Statistics & Probability Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Statistics & Probability with step-by-step solutions covering mean, median, mean deviation, variance, standard deviation, Bayes' theorem and combinatorial probability.
A curated set of JEE Main 2026 previous-year questions on Statistics & Probability, each solved step by step so you can check both the final answer and the full reasoning.
Solutions are AI-generated and pending review.
Solution
There are $9$ numbers, so the mean gives the total:
$$\sum x_i = 40 \times 9 = 360.$$The seven known numbers sum to $21+8+17+51+103+13+67 = 280$, so
$$a + b = 80.$$Median is the $5^{\text{th}}$ value when sorted, given as $21$.
Mean deviation about the median $= 26$:
$$\sum |x_i - 21| = 26 \times 9 = 234.$$The known deviations are
$$0+13+4+30+82+8+46 = 183,$$so
$$|a-21| + |b-21| = 234 - 183 = 51.$$With $a+b=80$ and $a>b$, take $b < 21 \le a$:
$$(a-21) + (21-b) = a - b = 51.$$Solving $a+b=80,\ a-b=51$:
$$a = \dfrac{131}{2},\qquad b = \dfrac{29}{2}.$$The sorted data $8,13,14.5,17,\mathbf{21},51,65.5,67,103$ confirms median $21$ and mean deviation $26$.
$$2a = 131.$$Answer: D
Solution
Label the tosses $1$–$8$. “First six” means tosses $1$–$6$; “last five” means tosses $4$–$8$. They overlap on tosses $4,5,6$. Let $j$ be the number of heads among tosses $4,5,6$.
Tosses $1$–$3$ must supply $4 - j$ heads, and tosses $7$–$8$ must supply $3 - j$ heads.
For each valid $j$, the count of favourable outcomes is
$$\binom{3}{\,4-j\,}\binom{3}{j}\binom{2}{\,3-j\,}.$$The binomials force $4-j \le 3$ and $3-j \le 2$, i.e. $j \ge 1$; also $j \le 3$. Checking each:
$$j=1:\ \binom{3}{3}\binom{3}{1}\binom{2}{2} = 1\cdot 3\cdot 1 = 3,$$$$j=2:\ \binom{3}{2}\binom{3}{2}\binom{2}{1} = 3\cdot 3\cdot 2 = 18,$$$$j=3:\ \binom{3}{1}\binom{3}{3}\binom{2}{0} = 3\cdot 1\cdot 1 = 3.$$Total favourable $= 3 + 18 + 3 = 24$ (confirmed by direct enumeration of all $2^8$ outcomes).
$$p = \dfrac{24}{2^8} = \dfrac{24}{256} = \dfrac{3}{32}.$$$$96p = 96 \times \dfrac{3}{32} = 9.$$Answer: 9
Solution
Subtract the two sums:
$$\sum\big[(x_i+5)^2 - (x_i-5)^2\big] = \sum 20x_i = 2500 - 100 = 2400,$$$$\Rightarrow \sum x_i = 120 \Rightarrow \bar{x} = \dfrac{120}{20} = 6.$$Add the two sums:
$$\sum\big[(x_i+5)^2 + (x_i-5)^2\big] = \sum\big(2x_i^2 + 50\big) = 2\sum x_i^2 + 1000 = 2600,$$$$\Rightarrow \sum x_i^2 = 800.$$Variance:
$$\sigma^2 = \dfrac{\sum x_i^2}{20} - \bar{x}^2 = \dfrac{800}{20} - 36 = 40 - 36 = 4 \Rightarrow \sigma = 2.$$$$\dfrac{\bar{x}}{\sigma} = \dfrac{6}{2} = 3 = 3:1.$$Answer: B
Solution
Each coin is chosen with probability $\dfrac{1}{N+1}$. A fair coin gives Head with probability $\tfrac{1}{2}$; the double-headed coin gives Head with probability $1$.
$$P(\text{Head}) = \dfrac{N\cdot \tfrac{1}{2} + 1\cdot 1}{N+1} = \dfrac{\tfrac{N}{2} + 1}{N+1} = \dfrac{N+2}{2(N+1)}.$$Set equal to $\dfrac{9}{16}$:
$$\dfrac{N+2}{2(N+1)} = \dfrac{9}{16} \Rightarrow 16(N+2) = 18(N+1)$$$$32 + 16N = 18N + 18 \Rightarrow 2N = 14 \Rightarrow N = 7.$$Answer: B
Solution
Since $a \in \{1,2,3,4\}$, the leading coefficient is always positive. The quadratic is positive for all real $x$ iff the discriminant is negative:
$$(2\sqrt{2}\,b)^2 - 4ac < 0 \Rightarrow 8b^2 - 4ac < 0 \Rightarrow 2b^2 < ac.$$Total ordered triples $(a,b,c)$: $4^3 = 64$.
Count triples with $2b^2 < ac$ (note $ac \le 16$, so $b=1$ or $b=2$ only):
- $b = 1$: need $ac > 2$. All $16$ pairs except $ac = 1$ (i.e. $a=c=1$) and $ac=2$ (i.e. $(1,2),(2,1)$). Favourable $= 16 - 3 = 13$.
- $b = 2$: need $ac > 8$. Pairs with $ac \in \{9,12,16\}$: $(3,3),(3,4),(4,3),(4,4)$ give $9,12,12,16$; so $4$ pairs.
Total favourable $= 13 + 4 = 17$.
$$P = \dfrac{17}{64},\qquad \gcd(17,64)=1 \Rightarrow m+n = 17 + 64 = 81.$$Answer: 81
Solution
Let $S = \sum x_i$ and $Q = \sum x_i^2$. Expanding:
$$\sum(x_i+2)^2 = Q + 4S + 40 = 180 \Rightarrow Q + 4S = 140,$$$$\sum(x_i-1)^2 = Q - 2S + 10 = 90 \Rightarrow Q - 2S = 80.$$Subtracting: $6S = 60 \Rightarrow S = 10$, hence $\bar{x} = 1$ and $Q = 80 + 2(10) = 100$.
Variance:
$$\sigma^2 = \dfrac{Q}{10} - \bar{x}^2 = \dfrac{100}{10} - 1 = 10 - 1 = 9 \Rightarrow \sigma = 3.$$Answer: D
Solution
Total ways to choose $3$ dates from $31$:
$$\binom{31}{3} = 4495.$$Increasing A.P. $\{t, t+d, t+2d\}$ with $d \ge 1$ and $t + 2d \le 31$. For a fixed common difference $d$, the first term $t$ ranges over $1 \le t \le 31 - 2d$, giving $31 - 2d$ choices. Summing over $d = 1$ to $15$:
$$\sum_{d=1}^{15}(31 - 2d) = 29 + 27 + \cdots + 1 = 15^2 = 225.$$Probability:
$$P = \dfrac{225}{4495} = \dfrac{45}{899}.$$Since $\gcd(45,899)=1$ (as $899 = 29\cdot 31$):
$$a + b = 45 + 899 = 944.$$Answer: 944
Solution
Each throw adds $5$ (tail) or $10$ (head), each with probability $\tfrac{1}{2}$. Work in units of $5$: the running total moves by $+1$ (tail) or $+2$ (head). We want the probability of the running total ever landing exactly on $30/5 = 6$.
Let $f(k)$ be the probability of ever hitting level $k$. To reach $k$, the last step comes either from $k-1$ (a tail) or from $k-2$ (a head):
$$f(k) = \tfrac{1}{2}\,f(k-1) + \tfrac{1}{2}\,f(k-2),$$with $f(0) = 1$ and $f(1) = \tfrac{1}{2}$ (level $1$ is reached only by a first tail).
$$f(2) = \tfrac{1}{2}\!\left(\tfrac{1}{2}+1\right) = \tfrac{3}{4},\quad f(3) = \tfrac{1}{2}\!\left(\tfrac{3}{4}+\tfrac{1}{2}\right) = \tfrac{5}{8},$$$$f(4) = \tfrac{1}{2}\!\left(\tfrac{5}{8}+\tfrac{3}{4}\right) = \tfrac{11}{16},\quad f(5) = \tfrac{1}{2}\!\left(\tfrac{11}{16}+\tfrac{5}{8}\right) = \tfrac{21}{32},$$$$f(6) = \tfrac{1}{2}\!\left(\tfrac{21}{32}+\tfrac{11}{16}\right) = \tfrac{43}{64}.$$So $P = \dfrac{43}{64}$, $\gcd(43,64)=1$:
$$m + n = 43 + 64 = 107.$$Answer: C
Solution
Mean $= 8 \Rightarrow \sum x_i = 8n$. Variance $= 16$:
$$\dfrac{\sum x_i^2}{n} - 8^2 = 16 \Rightarrow \sum x_i^2 = 80n.$$The $n$-th observation:
$$x_n = 8n - 48.$$Sum of squares:
$$496 + (8n - 48)^2 = 80n.$$Expand: $(8n-48)^2 = 64n^2 - 768n + 2304$, so
$$64n^2 - 768n + 2304 + 496 = 80n$$$$64n^2 - 848n + 2800 = 0 \Rightarrow 4n^2 - 53n + 175 = 0.$$$$n = \dfrac{53 \pm \sqrt{2809 - 2800}}{8} = \dfrac{53 \pm 3}{8} \Rightarrow n = 7 \text{ or } n = \dfrac{50}{8}.$$Only the integer is valid: $n = 7$.
Answer: D
Solution
Mean $= 8$:
$$2+4+\alpha+8+\beta+12+14 = 56 \Rightarrow \alpha + \beta = 16.$$Variance $= 16$: $\ \dfrac{\sum x_i^2}{7} - 64 = 16 \Rightarrow \sum x_i^2 = 560.$
$$4+16+\alpha^2+64+\beta^2+144+196 = 560 \Rightarrow \alpha^2 + \beta^2 = 136.$$From $\alpha+\beta=16$: $\alpha^2+\beta^2 = 256 - 2\alpha\beta = 136 \Rightarrow \alpha\beta = 60.$ Roots of $t^2 - 16t + 60 = 0$ are $6$ and $10$. With $\alpha < \beta$: $\alpha = 6,\ \beta = 10.$
New roots:
$$3\alpha + 2 = 20,\qquad 2\beta + 1 = 21.$$Sum $= 41$, product $= 420$:
$$x^2 - 41x + 420 = 0.$$Answer: B
Solution
Treat the $12$ balls as distinct and count perfect matchings (unordered partitions into $6$ pairs), since the order of drawing pairs does not affect whether each pair is mixed.
Total perfect matchings of $12$ distinct balls:
$$11!! = 11\cdot 9\cdot 7\cdot 5\cdot 3\cdot 1 = 10395.$$Favourable matchings — each pair one blue and one green — correspond to bijections between the $6$ blue and $6$ green balls:
$$6! = 720.$$Probability:
$$P = \dfrac{720}{10395} = \dfrac{16}{231}.$$Answer: B
Solution
Each origin is equally likely a priori: $P(\text{KANPUR}) = P(\text{ANANTPUR}) = \tfrac{1}{2}$. Given a word, the two visible consecutive letters are one of its consecutive letter-pairs, chosen uniformly.
KANPUR ($6$ letters $\Rightarrow 5$ consecutive pairs): KA, AN, NP, PU, UR. “AN” appears once, so $P(\text{AN}\mid \text{KANPUR}) = \tfrac{1}{5}.$
ANANTPUR ($8$ letters $\Rightarrow 7$ pairs): AN, NA, AN, NT, TP, PU, UR. “AN” appears twice, so $P(\text{AN}\mid \text{ANANTPUR}) = \tfrac{2}{7}.$
Bayes’ theorem:
$$P(\text{ANANTPUR}\mid \text{AN}) = \dfrac{\tfrac{1}{2}\cdot \tfrac{2}{7}}{\tfrac{1}{2}\cdot \tfrac{1}{5} + \tfrac{1}{2}\cdot \tfrac{2}{7}} = \dfrac{\tfrac{2}{7}}{\tfrac{1}{5} + \tfrac{2}{7}} = \dfrac{\tfrac{2}{7}}{\tfrac{17}{35}} = \dfrac{10}{17}.$$Answer: B
Solution
Total frequency: $N = 8+6+2+2+2+6 = 26.$
Mean:
$$\bar{x} = \dfrac{\sum f_i x_i}{N} = \dfrac{40 + 42 + 18 + 20 + 24 + 90}{26} = \dfrac{234}{26} = 9.$$Absolute deviations $|x_i - 9|$: $4, 2, 0, 1, 3, 6$, weighted by frequency:
$$\sum f_i |x_i - 9| = 8(4) + 6(2) + 2(0) + 2(1) + 2(3) + 6(6)$$$$= 32 + 12 + 0 + 2 + 6 + 36 = 88.$$Mean deviation:
$$\text{MD} = \dfrac{88}{26} = \dfrac{44}{13}.$$Answer: C
Solution
By the total probability theorem:
$$P(\text{win}) = P(A)\,P(\text{win}\mid A) + P(B)\,P(\text{win}\mid B).$$$$P(\text{win}) = (0.6)(0.8) + (0.4)(0.7) = 0.48 + 0.28 = 0.76.$$Answer: B
Solution
The $k$-th value is $k(k-1)$ with frequency $\binom{n}{k}$, for $k = 0,1,\ldots,n$.
Total frequency: $\sum_{k=0}^{n}\binom{n}{k} = 2^n.$
Weighted sum uses $\sum_{k=0}^{n} k(k-1)\binom{n}{k} = n(n-1)\,2^{n-2}$:
$$\bar{X} = \dfrac{n(n-1)\,2^{n-2}}{2^n} = \dfrac{n(n-1)}{4} = 60.$$$$n(n-1) = 240 \Rightarrow n = 16.$$Median of $2^{16} = 65536$ observations: locate the $\big(\tfrac{N}{2}\big)$-th and $\big(\tfrac{N}{2}+1\big)$-th values, i.e. positions $32768$ and $32769$, using cumulative frequencies (values are increasing in $k$):
$$\text{cum through } k=7:\ 1+16+120+560+1820+4368+8008+11440 = 26333,$$$$\text{cum through } k=8:\ 26333 + \binom{16}{8} = 26333 + 12870 = 39203.$$Both positions $32768$ and $32769$ lie in the $k = 8$ group, whose value is $8\cdot 7 = 56$.
$$\text{Median} = 56.$$Answer: A
Solution
Bayes’ theorem. Let $Bu, Sc, Ca$ denote the travel modes and $L$ the event of being late.
$$P(Bu)P(L\mid Bu) = \dfrac{2}{5}\cdot \dfrac{1}{5} = \dfrac{2}{25},$$$$P(Sc)P(L\mid Sc) = \dfrac{1}{5}\cdot \dfrac{1}{3} = \dfrac{1}{15},$$$$P(Ca)P(L\mid Ca) = \dfrac{2}{5}\cdot \dfrac{1}{4} = \dfrac{1}{10}.$$Common denominator $150$:
$$\dfrac{2}{25} = \dfrac{12}{150},\quad \dfrac{1}{15} = \dfrac{10}{150},\quad \dfrac{1}{10} = \dfrac{15}{150}.$$Total $= \dfrac{12+10+15}{150} = \dfrac{37}{150}.$
$$P(Bu\mid L) = \dfrac{12/150}{37/150} = \dfrac{12}{37}.$$Answer: B
Solution
Let the sets have sizes $n_1 = 4,\ n_2 = 6$, means $\bar{x}_1 = 1,\ \bar{x}_2 = 2$, variances $\sigma_1^2 = 13,\ \sigma_2^2 = 1$.
Combined mean:
$$\bar{x} = \dfrac{4(1) + 6(2)}{10} = \dfrac{16}{10} = \dfrac{8}{5} = 1.6.$$Combined variance (with $d_1 = \bar{x}_1 - \bar{x} = -0.6$, $d_2 = \bar{x}_2 - \bar{x} = 0.4$):
$$\sigma^2 = \dfrac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}.$$$$= \dfrac{4(13 + 0.36) + 6(1 + 0.16)}{10} = \dfrac{4(13.36) + 6(1.16)}{10}$$$$= \dfrac{53.44 + 6.96}{10} = \dfrac{60.4}{10} = 6.04.$$Answer: C