NDA (National Defence Academy) · Mathematics (JEE & NDA)
Statistics and Probability
Measures of dispersion, probability of events, conditional probability, Bayes theorem and probability distributions.
Six concepts covering dispersion through Bayes and the binomial — the JEE Main slice of statistics and probability. Two ledgers are the chapter's own worked problems; the rest are identities and criteria, so they decline a fabricated numeric drill.
- NDA (National Defence Academy)
- Medium level
- 6 concepts
- 5 practice questions
1Mean, variance and standard deviation
For n observations the mean is \bar{x}=\dfrac{\sum x_i}{n}. Variance is \sigma^2=\dfrac{\sum(x_i-\bar{x})^2}{n}=\dfrac{\sum x_i^2}{n}-\bar{x}^2, and the standard deviation is its positive square root \sigma. The computational form on the right avoids subtracting the mean from every data point.
Shifting every value by the same constant leaves \sigma^2 unchanged; multiplying every value by k multiplies the variance by k^2. That is the whole effect of a linear change of origin and scale.
Mean and variance are scalar formulas on a list of values; the definition lines in this concept carry them.
Computing variance
- Sum and sum of squaresAccumulate \sum x_i and \sum x_i^2 in one pass over the data.
- MeanDivide: \bar{x}=(\sum x_i)/n.
- Computational formReturn \sigma^2=(\sum x_i^2)/n-\bar{x}^2 rather than forming every (x_i-\bar{x})^2.
| Change | Mean | Variance |
|---|---|---|
| Add constant c to every x_i | shifts by c | unchanged |
| Multiply every x_i by k | scales by k | scales by k^2 |
| Both: y_i = kx_i + c | k\bar{x}+c | k^2\sigma^2 |
Every observation in a data set is increased by 5. The variance
- Increases by 5
- Stays the same
- Is multiplied by 25
Adding a constant shifts the mean but leaves every deviation x_i-\bar{x} unchanged, so \sigma^2 is unchanged. Multiplying by 25 would be the effect of scaling every value by 5, not of adding 5.
2Classical probability and the addition rule
When outcomes are equally likely, P(E)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}. The addition rule for two events is P(A\cup B)=P(A)+P(B)-P(A\cap B), and the complement is P(A')=1-P(A).
For an "at least one" event, expand the union into many overlapping cases only as a last resort — compute 1-P(\text{none}) instead. That single complement move is how most JEE "at least one" stems collapse.
Figure. Addition rule for two events: add each probability and subtract the overlap once so A\cap B is not double-counted.
Reading a union
- Name the piecesWrite P(A), P(B) and P(A\cap B) from the equally-likely count, or from independence if it is given.
- Subtract the overlapApply P(A\cup B)=P(A)+P(B)-P(A\cap B); forgetting the intersection double-counts.
- Prefer the complementIf the stem says "at least one", switch to 1-P(\text{none}) before enumerating cases.
For events with P(A)=0.4, P(B)=0.5 and P(A\cap B)=0.2, the value of P(A\cup B) is
- 0.7
- 0.9
- 0.2
P(A\cup B)=0.4+0.5-0.2=0.7. Adding without subtracting the intersection gives 0.9; quoting only the intersection gives 0.2.
3Conditional probability
The probability of A given that B has occurred is P(A|B)=\dfrac{P(A\cap B)}{P(B)}, provided P(B)>0. Conditioning replaces the sample space by B: only the outcomes inside B still count, and A\cap B is the favourable part of that restricted space.
Rearranged, the definition is the multiplication rule P(A\cap B)=P(A|B)P(B). That product form is what Bayes and every tree diagram start from.
Figure. conditioning replaces the sample space by B; favourable outcomes are only A\cap B inside that restricted space, so P(A|B)=P(A\cap B)/P(B).
Applying the definition
- Identify the givenThe event after "given" or "if" is B — it goes in the denominator.
- Form the intersectionThe numerator is P(A\cap B), not P(A) alone.
- DivideReturn P(A|B)=P(A\cap B)/P(B).
If P(A\cap B)=0.12 and P(B)=0.4, then P(A|B) equals
- 0.3
- 0.48
- 0.12
P(A|B)=0.12/0.4=0.3. Multiplying instead of dividing gives 0.48; leaving the intersection alone forgets to condition.
4Independent events
Events A and B are independent when P(A\cap B)=P(A)P(B). Equivalently, conditioning on one does not change the other: P(A|B)=P(A) whenever P(B)>0. Independence is a statement about probabilities, not about whether the events "look related" in the story.
Mutual exclusivity is a different claim — A\cap B=\emptyset — and for non-trivial events it is incompatible with independence: if P(A)>0 and P(B)>0 then P(A)P(B)>0, so the intersection cannot be empty.
Independence is the equality P(A\cap B)=P(A)P(B); the comparison table in this concept is the classification.
| Claim | Criterion | Consequence |
|---|---|---|
| Independent | P(A\cap B)=P(A)P(B) | P(A|B)=P(A) |
| Mutually exclusive | A\cap B=\emptyset | P(A\cup B)=P(A)+P(B) |
| Both, with P(A),P(B)>0 | impossible | product positive, intersection empty — contradiction |
If P(A)=0.5, P(B)=0.4 and P(A\cap B)=0.2, then A and B are
- Independent, since 0.5\times 0.4=0.2
- Mutually exclusive, since the intersection is small
- Dependent, because 0.2\neq 0.5+0.4
Independence is exactly P(A\cap B)=P(A)P(B). Here 0.5\times 0.4=0.2, so they are independent. Mutual exclusivity would require intersection 0, and comparing the intersection to the sum is the wrong test.
5Bayes' theorem
If B_1,\dots,B_n partition the sample space, Bayes' theorem updates a prior using observed evidence A: P(B_i|A)=\dfrac{P(A|B_i)P(B_i)}{\sum_j P(A|B_j)P(B_j)}. The denominator is the total probability of the evidence — compute it once and reuse it for every hypothesis.
In a two-bag problem the partition is "bag A or bag B", the evidence is the colour drawn, and the posterior answers "which bag did this ball come from?".
Figure. Each path multiplies a prior by a likelihood. The two red paths are the joints; their sum is P(R), and the left joint over that sum is the posterior that the red ball came from bag A.
Running Bayes
- Priors and likelihoodsList P(B_i) and P(A|B_i) for every piece of the partition — a small table is enough.
- Total probabilityForm P(A)=\sum_j P(A|B_j)P(B_j); this is the denominator for every posterior.
- PosteriorReturn P(B_i|A)=P(A|B_i)P(B_i)/P(A) for the hypothesis asked.
| Hypothesis | Prior P(B_i) | Likelihood P(R|B_i) | Joint P(R\cap B_i) |
|---|---|---|---|
| Bag A | 1/2 | 3/5 | 3/10 |
| Bag B | 1/2 | 1/5 | 1/10 |
| Total P(R) | 4/10 |
Bayes' theorem with two bags
Bag A has 3 red and 2 black balls; bag B has 1 red and 4 black. A bag is chosen at random and a red ball is drawn. Find the probability it came from bag A.
- Priors P(A)=P(B)=\tfrac12; likelihoods P(R|A)=\tfrac35, P(R|B)=\tfrac15joints \tfrac{3}{10} and \tfrac{1}{10}
- P(R)=\tfrac12\cdot\tfrac35+\tfrac12\cdot\tfrac15\tfrac{4}{10}
- P(A|R)=(\tfrac12\cdot\tfrac35)/P(R)\dfrac{3/10}{4/10}=\dfrac{3}{4}
Pro tip. The Bayes denominator is the total probability of the evidence; compute it once and reuse it for every hypothesis.
In the two-bag problem above, P(B|R) equals
- 1/4
- 1/2
- 1/5
The same denominator 4/10 with numerator 1/10 gives P(B|R)=1/4. The prior 1/2 ignores the red evidence; the likelihood 1/5 never updates.
6Binomial distribution
For n independent Bernoulli trials with success probability p and q=1-p, the number of successes X satisfies P(X=r)={}^{n}C_{r}\,p^{r}q^{n-r}. The mean is np and the variance is npq.
When the trial is a fair coin, p=q=\tfrac12 and every specific sequence has probability (1/2)^{n}, so P(X=r) collapses to \dfrac{{}^{n}C_{r}}{2^{n}}.
Figure. Fair-coin binomial: n=5, r=3, p=1/2 gives P(X=3)={}^5C_3(1/2)^5=10/32=5/16. The mean is np=5/2.
Evaluating a binomial probability
- Identify n, r, pTrials, required successes, and success probability on one trial; set q=1-p.
- Write the massP(X=r)={}^{n}C_{r}\,p^{r}q^{n-r}.
- SimplifyFor p=\tfrac12, cancel to {}^{n}C_{r}/2^{n}; otherwise leave exact powers of p and q.
Binomial probability
A fair coin is tossed 5 times. Find the probability of getting exactly 3 heads.
- n=5, r=3, p=q=\tfrac12fair-coin case
- P(X=3)={}^{5}C_{3}\left(\tfrac12\right)^{3}\left(\tfrac12\right)^{2}{}^{5}C_{3}\left(\tfrac12\right)^{5}
- {}^{5}C_{3}=10, so 10/32\dfrac{5}{16}
Pro tip. When p=q=\tfrac12, every outcome has probability (1/2)^{n}, so the answer is just \dfrac{{}^{n}C_{r}}{2^{n}}.
For a binomial random variable with n=5 and p=\tfrac12, the mean np equals
- 5/2
- 5/4
- 5
Mean is np=5\cdot\tfrac12=\tfrac52. Variance npq=5/4 is the distractor that swaps mean for variance; 5 forgets to multiply by p.
Notes
- Measures of dispersion: The mean is \bar{x}=\dfrac{\sum x_i}{n}; the variance is \sigma^2=\dfrac{\sum(x_i-\bar{x})^2}{n}=\dfrac{\sum x_i^2}{n}-\bar{x}^2, and the standard deviation is \sigma. Variance is unaffected by shifting data but scales by k^2 under multiplication by k.
- Probability basics: For equally likely outcomes, P(E)=\dfrac{\text{favourable}}{\text{total}}. The addition rule is P(A\cup B)=P(A)+P(B)-P(A\cap B), and P(A')=1-P(A).
- Conditional probability and independence: P(A|B)=\dfrac{P(A\cap B)}{P(B)}. Events are independent when P(A\cap B)=P(A)P(B), in which case P(A|B)=P(A).
- Bayes' theorem: If B_1,\dots,B_n partition the sample space, P(B_i|A)=\dfrac{P(A|B_i)P(B_i)}{\sum_j P(A|B_j)P(B_j)}; it updates prior probabilities using observed evidence.
- Binomial distribution: For n independent trials with success probability p, P(X=r)={}^nC_r p^r q^{n-r} (with q=1-p), mean np and variance npq.
Formulas
- Mean / variance: \bar{x}=\dfrac{\sum x_i}{n},\quad \sigma^2=\dfrac{\sum x_i^2}{n}-\bar{x}^2
- Addition rule: P(A\cup B)=P(A)+P(B)-P(A\cap B)
- Conditional: P(A|B)=\dfrac{P(A\cap B)}{P(B)}
- Independence: P(A\cap B)=P(A)P(B)
- Bayes: P(B_i|A)=\dfrac{P(A|B_i)P(B_i)}{\sum_j P(A|B_j)P(B_j)}
- Binomial: P(X=r)={}^nC_r p^r q^{n-r},\ \text{mean }np,\ \text{var }npq
Exam traps & shortcuts
- For 'at least one' events, compute 1-P(\text{none}) rather than summing many cases.
- Use variance =\dfrac{\sum x^2}{n}-\bar{x}^2 instead of the deviation form to avoid subtracting the mean from every data point.
- In Bayes problems, first write all prior probabilities and likelihoods in a small table; the denominator is just their weighted sum (total probability).
Reference tables
| Idea | Formula | Exam trap |
|---|---|---|
| Mean / variance | \bar{x}=(\sum x_i)/n, \sigma^2=(\sum x_i^2)/n-\bar{x}^2 | Shift leaves \sigma^2 unchanged; scale by k multiplies it by k^2 |
| Addition rule | P(A\cup B)=P(A)+P(B)-P(A\cap B) | "At least one" → 1-P(\text{none}) |
| Conditional | P(A|B)=P(A\cap B)/P(B) | Denominator is the given event |
| Independence | P(A\cap B)=P(A)P(B) | Not the same as mutually exclusive |
| Bayes | P(B_i|A)=P(A|B_i)P(B_i)/\sum_j P(A|B_j)P(B_j) | Denominator is total probability of the evidence |
| Binomial | P(X=r)={}^{n}C_{r}p^{r}q^{n-r}, mean np, var npq | Fair coin → {}^{n}C_{r}/2^{n} |
Recap
Read only this the night before.
- Dispersion
- Use (\sum x^2)/n-\bar{x}^2. Shifts ignore variance; a factor k multiplies it by k^2.
- Union
- Add and subtract the intersection. "At least one" is 1-P(\text{none}).
- Conditional
- P(A|B)=P(A\cap B)/P(B) — the given event is the new sample space.
- Independence
- Product of probabilities. Mutually exclusive with positive probabilities cannot also be independent.
- Bayes
- Prior × likelihood over total probability of the evidence. Two-bag red draw: posterior 3/4 for bag A.
- Binomial
- {}^{n}C_{r}p^{r}q^{n-r}, mean np. Fair coin: {}^{n}C_{r}/2^{n} — five tosses, three heads → 5/16.
Practise Statistics and Probability
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