Probability and Statistics Previous Year Questions with Solutions

1.0Introduction

Probability and Statistics previous year questions typically cover basic probability, conditional probability, Bayes' theorem, random variables, binomial distribution, mean, median, variance, and standard deviation. Examples include finding the probability of drawing certain colored balls, calculating conditional probability, determining expected values, and analyzing binomial experiments like coin tosses. Solutions involve applying formulas such as $P(E)=\frac{\text{favorable outcomes}}{\text{total outcomes}}$ , Bayes' theorem, and using binomial coefficients. Questions on statistics require calculating mean, variance, and standard deviation from given data. Regular practice of these problems strengthens understanding, sharpens calculation skills, and builds speed and accuracy for JEE exams.

2.0Probability and Statistics: Previous Year Questions for JEE with Solutions

Probability and Statistics is a vital part of the JEE Mathematics syllabus. It combines logical reasoning with numerical calculations to predict outcomes and understand data. Questions from this chapter often appear in both JEE Main and JEE Advanced, sometimes directly formula-based and sometimes conceptual.

3.0Key Concepts to Remember

Probability

• Trial: Performing an experiment.
• Sample Space (S): Set of all possible outcomes.
• Event: A subset of the sample space.
• Classical Probability:

$P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

• Conditional Probability:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$

• Independent Events: Events where occurrence of one does not affect the other.
• Bayes' Theorem: A formula that describes the probability of an event, based on prior knowledge of conditions that might be related to the event. 

4.0Probability Formula for JEE 

1. Basic Probability 

$P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

where:

• P(E) = probability of event E
• 0 ≤ P(E) ≤ 1

2. Complementary Event

P(E′) = 1 − P(E) 

where E′ is the event "E does not happen."

3. Addition Theorem

• For any two events A and B:

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

• If A and B are mutually exclusive (they cannot happen together):

$P(A\cup B)=P(A)+P(B)$

4. Conditional Probability

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}\text{(provided }P(B)\ne 0\text{)}$

Meaning: Probability of A happening given that B has already happened.

5. Multiplication Theorem

• For any two events A and B:

$P(A\cap B)=P(A)\times P(B\mid A)$

or

$P(A\cap B)=P(B)\times P(A\mid B)$

• If A and B are independent events:

$P(A\cap B)=P(A)\times P(B)$

6. Bayes' Theorem

(important for conditional probability problems) 

$P(A\mid B)=\frac{P(B\mid A)\times P(A)}{P(B)}$

where:

$P(B)=P(B\mid A)P(A)+P(B\mid A^{\prime })P(A^{\prime })$

7. Independent Events

• Two events A and B are independent if:

$P(A\cap B)=P(A)\times P(B)$

• Also:

$P(A\mid B)=P(A),P(B\mid A)=P(B)$

8. Total Probability Theorem

If $E_1,E_2,E_3,\dots ,E_n$ are mutually exclusive and exhaustive events:

$P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+\dots +P(E_n)P(A\mid E_n)$

9. Odds in Favour and Against

$$\begin{gathered} \text{Odds in favour of E =}\frac{P(E)}{1-P(E)} \\ \text{Odds against E =}\frac{1-P(E)}{P(E)} \end{gathered}$$

10. Important Shortcuts for Problems

Tossing a coin:

• Sample space: $2^n$ outcomes for n tosses

Throwing a die:

• Sample space: $6^n$ outcomes for n dice

Drawing cards:

• 52 cards total
• 13 cards per suit (hearts, diamonds, clubs, spades)

Statistics

• Mean (Average): Sum of observations divided by the number of observations.
• Median: Middle value in ordered data.
• Mode: Most frequently occurring value.
• Variance: Measure of how data points differ from the mean.
• Standard Deviation (σ):

$\sigma =\sqrt{\text{Variance}}$

• Binomial Distribution: A probability distribution that summarizes the likelihood that a value will take one of two independent states under a given number of parameters.

5.0Statistics Formula for JEE

1. Range

Range = Largest observation − Smallest observation 

2. Mean Deviation (M.D.)

a) For Ungrouped Data

• About Mean $(\bar{X})$:

$\text{M.D.}(\bar{X})=\frac{1}{n}{\sum }_{i=1}^n|X_i-\bar{X}|$

• About Median (M):

$\text{M.D.}(M)=\frac{1}{n}{\sum }_{i=1}^n|X_i-M|$

b) For Discrete Frequency Distribution

• About Mean $(\bar{X})$:

$\text{M.D.}(\bar{X})=\frac{1}{N}{\sum }_{i=1}^n{f}_i\left|x_i-\bar{X}\right|$

• About Median (M):

$\text{M.D.}(M)=\frac{1}{N}{\sum }_{i=1}^n{f}_i\left|x_i-M\right|$

c) For Continuous Frequency Distribution

• About Mean $(\bar{X})$:

$\text{M.D.}(\bar{X})=\frac{1}{N}{\sum }_{i=1}^n{f}_i\left|x_i-\bar{X}\right|$

• About Median (M): 

$\text{M.D.}(M)=\frac{1}{N}{\sum }_{i=1}^n{f}_i\left|x_i-M\right|$

Note: Here, $x_i$ represents the midpoints of the class intervals.

3. Variance (σ²) and Standard Deviation (σ)

a) For Ungrouped Data

• Variance:

${\sigma }^2=\frac{1}{n}{\sum }_{i=1}^n(X_i-\bar{X}{)}^2$

• Standard Deviation:

$\sigma =\sqrt{{\sigma }^2}$

b) For Discrete Frequency Distribution

• Variance:

${\sigma }^2=\frac{1}{N}{\sum }_{i=1}^n{f}_i(X_i-\bar{X}{)}^2$

• Standard Deviation:

$\sigma =\sqrt{{\sigma }^2}$

c) For Continuous Frequency Distribution

• Variance:

${\sigma }^2=\frac{1}{N}{\sum }_{i=1}^n{f}_i(x_i-\bar{X}{)}^2$

• Standard Deviation:

$\sigma =\sqrt{{\sigma }^2}$

Note: An alternative formula for variance:

${\sigma }^2=\frac{1}{N}{\sum }_{i=1}^n{f}_i{x}_i^2-(\bar{X}{)}^2$

4. Standard Deviation Using Assumed Mean Method

• Formula:

$\sigma =h\times \sqrt{\left(\frac{1}{N}{\sum }_{i=1}^n{f}_i{y}_i^2\right)-{\left(\frac{1}{N}{\sum }_{i=1}^n{f}_i{y}_i\right)}^2}$

 Where:

• h = class width

$y_i=\frac{x_i-A}{h}$

• A = assumed mean

5. Coefficient of Variation (C.V.)

Formula:

$\text{C.V.}=\frac{\sigma }{\bar{X}}\times 100$

6.0Previous Year Questions with Solution on Probability

1. In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $\frac{1}{3}and\frac{2}{3}$ respectively. Let x be the number of matches that the team wins, and y be the number of matches that team loses. If the probability $P(|x–y|<2)$ is p, then 3⁹p equals……

Ans. (8288)

Sol.

$$\begin{gathered} P(W)=\frac{1}{3} \\ P(L)=\frac{2}{3} \end{gathered}$$

x = number of matches that team wins

y = number of matches that team loses

and x + y = 10

x, y ∈ N

Case-I :

$$\begin{gathered} |x-y|\le 2\text{and}x+y=10 \\ |x-y|=0,1,2\text{where}x,y\in \mathbb{N} \\ Case-I:|x-y|=0\Rightarrow x=y \\ x+y=10\Rightarrow x=y=5 \\ P(|x-y|=0)={}^{10}{C}_5{\left(\frac{1}{3}\right)}^5{\left(\frac{2}{3}\right)}^5 \end{gathered}$$

Case-II :

$|x-y|=1\Rightarrow x-y=\pm 1$

Case-III :

$$\begin{gathered} |x-y|=2\Rightarrow x-y=\pm 2 \\ x-y=2\Rightarrow x+y=10\Rightarrow x=6,y=4 \\ x-y=-2\Rightarrow x+y=10\Rightarrow x=4,y=6 \\ P(|x-y|=2)={}^{10}{C}_6{\left(\frac{1}{3}\right)}^6{\left(\frac{2}{3}\right)}^4+{}^{10}{C}_4{\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^6 \\ p={}^{10}{C}_5\frac{2^5}{3^{10}}+{}^{10}{C}_6\frac{2^4}{3^{10}}+{}^{10}{C}_4\frac{2^6}{3^{10}} \\ {3}^{9}p=\frac{1}{3}\left({}^{10}{C}_5{2}^5+{}^{10}{C}_6{2}^4+{}^{10}{C}_4{2}^6\right) \\ =8288 \end{gathered}$$

2. Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

$(1)\frac{4}{17}(2)\frac{5}{18}(3)\frac{7}{18}(4)\frac{5}{16}$

Ans. (2)

Sol.

$$\begin{gathered} \begin{array}{c}A\\ 7R5B\end{array}\begin{array}{c}B\\ 5R7B\end{array}\begin{array}{c}C\\ 5R7B\end{array} \\ P(B)=\frac{1}{3}\cdot \frac{5}{12}+\frac{1}{3}\cdot \frac{7}{12}+\frac{1}{3}\cdot \frac{6}{12} \\ \text{Required probability}=\frac{\frac{1}{3}\cdot \frac{5}{12}}{\frac{1}{3}\left(\frac{5}{12}+\frac{7}{12}+\frac{6}{12}\right)}=\frac{5}{18} \end{gathered}$$

3. A company has two plants A and B to manufacture motorcycles.  60% motorcycles are manufactured at plant A and the remaining are manufactured at plant B. 80% of the motorcycles manufactured at plant A are rated of the standard quality, while 90% of the motorcycles manufactured at plant B are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If p is the probability that it was manufactured at plant B, then 126p is    

(1) 54

(2) 64

(3) 66

(4) 56

Ans. (1)

Sol.

P(Manufactured at B / found standard quality) = ?

A : Found S.Q

B : Manufacture B

C : Manufacture A

$$\begin{gathered} P(E_1)=\frac{40}{100},P(E_2)=\frac{60}{100} \\ P(A\mid E_1)=\frac{90}{100},P(A\mid E_2)=\frac{80}{100} \\ P(E_1\mid A)=\frac{P(A\mid E_1)\cdot P(E_1)}{P(A\mid E_1)\cdot P(E_1)+P(A\mid E_2)\cdot P(E_2)}=\frac{3}{7} \\ \therefore 126P=54 \end{gathered}$$

4. Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables X and Y respectively denote the number of blue and Yellow balls. If $\bar{X}and\bar{Y}$ are the means of X and Y respectively, then $7\bar{X}+4\bar{Y}$ is equal to ______.

Ans. (17)

Sol.

$$\begin{gathered} 7\bar{X}=\left(\frac{{}^5{C}_1\cdot {}^4{C}_2{+}^5{C}_2\cdot {}^4{C}_1\cdot 2{+}^5{C}_3\cdot {}^4{C}_0\cdot 3}{{}^9{C}_3}\right)\cdot 7 \\ =\frac{30+80+30}{84}\cdot 7=\frac{140}{12}=\frac{70}{6}=\frac{35}{3} \end{gathered}$$

$4\bar{Y}=\left(\frac{40+60+12}{84}\right)\times 4=\frac{112}{21}=\frac{16}{3}$

5. Let a, b and c denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that ax² + bx + c = 0 has all real roots is $\frac{m}{n}$, gcd(m, n) = 1, then m + n is equal to ________.

Ans. (19)

Sol. a, b, c ∈ {1, 2, 3, 4}

Tetrahedral dice

$$\begin{gathered} a{x}^2+bx+c=0 \\ \text{has all real roots} \\ \Rightarrow D\ge 0 \\ \Rightarrow b^2-4ac\ge 0 \\ \text{Let }b=1\Rightarrow 1-4ac\ge 0\text{(Not feasible)} \\ b=2\Rightarrow 4-4ac\ge 0\Rightarrow 1\ge ac\Rightarrow a=1,c=1 \\ b=3\Rightarrow 9-4ac\ge 0\Rightarrow \frac{9}{4}\ge ac \\ \Rightarrow a=1,c=1 \\ \Rightarrow a=1,c=2 \\ \Rightarrow a=2,c=1 \\ \mathbf{b}=4\Rightarrow 16-4ac\ge 0 \\ \Rightarrow 4\ge ac \\ \Rightarrow a=1,c=1 \\ \Rightarrow a=1,c=2\Rightarrow a=2,c=1 \\ \Rightarrow a=1,c=3\Rightarrow a=3,c=1 \\ \Rightarrow a=1,c=4\Rightarrow a=4,c=1 \\ \Rightarrow a=2,c=2 \\ \text{Probability}=\frac{12}{(4)(4)(4)}=\frac{3}{16}=\frac{m}{n} \\ \Rightarrow m+n=3+16=\boxed{19} \end{gathered}$$

6. A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

$(1)\frac{2}{5}(2)\frac{2}{7}(3)\frac{1}{3}(4)\frac{1}{5}$

Ans. (2)

Sol.

$$\begin{gathered} P(4W4B\mid 2W2B)=\frac{P(4W4B)\times P(2W2B\mid 4W4B)}{P(2W6B)\times P(2W2B\mid 2W6B)+P(3W5B)\times P(2W2B\mid 3W5B)+\dots +P(6W2B)\times P(2W2B\mid 6W2B)} \\ =\frac{\frac{1}{5}\times \frac{{}^4{C}_2\cdot {}^4{C}_2}{{}^8{C}_4}}{\frac{1}{5}\times \frac{{}^2{C}_2\cdot {}^6{C}_2}{{}^8{C}_4}+\frac{1}{5}\times \frac{{}^3{C}_2\cdot {}^5{C}_2}{{}^8{C}_4}+\dots +\frac{1}{5}\times \frac{{}^6{C}_2\cdot {}^2{C}_2}{{}^8{C}_4}} \\ =\frac{2}{7} \end{gathered}$$

7. A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required and let a = P(X = 3), b = P(X ≥ 3) and c = P(X ≥ 6 |X > 3). Then $\frac{b+c}{a}$ is equal to _____.

Ans. (12)

Sol.

$$\begin{gathered} a=P(X=3)=\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=\frac{25}{216} \\ b=P(X\ge 3)=\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^3\cdot \frac{1}{6}+{\left(\frac{5}{6}\right)}^4\cdot \frac{1}{6}+\dots \\ =\frac{25}{216}\cdot \frac{1}{1-\frac{5}{6}}=\frac{25}{216}\cdot 6=\frac{25}{36} \\ P(X\ge 6)={\left(\frac{5}{6}\right)}^5\cdot \frac{1}{6}+{\left(\frac{5}{6}\right)}^6\cdot \frac{1}{6}+{\left(\frac{5}{6}\right)}^7\cdot \frac{1}{6}+\dots \\ ={\left(\frac{5}{6}\right)}^5\cdot \frac{1}{6}\cdot \left(\frac{1}{1-\frac{5}{6}}\right) \\ ={\left(\frac{5}{6}\right)}^5 \\ c=\frac{{\left(\frac{5}{6}\right)}^5}{{\left(\frac{5}{6}\right)}^3}={\left(\frac{5}{6}\right)}^2=\frac{25}{36} \\ \frac{b+c}{a}=\frac{{\left(\frac{5}{6}\right)}^2+{\left(\frac{5}{6}\right)}^2}{{\left(\frac{5}{6}\right)}^2\cdot \frac{1}{6}}=12 \end{gathered}$$

8. A fair die is thrown until 2 appears. Then the probability that 2 appears in an even number of throws is  

$(1)\frac{5}{6}(2)\frac{1}{6}(3)\frac{5}{11}(4)\frac{6}{11}$

Ans. (3)

Sol. Required probability =

$$\begin{gathered} \frac{5}{6}\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^3\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^5\times \frac{1}{6}+\dots \\ =\frac{1}{6}\times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11} \end{gathered}$$

9. Bag A contains 3 white, 7 red balls, and Bag B contains 3 white, 2 red balls. One bag is selected at random, and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

$$\begin{gathered} (1)\frac{1}{4} \\ (2)\frac{1}{9} \\ (3)\frac{1}{3} \\ (4)\frac{3}{10} \end{gathered}$$

Ans. (3)

Sol.

$$\begin{gathered} E_1:\text{A is selected} \\ {E}_2:\text{B is selected} \\ E:\text{white ball is drawn} \\ P(E_1|E)= \\ P(E_1|E)=\frac{P(E_1)\cdot P(E|E_1)}{P(E_1)\cdot P(E|E_1)+P(E_2)\cdot P(E|E_2)} \\ =\frac{\frac{1}{2}\times \frac{3}{10}}{\frac{1}{2}\times \frac{3}{10}+\frac{1}{2}\times \frac{3}{5}} \\ =\frac{3}{3+6}=\frac{1}{3} \end{gathered}$$

10. A coin is based so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-

$$\begin{gathered} (1)\frac{2}{9} \\ (2)\frac{1}{9} \\ (3)\frac{2}{27} \\ (4)\frac{1}{27} \end{gathered}$$

Ans. (1)

Sol.

$$\begin{gathered} \text{Let probability of tail}=\frac{1}{3} \\ \Rightarrow \text{Probability of head}=\frac{2}{3} \\ \text{Probability of getting 2 tails and 1 head} \\ =\left(\frac{1}{3}\times \frac{2}{3}\times \frac{1}{3}\right)\times 3 \\ =\frac{2}{27}\times 3=\frac{2}{9} \end{gathered}$$

11. Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable x to be the number of rotten apples in a draw of two apples, the variance of x is 

$$\begin{gathered} (1)\frac{37}{153} \\ (2)\frac{57}{153} \\ (3)\frac{47}{153} \\ (4)\frac{40}{153} \end{gathered}$$

Ans. (4)

Sol.

$$\begin{gathered} \text{3 bad apples, 15 good apples} \\ \text{Let }X\text{ be the number of bad apples} \\ P(X=0)=\frac{{}^{15}{C}_2}{{}^{18}{C}_2}=\frac{105}{153} \\ P(X=1)=\frac{{}^3{C}_1\times {}^{15}{C}_1}{{}^{18}{C}_2}=\frac{45}{153} \\ P(X=2)=\frac{{}^3{C}_2}{{}^{18}{C}_2}=\frac{3}{153}E(X)=0\times \frac{105}{153}+1\times \frac{45}{153}+2\times \frac{3}{153}=\frac{51}{153} \\ =\frac{1}{3} \\ \text{Var}(X)=E(X^2)-(E(X){)}^2 \\ =0\times \frac{105}{153}+1\times \frac{45}{153}+4\times \frac{3}{153}-{\left(\frac{1}{3}\right)}^2 \\ =\frac{57}{153}-\frac{1}{9}=\frac{40}{153} \end{gathered}$$

7.0Previous Year Questions with Solutions on Statistics

1. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On respectively, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is  

(1) $\sqrt{3.86}$

(2) 1.8

(3) $\sqrt{3.96}$

(4) 1.94

Ans. (3)

Sol.

$$\begin{gathered} Mean\bar{x}=10 \\ \Rightarrow \frac{\sum x_i}{20}=10 \\ \Rightarrow \sum x_i=10\times 20=200 \\ \text{If 8 is replaced by 12, then:} \\ \sum x_i=200-8+12=204 \\ \text{Correct mean }\bar{x}=\frac{204}{20}=10.2 \\ \text{Standard deviation}=2\Rightarrow \text{Variance}=(S.D.{)}^2=2^2=4 \\ \Rightarrow \frac{\sum x_i^2}{20}-{\left(\frac{\sum x_i}{20}\right)}^2=4 \\ \Rightarrow \frac{\sum x_i^2}{20}-(10{)}^2=4 \\ \Rightarrow \frac{\sum x_i^2}{20}=104\Rightarrow \sum x_i^2=2080 \\ \text{Now, replace ’8’ with ’12:} \\ \sum x_i^2=2080-8^2+12^2=2080-64+144=2160 \\ \text{Variance of corrected data}=\frac{2160}{20}-(10.2{)}^2 \\ =108-104.04=3.96 \\ \text{Correct standard deviation:} \\ =\sqrt{3.96} \end{gathered}$$

2. Let a, b, c ∈ N and a < b < c. Let the mean, the mean deviation about the mean and the variance of the 5 observations 9, 25, a, b, c be 18, 4 and $\frac{136}{5}$, respectively. Then 2a + b – c is equal to _______ .

Ans. (33)

Sol. a, b, c ∈ N a < b < c

$\bar{x}=\text{mean}=\frac{9+25+a+b+c}{5}=18$

a + b + c = 56

$\text{Mean deviation}=\frac{\sum |x_i-\bar{x}|}{n}=4$

= 9 + 7 + |18 – a| + |18 – b| + |18 – c| = 20

= |18 – a| + |18 – b| + |18 – c| = 4

$\text{Variance}=\frac{\sum (x_i-\bar{x}{)}^2}{n}=\frac{136}{5}$

= 81 + 49 + |18 – a|² + |18 – b|² + |18 – c|² = 136

= (18 – a)² + (18 – b)² + (18 – c)² = 6

Possible values (18 – a)² = 1, (18 – b)² = 1, (18 – c)² = 4

a < b < c

so 18 – a = 1     18 – b = –1       18 – c = –2

a = 17   b = 19 c = 20

a + b + c = 56

2a + b – c = 34 + 19 – 20 = 33

3. The frequency distribution of the age of students in a class of 40 students is given below.

If the mean deviation about the median is 1.25, then 4x + 5y is equal to :

(1) 43

(2) 44

(3) 47

(4) 46

Ans. (2)

Sol. x + y = 10 .........(1)

Median = 18 = M

$$\begin{gathered} \text{M.D.}=\frac{\sum f_i|x_i-M|}{\sum f_i} \\ 1.25=\frac{36+x+2y}{40} \end{gathered}$$

x + 2y = 14 .........(2)

by (1) & (2)

x = 6, y = 4

⇒ 4x + 5y = 24 + 20 = 44

4. Consider 10 observation x₁, x_2,…., x₁₀. such that ${\sum }_{i=1}^{10}(x_i-\alpha )=2$  and ${\sum }_{i=1}^{10}(x_i-\beta {)}^2=40,$ where α, β are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}and\frac{84}{25}$ respectively. The $\frac{\beta }{\alpha }$ is equal to :  

(1) 2

(2) $\frac{3}{2}$

(3) $\frac{5}{2}$

4) 1

Ans. (1)

Sol.

$$\begin{gathered} x_1,x_2,\dots ,x_{10} \\ {\sum }_{i=1}^{10}(x_i-\alpha )=2\Rightarrow {\sum }_{i=1}^{10}{x}_i-10\alpha =2 \\ Mean\mu =\frac{6}{5}=\frac{\sum x_i}{10}\Rightarrow \sum x_i=12 \\ 12-10\alpha =2\Rightarrow 10\alpha =10\Rightarrow \alpha =1 \\ Now,consider: \\ {\sum }_{i=1}^{10}(x_i-\beta {)}^2=40 \\ Let{y}_i=x_i-\mu \\ {\sigma }_x^2=\frac{1}{10}\sum y_i^2=\frac{1}{10}\sum (x_i-\beta {)}^2-{\left(\frac{\sum (x_i-\beta )}{10}\right)}^2 \\ {\sigma }_x^2=\frac{84}{25}=4-{\left(\frac{12-10\beta }{10}\right)}^2 \\ \Rightarrow {\left(\frac{6-5\beta }{5}\right)}^2=4-\frac{84}{25}=\frac{16}{25} \\ \Rightarrow 6-5\beta =\pm 4\Rightarrow \beta =\frac{2}{5}\text{ (not possible)},\beta =2\text{ (valid)} \\ \Rightarrow \frac{\beta }{\alpha }=\frac{2}{1}=2 \end{gathered}$$

5. The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If µ and σ² denote the mean and variance of the correct observations respectively, then $15(\mu +{\mu }^2+{\sigma }^2)$ is equal to ……………………..

Ans. (2521)

Sol. Let the incorrect mean be $\mu$ and standard deviation be $\sigma$

$$\begin{gathered} {\mu }^{\prime }=\frac{\sum x_i}{15}=12\Rightarrow \sum x_i=180 \\ \text{As per given information, correct}\sum x_i=180-10+12=182 \\ \Rightarrow \mu =\text{(correct mean)}=\frac{182}{15} \\ \text{Also,} \\ {\sigma }^{\prime }=\sqrt{\frac{\sum x_i^2}{15}-144}=3\Rightarrow \frac{\sum x_i^2}{15}=9+144=153\Rightarrow \sum x_i^2=2295 \\ \text{Correct value:} \\ \sum x_i^2=2295-100+144=2339 \\ \text{Therefore,} \\ {\sigma }^2=\text{(correct variance)}=\frac{2339}{15}-\frac{182\times 182}{15\times 15} \\ \text{Required value:} \\ =15(\mu +{\mu }^2+{\sigma }^2) \\ =15\left(\frac{182}{15}+\frac{182\times 182}{15\times 15}+\frac{2339}{15}-\frac{182\times 182}{15\times 15}\right) \\ =15\left(\frac{182}{15}+\frac{2339}{15}\right)=2521 \end{gathered}$$

6. Let the mean and the variance of 6 observation a, b, 68, 44, 48, 60 be 55 and 194, respectively if a > b, then a + 3b is

(1) 200

(2) 190

(3) 180

(4) 210

Ans. (3)

Sol.-

$$\begin{gathered} \text{Data: }a,b,68,44,48,60 \\ \text{Mean}=55,\text{Variance}=194,a>b,a+3b \\ \text{Mean: }\frac{a+b+68+44+48+60}{6}=55 \\ a+b+220=55\times 6=330\Rightarrow a+b=110 \\ \text{Variance}=\frac{\sum x_i^2}{n}-(\bar{x}{)}^2=194 \\ \frac{a^2+b^2+68^2+44^2+48^2+60^2}{6}-55^2=194 \\ {a}^2+b^2+4624+1936+2304+3600=6\times (3025+194) \\ {a}^2+b^2+12484=6\times 3219=19314\Rightarrow a^2+b^2=19314-12484=6850 \\ Now, \\ (a+b{)}^2=110^2=12100 \\ {a}^2+b^2+2ab=12100\Rightarrow 6850+2ab=12100 \\ 2ab=12100-6850=5250\Rightarrow ab=2625 \\ \Rightarrow (110-b)b=2625(a=110-b) \\ {b}^2=110b+2625=0 \\ b=35,75 \\ buta>b \\ a+b=110 \\ Sob=35anda=75 \\ Now \\ \Rightarrow a+3b \\ \Rightarrow 75+3(35) \\ \Rightarrow 180 \end{gathered}$$