Statistics & Probability
Complete JEE preparation for statistics and probability — measures of central tendency, dispersion, classical and conditional probability, Bayes' theorem, binomial distribution, and random variables.
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01
Mean, Median & Mode
Measures of central tendency — arithmetic mean, median, and mode for ungrouped and grouped data, with JEE-specific shortcuts.
Measures of Central Tendency
Arithmetic Mean
For data $x_1, x_2, \ldots, x_n$: $\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i$.
For grouped data with frequencies: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
JEE Trick — Assumed Mean Method: If $d_i = x_i - a$, then $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$ — reduces computation significantly.
For grouped data with frequencies: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
JEE Trick — Assumed Mean Method: If $d_i = x_i - a$, then $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$ — reduces computation significantly.
Median
For sorted data of $n$ values: median is $x_{(n+1)/2}$ if $n$ is odd, or $\frac{x_{n/2}+x_{n/2+1}}{2}$ if $n$ is even.
For grouped data: $\text{Median} = l + \frac{(N/2 - F)}{f} \times h$, where $l$ = lower boundary of median class, $F$ = cumulative frequency before median class, $f$ = frequency of median class, $h$ = class width.
For grouped data: $\text{Median} = l + \frac{(N/2 - F)}{f} \times h$, where $l$ = lower boundary of median class, $F$ = cumulative frequency before median class, $f$ = frequency of median class, $h$ = class width.
Mode
The value with highest frequency. For grouped data: $\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$.
Empirical Relation
$\text{Mode} \approx 3\,\text{Median} - 2\,\text{Mean}$ (for moderately skewed distributions).
JEE Shortcut: If two of the three are known, quickly estimate the third using this relation.
JEE Shortcut: If two of the three are known, quickly estimate the third using this relation.
📝 Example
The mean of 5 observations is 4.4. If three of them are 1, 2, and 6, and two observations are equal, find them.
Sum $= 5 \times 4.4 = 22$. Let the two equal observations be $x$. Then $1+2+6+2x=22 \Rightarrow 2x=13 \Rightarrow x=6.5$. The observations are $6.5$ and $6.5$.
02
Variance & Standard Deviation
Measures of dispersion — variance, standard deviation, and their properties. Heavily tested in JEE Mains.
Variance & Standard Deviation
Variance
$\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$
Standard Deviation: $\sigma = \sqrt{\text{Variance}}$.
Standard Deviation: $\sigma = \sqrt{\text{Variance}}$.
JEE Shortcut — Variance Formula
$\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \left(\frac{\sum f_i x_i}{\sum f_i}\right)^2 = \overline{x^2} - (\bar{x})^2$.
Key: Variance = (mean of squares) $-$ (square of mean).
Key: Variance = (mean of squares) $-$ (square of mean).
Effect of Linear Transformation
If $y_i = ax_i + b$, then $\bar{y} = a\bar{x}+b$, $\sigma_y^2 = a^2\sigma_x^2$, $\sigma_y = |a|\sigma_x$.
JEE Trick: Adding a constant does not change variance. Multiplying by $a$ multiplies SD by $|a|$.
JEE Trick: Adding a constant does not change variance. Multiplying by $a$ multiplies SD by $|a|$.
Combined Mean & Variance
For two groups of sizes $n_1, n_2$ with means $\bar{x}_1, \bar{x}_2$ and variances $\sigma_1^2, \sigma_2^2$:
Combined mean: $\bar{x} = \frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}$.
Combined variance: $\sigma^2 = \frac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}$, where $d_i = \bar{x}_i-\bar{x}$.
Combined mean: $\bar{x} = \frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}$.
Combined variance: $\sigma^2 = \frac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}$, where $d_i = \bar{x}_i-\bar{x}$.
📝 Example
If the variance of $x_1, x_2, \ldots, x_n$ is $\sigma^2$, find the variance of $3x_1+5, 3x_2+5, \ldots, 3x_n+5$.
Using the transformation $y = 3x + 5$: $\sigma_y^2 = 3^2\sigma^2 = 9\sigma^2$. Adding 5 does not affect variance.
03
Probability — Classical, Addition & Multiplication
Classical definition, axioms, addition and multiplication rules — the core probability toolkit for JEE.
Classical Probability & Rules
Classical Probability
$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$, where all outcomes are equally likely. $0 \leq P(E) \leq 1$.
Addition Rule
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
For mutually exclusive events: $P(A \cup B) = P(A) + P(B)$.
Complement: $P(A') = 1 - P(A)$.
For mutually exclusive events: $P(A \cup B) = P(A) + P(B)$.
Complement: $P(A') = 1 - P(A)$.
Multiplication Rule
$P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)$.
For independent events: $P(A \cap B) = P(A) \cdot P(B)$.
JEE Trick: "At least one" problems — use complement: $P(\text{at least one}) = 1 - P(\text{none})$.
For independent events: $P(A \cap B) = P(A) \cdot P(B)$.
JEE Trick: "At least one" problems — use complement: $P(\text{at least one}) = 1 - P(\text{none})$.
De Morgan's Laws in Probability
$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
$P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)$.
$P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)$.
📝 Example
A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find $P(\text{both red})$.
$P = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \boxed{\frac{5}{14}}$.
04
Conditional Probability & Bayes' Theorem
Conditional probability, independence, and Bayes' theorem — JEE Advanced favourites with real-world applications.
Conditional Probability & Bayes' Theorem
Conditional Probability
$P(A|B) = \frac{P(A \cap B)}{P(B)}$, provided $P(B) > 0$.
Independence: $A$ and $B$ are independent iff $P(A|B) = P(A)$, equivalently $P(A \cap B) = P(A)\cdot P(B)$.
Independence: $A$ and $B$ are independent iff $P(A|B) = P(A)$, equivalently $P(A \cap B) = P(A)\cdot P(B)$.
Total Probability Theorem
If $B_1, B_2, \ldots, B_n$ partition the sample space:
$$P(A) = \sum_{i=1}^{n} P(B_i)\,P(A|B_i)$$
Bayes' Theorem
$$P(B_k|A) = \frac{P(B_k)\,P(A|B_k)}{\sum_{i=1}^{n} P(B_i)\,P(A|B_i)}$$
JEE Trick: Draw a tree diagram — multiply along branches for numerator, sum all branches reaching event $A$ for denominator.
📝 Example
Box I has 3 red and 4 black balls; Box II has 5 red and 6 black balls. A box is chosen at random and a ball drawn is red. Find $P(\text{Box I | red})$.
$P(I)=\frac{1}{2}$, $P(R|I)=\frac{3}{7}$; $P(II)=\frac{1}{2}$, $P(R|II)=\frac{5}{11}$.
$P(R) = \frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{5}{11} = \frac{3}{14}+\frac{5}{22} = \frac{33+35}{154} = \frac{68}{154} = \frac{34}{77}$.
$P(I|R) = \frac{\frac{1}{2}\cdot\frac{3}{7}}{\frac{34}{77}} = \frac{\frac{3}{14}}{\frac{34}{77}} = \frac{3}{14}\cdot\frac{77}{34} = \frac{231}{476} = \frac{33}{68}$.
$P(R) = \frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{5}{11} = \frac{3}{14}+\frac{5}{22} = \frac{33+35}{154} = \frac{68}{154} = \frac{34}{77}$.
$P(I|R) = \frac{\frac{1}{2}\cdot\frac{3}{7}}{\frac{34}{77}} = \frac{\frac{3}{14}}{\frac{34}{77}} = \frac{3}{14}\cdot\frac{77}{34} = \frac{231}{476} = \frac{33}{68}$.
05
Binomial Distribution & Random Variables
Discrete random variables, expectation, variance, and the binomial distribution — a must-know for JEE Mains.
Random Variables & Expectation
Random Variable
A random variable $X$ assigns a real number to each outcome. The probability distribution $P(X=x_i)=p_i$ satisfies $\sum p_i = 1$, $p_i \geq 0$.
Expectation & Variance
$E(X) = \sum x_i\,p_i = \mu$.
$\text{Var}(X) = E(X^2) - [E(X)]^2 = \sum x_i^2\,p_i - \mu^2$.
Properties: $E(aX+b)=aE(X)+b$; $\text{Var}(aX+b)=a^2\text{Var}(X)$.
$\text{Var}(X) = E(X^2) - [E(X)]^2 = \sum x_i^2\,p_i - \mu^2$.
Properties: $E(aX+b)=aE(X)+b$; $\text{Var}(aX+b)=a^2\text{Var}(X)$.
Binomial Distribution
Binomial Distribution $B(n, p)$
For $n$ independent trials, each with success probability $p$ (and $q = 1-p$):
$$P(X = r) = \binom{n}{r}p^r q^{n-r},\quad r = 0, 1, 2, \ldots, n$$
Mean, Variance & Mode
$E(X) = np$, $\text{Var}(X) = npq$, $\sigma = \sqrt{npq}$.
Mode: The most probable value of $X$ lies in $[(n+1)p - 1,\; (n+1)p]$.
JEE Trick: If $(n+1)p$ is an integer, there are two modes: $(n+1)p$ and $(n+1)p-1$.
Mode: The most probable value of $X$ lies in $[(n+1)p - 1,\; (n+1)p]$.
JEE Trick: If $(n+1)p$ is an integer, there are two modes: $(n+1)p$ and $(n+1)p-1$.
JEE Shortcut — Ratio of Consecutive Terms
$\frac{P(X=r+1)}{P(X=r)} = \frac{(n-r)p}{(r+1)q}$. The probabilities increase while this ratio $> 1$ and decrease after. Use this to quickly find the mode.
📝 Example
A coin is tossed 6 times. Find $P(\text{at least 2 heads})$.
$P(\text{at least 2}) = 1 - P(0) - P(1) = 1 - \binom{6}{0}\left(\frac{1}{2}\right)^6 - \binom{6}{1}\left(\frac{1}{2}\right)^6 = 1 - \frac{1}{64} - \frac{6}{64} = 1 - \frac{7}{64} = \boxed{\frac{57}{64}}$.
★ Key Takeaways
- Variance = (mean of squares) $-$ (square of mean) — the most useful computational formula for JEE.
- Adding a constant to all data does not change variance or SD; multiplying by $a$ multiplies SD by $|a|$.
- "At least one" problems are almost always easier with the complement: $1 - P(\text{none})$.
- For Bayes' theorem, draw a tree diagram — it makes the computation systematic and error-free.
- In binomial distribution, $E(X) = np$ and $\text{Var}(X) = npq$ — these are the two most tested formulas in JEE Mains.
✎ Practice Problems
Problem 1
The mean and variance of 8 observations are 9 and 9.25 respectively. If 6 of them are $6, 7, 10, 12, 12, 13$, find the remaining two observations.
Show Solution ▼
Let the remaining be $a, b$. Sum $= 72$, so $60+a+b=72 \Rightarrow a+b=12$. $\sum x_i^2 = 8(9.25+81) = 8(90.25) = 722$. Given six: $36+49+100+144+144+169=642$. So $a^2+b^2=80$. From $a+b=12$: $(a+b)^2 = a^2+2ab+b^2 = 144$, so $ab = 32$. $a, b$ are roots of $t^2-12t+32=0 \Rightarrow (t-4)(t-8)=0$. The observations are $4$ and $8$.
Problem 2
Three dice are thrown. Find the probability that the sum is at least 5.
Show Solution ▼
Total $= 6^3 = 216$. $P(\text{sum} < 5)$: sum $= 3$ (only $(1,1,1)$, 1 way); sum $= 4$ ($\binom{3}{1}$ ways: one die shows 2, rest show 1, $= 3$ ways). $P(\text{sum}<5) = \frac{4}{216}$. $P(\text{sum}\geq5)=1-\frac{4}{216}=\frac{212}{216}=\boxed{\frac{53}{54}}$.
Problem 3
A and B independently try to solve a problem. $P(A)=\frac{1}{3}$, $P(B)=\frac{1}{4}$. Find $P(\text{exactly one solves it})$.
Show Solution ▼
$P(\text{exactly one}) = P(A)P(B')+P(A')P(B) = \frac{1}{3}\cdot\frac{3}{4}+\frac{2}{3}\cdot\frac{1}{4} = \frac{3}{12}+\frac{2}{12} = \frac{5}{12}$.
Problem 4
In a binomial distribution $B(n,p)$, the mean is 4 and variance is 3. Find $n$ and $P(X \geq 1)$.
Show Solution ▼
$np=4$, $npq=3 \Rightarrow q=\frac{3}{4}$, $p=\frac{1}{4}$, $n=16$. $P(X\geq1)=1-P(0)=1-\binom{16}{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{16}=1-\left(\frac{3}{4}\right)^{16}$.
Problem 5
A factory has machines A, B, C producing 50%, 30%, 20% of output respectively. Defective rates are 3%, 4%, 5%. A product is found defective. Find $P(\text{from machine A})$.
Show Solution ▼
By Bayes' theorem: $P(A|D) = \frac{0.50\times0.03}{0.50\times0.03+0.30\times0.04+0.20\times0.05} = \frac{0.015}{0.015+0.012+0.010} = \frac{0.015}{0.037} = \frac{15}{37}$.
🎯 Interactive MCQs
1. If the variance of $1, 2, 3, \ldots, n$ is $\frac{n^2-1}{12}$, then the SD of $2, 4, 6, \ldots, 2n$ is:
A $\sqrt{\frac{n^2-1}{12}}$
B $\sqrt{\frac{n^2-1}{3}}$
C $\frac{n^2-1}{3}$
D $\frac{2(n^2-1)}{12}$
2. If $P(A) = 0.4$, $P(B) = 0.3$, and $P(A \cup B) = 0.6$, then $P(A|B)$ is:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{2}{3}$
3. The probability of a man hitting a target is $\frac{1}{4}$. If he fires 7 times, the probability of hitting at least twice is:
A $1 - \frac{7290}{16384}$
B $1 - \left(\frac{3}{4}\right)^7$
C $\binom{7}{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^5$
D $\frac{21 \cdot 9}{1024}$
4. If $E$ and $F$ are events with $P(E) = \frac{1}{4}$, $P(F) = \frac{1}{2}$, and $P(E \cap F) = \frac{1}{8}$, then $P(E' \cap F')$ equals:
A $\frac{1}{4}$
B $\frac{3}{8}$
C $\frac{5}{8}$
D $\frac{1}{2}$
5. A random variable $X$ has $P(X=0)=\frac{1}{4}$, $P(X=1)=\frac{1}{2}$, $P(X=2)=\frac{1}{4}$. Then $\text{Var}(X)$ is:
A $\frac{1}{2}$
B $1$
C $\frac{3}{2}$
D $\frac{1}{4}$