).

Concept

If X and Y are independent, then E(XY) = E(X)E(Y)

Covariance Cov(X, Y) of X and Y = E(XY) – E(X)E(Y)

\(Correlation~coefficient = \frac{{{\bf{Cov}}\left( {{\bf{X}},;{\bf{Y}}} \right)}}{{{{\bf{\sigma }}_{\bf{X}}}{{\bf{\sigma }}_{\bf{Y}}}}}{\rm{;}}\)

Calculation

Given that X and Y are independent, then the Covariance Cov(X, Y) = 0

Hence, the correlation coefficient = 0.

The correct option is (3).

Concept

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

If A and B are independent, then P(A ∩ B) = P(A)P(B).

Calculation

\({\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right) = \frac{2}{3},{\rm{;P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = \frac{1}{6}\)

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(A) + P(B) = P(A ∪ B) + P(A ∩ B)

\( \Rightarrow {\rm{P}}\left( {\rm{A}} \right) + {\rm{;P}}\left( {\rm{B}} \right) = \frac{5}{6}\)

Given A and B are independent so, P(A ∩ B) = P(A)P(B)

\( \Rightarrow {\rm{P}}\left( {\rm{A}} \right) = \frac{1}{{6{\rm{P}}\left( {\rm{B}} \right)}}\)

As, \({\rm{P}}\left( {\rm{A}} \right) + {\rm{;P}}\left( {\rm{B}} \right) = \frac{5}{6}\)

\( \Rightarrow ;\frac{1}{{6{\rm{P}}\left( {\rm{B}} \right)}} + {\rm{;P}}\left( {\rm{B}} \right) = \frac{5}{6}\)

⇒ 6P(B)² - 5P(B) + 1 = 0

\(\Rightarrow {\rm{P}}\left( {\rm{B}} \right) = \frac{{5 \pm \sqrt {{{\left( { - 5} \right)}^2} - 4 \times 6} }}{{2 \times 6}}\)

\(\Rightarrow {\rm{P}}\left( {\rm{B}} \right) = {\rm{;}}\frac{1}{2},{\rm{;}}\frac{1}{3}\)

\({\rm{P}}\left( {\rm{A}} \right) = \frac{5}{6} - {\rm{P}}\left( {\rm{B}} \right)\)

\(\Rightarrow {\rm{P}}\left( {\rm{A}} \right) = \frac{1}{3},{\rm{;}}\frac{1}{2}\)

Given that P(B) < P(A) so the possible value of P(B) is \(\frac{1}{3}\)

Correct option is (2).

Concept

Mean of n observations = \(\frac{1}{{\bf{n}}}\mathop \sum \limits_{{\bf{i}} = 1}^{{\bf{i}} = {\bf{n}}} {{\bf{X}}_{\bf{i}}}\)

Standard deviation = \(\sqrt {\frac{1}{{\rm{n}}}\mathop \sum \limits_{{\bf{i}} = 1}^{{\bf{i}} = {\bf{n}}} {{\left( {{{\bf{X}}_{\bf{i}}} - {\bf{mean}}} \right)}^2}} \)

where, X_i = observation

Calculation

The mean of 100 observations is 50

\( \Rightarrow \frac{1}{{100}}\mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = 100} {{\rm{X}}_{\rm{i}}} = 50\)

\( \Rightarrow \mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = 100} {{\rm{X}}_{\rm{i}}} = 5000\)

Standard deviation is 10

\(\Rightarrow \sqrt {\frac{1}{{100}}\mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {{\left( {{{\rm{X}}_{\rm{i}}} - 50} \right)}^2}} = 10\)

5 is subtracted from each X_i and the resultant is divided by 4 so, the new observation becomes \(\frac{{{{\bf{X}}_{\bf{i}}} - 5}}{4}\).

Mean of new 100 observations is

\(= \frac{1}{{100}}\mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = 100} \left( {\frac{{{{\rm{X}}_{\rm{i}}} - 5}}{4}} \right)\)

\(= \frac{1}{{100}}\left[ {\frac{1}{4}\mathop \sum \limits_{i = 1}^{i = 100} {{\rm{X}}_{\rm{i}}} - \frac{1}{4}\mathop \sum \limits_{i = 1}^{i = 100} 5} \right]\)

\(= \frac{1}{{100}}\left[ {\frac{{5000}}{4} - \frac{{100 \times 5}}{4}} \right]\)

= 11.25

Standard deviation of new observations is

\(= \sqrt {\frac{1}{{100}}\mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {{\left( {\frac{{{{\rm{X}}_{\rm{i}}} - 5}}{4} - 11.25} \right)}^2}} \)

\(= \sqrt {\frac{1}{{100}}\mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {{\left( {\frac{{{{\rm{X}}_{\rm{i}}} - 50}}{4}} \right)}^2}}\)

\(= \sqrt {\frac{1}{{1600}}\mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {{\left( {{{\rm{X}}_{\rm{i}}} - 50} \right)}^2}}\)

\(= \frac{1}{4}\sqrt {\frac{1}{{100}}\mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {{\left( {{{\rm{X}}_{\rm{i}}} - 50} \right)}^2}}\)

\(= \frac{{10}}{4}\)

= 2.5

Correct option is (3).

Concept

If A and B are two events, then the conditional probability \({\bf{P}}\left( {{\bf{A}}/{\bf{B}}} \right) = \frac{{{\bf{P}}\left( {{\bf{A}}; \cap ;{\bf{B}}} \right)}}{{{\bf{P}}\left( {\bf{B}} \right)}}\)

The sample space when a pair of dice is rolled is as follows:

(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)

Total outcomes = 36

Calculation

Let A be the set of all pairs in which 6 appears on the first dice and B be the set of all pairs sum of whose numbers is 8.

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

B = (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

A ∩ B = (6, 2)

\({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = \frac{1}{{36}}\)

\({\rm{P}}\left( {\rm{B}} \right) = \frac{5}{{36}}\)

Required probability is \({\bf{P}}\left( {{\bf{A}}/{\bf{B}}} \right) = \frac{{{\bf{P}}\left( {{\bf{A}}; \cap ;{\bf{B}}} \right)}}{{{\bf{P}}\left( {\bf{B}} \right)}} = \frac{1}{5}\)

Correct option is (3).

Concept

The sample space is as follows:

(R₁, r₁), (R₁, r₂), (R₁, b₁), (R₁, b₂), (R₁, y), (R₁, w)

(R₂, r₁), (R₂, r₂), (R₂, b₁), (R₂, b₂), (R₂, y), (R₂, w)

(B₁, r₁), (B₁, r₂), (B₁, b₁), (B₁, b₂), (B₁, y), (B₁, w)

(B₂, r₁), (B₂, r₂), (B₂, b₁), (B₂, b₂), (B₂, y), (B₂, w)

(Y, r₁), (Y, r₂), (Y, b₁), (Y, b₂), (Y, y), (Y, w)

(W, r₁), (W, r₂), (W, b₁), (W, b₂), (W, y), (W, w)

R₁, R₂, B₁, B₂, Y, W are black, red, yellow and white faces of dice 1.

r₁, r₂, b₁, b₂, y, w are black, red, yellow and white faces of dice 2.

Total outcomes = 36

Calculation

Pairs with same colours are as follows:

(R₁, r₁), (R₁, r₂), (R₂, r₁), (R₂, r₂), (B₁, b₁), (B₁, b₂), (B₂, b₁), (B₂, b₂), (Y, y), (W, w)

Number of Favourable outcomes = 10

Required Probability = \(\frac{{10}}{{36}} = \frac{5}{{18}}\)

Correct option is (3).

Concept

\({\bf{C}}\left( {{\bf{n}},;{\bf{r}}} \right) = ;\frac{{{\bf{n}}!}}{{{\bf{r}}!\left( {{\bf{n}} - {\bf{r}}} \right)!}}\)

Calculation

From n socks, 2 can be chosen in C(n, 2) ways.

From 3 red socks, 2 can be chosen in C(3, 2) ways

Probability that both socks selected are red

\( = \frac{{{\rm{C}}\left( {3,{\rm{;;;}}2} \right)}}{{{\rm{C}}\left( {{\rm{n}},{\rm{;;;}}2} \right)}}\)

\(= \frac{6}{{{\rm{n}}\left( {{\rm{n}} - 1} \right)}}\)

It is given that Probability that both socks selected are red = \(\frac{1}{2}\)

\( \Rightarrow ;\frac{6}{{{\rm{n}}\left( {{\rm{n}} - 1} \right)}} = \frac{1}{2}\)

⇒ n² – n – 12 = 0

⇒ (n - 4)(n + 3) = 0

⇒ n = 4, -3

n cannot be negative so, n = 4.

Correct option is (2).

Concept

\({\bf{C}}\left( {{\bf{n}},;{\bf{r}}} \right) = \frac{{{\bf{n}}!}}{{{\bf{r}}!\left( {{\bf{n}} - {\bf{r}}} \right)!}}\)

Calculation

From 52 cards, 2 can be chosen in C(52, 2) ways.

There are 4 suits in a deck of 52 cards and each number is present in all the 4 suits.

There are 13 numbers form 1(Ace card) to 13(King card).

If we consider the number 1, then from 4 Aces, 2 can be chosen in C(4, 2) ways.

There are 13 numbers in total so, total ways in which two same numbers can be chosen is 13 × C(4, 2)

Required probability

\( = \frac{{13; \times ;{\rm{C}}\left( {4,{\rm{;}}2} \right)}}{{{\rm{C}}\left( {52,{\rm{;}}2} \right)}}\)

\(= \frac{{13; \times ;12}}{{52; \times ;51}}\)

\(= \frac{1}{{17}}\)

Correct option is (1).

Concept

When a fair dice is rolled, probability of getting any number is \(\frac{1}{6}\).

Suppose an experiment is repeated n times. Probability of success be p.

If X denote the number of success, then

P(X = k) = C(n, k) × p^k (1 – p)^{n - k} , 0 ≤ k ≤ n

where

\({\bf{C}}\left( {{\bf{n}},;{\bf{k}}} \right) = ;\frac{{{\bf{n}}!}}{{{\bf{k}}!\left( {{\bf{n}} - {\bf{k}}} \right)!}}\)

Mean of X = np

Standard deviation of X \(= \sqrt {{\bf{np}}\left( {1 - {\bf{p}}} \right)}\)

Calculation

A dice is rolled 8 times so, n = 8.

In a single roll of dice, probability of getting 5 or 6 is

= probability of getting 5 + probability of getting 6

\(= \frac{1}{6} + \frac{1}{6}\)

\(= \frac{1}{3}\)

So probability of success \({\bf{p}} = \frac{1}{3}\)

\({\bf{Mean}} = {\bf{n}} \times {\bf{p}} = \frac{8}{3}\)

\({\bf{Standard}};{\bf{deviation}} = \sqrt {{\bf{np}}\left( {1 - {\bf{p}}} \right)} = \frac{4}{3}\)

Correct option is (2).

Concept

If X and Y are two mutually exclusive events, then P(X ∩ Y) = 0.

If X and Y are two events, then the conditional probability \({\bf{P}}\left( {{\bf{X}}/{\bf{Y}}} \right) = \frac{{{\bf{P}}\left( {{\bf{X}}; \cap ;{\bf{Y}}} \right)}}{{{\bf{P}}\left( {\bf{Y}} \right)}}\)

De Morgan’s Law

\(\overline {{\rm{A}} \cup {\rm{B}}} ; = {\rm{\bar A}} \cap {\rm{\bar B}}\)

\(\overline {{\rm{A}} \cap {\rm{B}}} = ;{\rm{\bar A}} \cup {\rm{\bar B}}\)

Calculation

Given A̅ and B̅ are mutually exclusive so P(A̅ ∩ B̅) = 0

Now,

P(A̅ ∪ B̅) = P(A̅) + P(B̅) - P(A̅ ∩ B̅)

\( \Rightarrow {\rm{P}}\left( {\overline {{\rm{A}} \cap {\rm{B}}} } \right) = 1 - {\rm{P}}\left( {\rm{A}} \right) + 1 - {\rm{P}}\left( {\rm{B}} \right)\)

⇒ 1 - P(A ∩ B) = 1  -P(A) + 1 - P(B)

⇒ P(A ∩ B) = P(A) + P(B) - 1

⇒ P(A ∩ B) = 0.1

The required conditional probability is \({\rm{P}}\left( {{\rm{A}}|{\rm{B}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A;}} \cap {\rm{;B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}} = \frac{1}{6}\)

Correct option is (2).

Calculation

Suppose we have n observations X₁, X₂, ….., X_n.

The arithmetic mean \({\rm{\bar X}} = \frac{1}{{\rm{n}}}\mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {{\rm{X}}_{\rm{i}}}\)

Sum of deviations from arithmetic mean

\(= \mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} \left( {{{\rm{X}}_{\rm{i}}} - {\rm{\bar X}}} \right)\)

\( = \mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {{\rm{X}}_{\rm{i}}} - \mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {\rm{\bar X}}\)

\(= {\rm{n\bar X}} - {\rm{n\bar X}}\)

Statement 1 is correct.

For a symmetric distribution mean, median and mode are equal i.e mean = median = mode. Hence, statement 2 is false.

Concept

Let Z = aX + bY.

Expectation of Z = E(Z) = aE(X) + bE(Y)

Variance of Z = Var(Z) = a²Var(X) + b²Var(Y) + 2abρ_XYσ_Xσ_Y

ρ_XY = Correlation coefficient of X and Y.

\({{\rm{\rho }}_{{\rm{XY}}}} = \frac{{{\rm{Cov}}\left( {{\rm{X}},{\rm{;Y}}} \right)}}{{{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}}}\)

Covariance of X and Y = Cov(X, Y) = E(XY) - E(X)E(Y)

\({{\rm{\sigma }}_{\rm{X}}} = \sqrt {{\rm{Var}}\left( {\rm{X}} \right)} \)

\({{\rm{\sigma }}_{\rm{Y}}} = \sqrt {{\rm{Var}}\left( {\rm{Y}} \right)}\)

If Z = X + a where a is some constant, then Var(Z) = Var(X)

Covariance of X and Y = Cov(X, Y) = E(XY) - E(X)E(Y)

Calculation

Z = X + 5

⇒ Var(Z) = Var(X)

⇒ σ_Z = σ_X

\({\rm{W}} = \frac{{\rm{Y}}}{3}\)

\( \Rightarrow {\rm{Var}}\left( {\rm{W}} \right) = \frac{{{\rm{Var}}\left( {\rm{Y}} \right)}}{9}\)

\(\Rightarrow {\rm{;}}{{\rm{\sigma }}_{\rm{W}}} = \frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{3}\)

Cov(Z, W)

= E(ZW) – E(Z)E(W)

\(= {\rm{E}}\left( {\frac{{{\rm{XY}}}}{3} + \frac{{5{\rm{Y}}}}{3}} \right) - {\rm{E}}\left( {{\rm{X}} + 5} \right){\rm{E}}\left( {\frac{{\rm{Y}}}{3}} \right)\)

\(= \frac{1}{3}{\rm{E}}\left( {{\rm{XY}}} \right) + \frac{5}{3}{\rm{E}}\left( {\rm{Y}} \right) - \frac{1}{3}{\rm{E}}\left( {\rm{X}} \right){\rm{E}}\left( {\rm{Y}} \right) - \frac{5}{3}{\rm{E}}\left( {\rm{Y}} \right)\)

\( = \frac{1}{3}\left( {{\rm{E}}\left( {{\rm{XY}}} \right) - {\rm{E}}\left( {\rm{X}} \right){\rm{E}}\left( {\rm{Y}} \right)} \right)\)

\(= {\rm{;}}\frac{1}{3}{\rm{Cov}}\left( {{\rm{X}},{\rm{;Y}}} \right)\)

Correlation coefficient of Z and W is

\({{\rm{\rho }}_{{\rm{ZW}}}} = \frac{{{\rm{Cov}}\left( {{\rm{Z}},{\rm{;W}}} \right)}}{{{{\rm{\sigma }}_{\rm{Z}}}{{\rm{\sigma }}_{\rm{W}}}}}\)

\(= \frac{{\frac{1}{3}{\rm{Cov}}\left( {{\rm{X}},{\rm{;Y}}} \right)}}{{{{\rm{\sigma }}_{\rm{X}}}\frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{3}}}\)

\(= \frac{{{\rm{Cov}}\left( {{\rm{X}},{\rm{;;Y}}} \right)}}{{{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}}}\)

= ρ_XY

So ρ_ZW = ρ_XY = 0.6

Correct option is (4).

Concept

Sum of first n natural numbers is = \(\frac{{{\bf{n}}\left( {{\bf{n}} + 1} \right)}}{2}\)

Sum of square of first n natural numbers = \(\frac{{{\bf{n}}\left( {{\bf{n}} + 1} \right)\left( {2{\bf{n}} + 1} \right)}}{6}\)

Mean of first n natural numbers = \(\frac{1}{{\rm{n}}} \times {\rm{Sum;of;first;n;natural;numbers}} = \frac{{{\rm{n}} + 1}}{2}\)

If a constant k is multiplied with each of the first n natural numbers, then

\({\rm{Sum;is;}} = {\rm{k}}\frac{{{\rm{n}}\left( {{\rm{n}}; + ;1} \right)}}{2}\)

\({\rm{Mean;}} = {\rm{k}}\frac{{{\rm{n}}; + ;1}}{2}\)

The Variance is given as = \(\frac{1}{n}\left( {\mathop \sum \limits_{i = 1}^{i = n} {{\left( {i - {\rm{Mean}}} \right)}^2}} \right)\)

Calculation

Given that each number between 1 and 20 is multiplied by 3 so, the mean is

\(= 3\left( {\frac{{20; + ;1}}{2}} \right)\)

\(= \frac{{63}}{2}\)

The variance will be \(\frac{1}{{20}}\left( {\mathop \sum \limits_{i = 1}^{i = 20} {{\left( {3i - \frac{{63}}{2}} \right)}^2}} \right)\)

\(= \frac{1}{{20}}\mathop \sum \limits_{i = 1}^{i = 20} {\left( {3i - \frac{{63}}{2}} \right)^2}\)

\(= \frac{1}{{20}}\mathop \sum \limits_{i = 1}^{i = 20} \left( {9{i^2} + \frac{{3969}}{4} - 189i} \right)\)

\(= \frac{1}{{20}}\left[ {9\mathop \sum \limits_{i = 1}^{i = 20} {i^2} + \mathop \sum \limits_{i = 1}^{i = 20} \frac{{3969}}{4} - 189\mathop \sum \limits_{i = 1}^{i = 20} i} \right] \)

\(= \frac{1}{{20}}\left[ {9 \times \frac{{20 \times 21 \times 41}}{6} + 20 \times \frac{{3969}}{4} - 189 \times \frac{{20 \times 21}}{2}} \right]$ \)

\(= \frac{1}{{20}}\left( {25830 + 19845 - 39690} \right)\)

= 299.25

Correct option is (3).$

Calculation

Consider two concentric circles as shown in the following figure:

Radius of the inner circle \(= {\rm{}}\frac{{\rm{r}}}{2}\)

Radius of the outer circle = r

Points lying inside the inner circle will definitely be closer to the centre than the circumference of the outer circle.

Possible region where these points lie = area of the inner circle = \(\frac{{{\rm{\pi }}{{\rm{r}}^2}}}{4}\)

Required probability \(= \frac{{{\rm{area;of;the;inner;circle}}}}{{{\rm{total;area}}}}\)

\(= \frac{{{\rm{\pi }}{{\rm{r}}^2}}}{{4; \times ;{\rm{\pi }}{{\rm{r}}^2}}}\)

\(= \frac{1}{4}\)

Correct option is (1).

Concept

If A and B are two mutually exclusive events, then P(A ∩ B) = 0.

Calculation

Probability of occurrence of either event A or event B = P(A ∪ B)

Now,

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(A ∪ B) = P(A) + P(B)

Correct option is (1).