JEE Main Maths Statistics & Probability 2027: Standard Deviation, Bayes' Theorem & Distributions

This chapter pairs two ideas that feel different but share the same core skill: summarising uncertainty with a single number. Statistics measures spread in data you already have; probability measures spread in outcomes you have not seen yet.

Bayes' theorem is really just asking: given what I now know, how should I update what I believed before?

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Chapter at a Glance

Priority reflects past-paper trends, not an official NTA weightage.

What the NTA Syllabus Covers

• Measures of dispersion, calculation of mean, median and mode of grouped and ungrouped data
• Calculation of standard deviation, variance and mean deviation for grouped and ungrouped data
• Probability of an event, addition and multiplication theorems of probability
• Bayes' theorem, probability distribution of a random variable

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Master These Topics

1. Standard Deviation and Variance for Grouped Data

For grouped data with frequencies $f_i$ and class marks $x_i$ (with $N = \sum f_i$), the mean is $\bar{x} = \dfrac{\sum f_ix_i}{N}$, and

$$\sigma^2 = \frac{\sum f_ix_i^2}{N} - \bar{x}^2 \qquad \sigma = \sqrt{\sigma^2}$$

Mean deviation about the mean is $\dfrac{\sum f_i\lvert x_i - \bar{x}\rvert}{N}$.

Worked example: For the values $2, 4, 4, 4, 5, 5, 7, 9$ (ungrouped, $N = 8$), $\bar{x} = \dfrac{40}{8} = 5$. Then $\sum(x_i-\bar{x})^2 = 9+1+1+1+0+0+4+16 = 32$, so $\sigma^2 = 32/8 = 4$ and $\sigma = 2$.

Trap: Variance uses squared deviations, while mean deviation uses absolute deviations. They are not interchangeable, and the two give different values for the same data.

2. Addition and Multiplication Theorems

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

For mutually exclusive events, $P(A \cap B) = 0$, so the formula simplifies to a plain sum.

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

For independent events, $P(A \cap B) = P(A)P(B)$.

Worked example: Two dice are rolled. $P(\text{sum} = 7 \text{ or sum} = 11)$: these are mutually exclusive, with $P(\text{sum}=7) = 6/36$ and $P(\text{sum}=11) = 2/36$, so the total is $8/36 = 2/9$.

3. Bayes' Theorem

$$P(A_i \mid B) = \frac{P(A_i)\,P(B \mid A_i)}{\sum_j P(A_j)\,P(B \mid A_j)}$$

Worked example: A factory has two machines. Machine 1 makes 60% of items with a 2% defect rate, and Machine 2 makes 40% with a 5% defect rate. Given that an item is defective, the probability it came from Machine 2 is

$$P(M_2 \mid D) = \frac{0.4 \times 0.05}{0.6 \times 0.02 + 0.4 \times 0.05} = \frac{0.02}{0.012 + 0.02} = \frac{0.02}{0.032} = 0.625$$

4. Probability Distribution of a Random Variable

A random variable $X$ has a probability distribution when each value $x_i$ is assigned $P(X = x_i)$, with $\sum P(X=x_i) = 1$. The mean (expectation) is $E(X) = \sum x_iP(X=x_i)$, and the variance is $E(X^2) - [E(X)]^2$.

Worked example: Let $X$ be the number of heads in 4 tosses of a fair coin, so $P(X=k) = \dbinom{4}{k}\left(\dfrac{1}{2}\right)^4$ for $k = 0,1,2,3,4$, giving probabilities $\tfrac{1}{16}, \tfrac{4}{16}, \tfrac{6}{16}, \tfrac{4}{16}, \tfrac{1}{16}$. The mean is $E(X) = 2$ and the variance is $1$.

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Common Traps to Avoid

• Using $P(A\cap B)=P(A)P(B)$ when the events are not independent.
• Confusing variance (squared deviations) with mean deviation (absolute deviations).
• Getting the numerator and denominator of Bayes' theorem the wrong way round.
• Forgetting that the probabilities of a random variable's distribution must sum to 1.

60-Second Revision Sheet

• $\sigma^2 = \dfrac{\sum f_ix_i^2}{N} - \bar{x}^2$
• $P(A\cup B) = P(A)+P(B)-P(A\cap B)$; $P(A\cap B) = P(A)P(B\mid A)$
• Bayes: $P(A_i\mid B) = \dfrac{P(A_i)P(B\mid A_i)}{\sum_j P(A_j)P(B\mid A_j)}$
• $E(X) = \sum x_iP(X=x_i)$; Var$(X) = E(X^2)-[E(X)]^2$

Your Study Plan

1. Day 1: mean, variance and standard deviation for grouped data.
2. Day 2: addition and multiplication theorems, independent and mutually exclusive events.
3. Day 3: Bayes' theorem with two- and three-cause problems.
4. Day 4: probability distributions, expectation, variance and a timed mixed set.

Practice Statistics & Probability Questions Free → (opens in a new tab)

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