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Probability

18 JEE Main questions · Maths

  1. MathsJEE Main 21 Jan 2026, Shift 2·Q10·MCQ·Moderate
    Open

    Mean and variance of distribution

    A random variable XX takes values 0, 1, 2, 3 with probabilities 2a+130,8a−130,4a+130\dfrac{2a+1}{30}, \dfrac{8a-1}{30}, \dfrac{4a+1}{30}, bb respectively, where a,b∈Ra, b \in \mathbb{R}. Let μ\mu and σ\sigma respectively be the mean and standard deviation of XX such that σ2+μ2=2\sigma^2 + \mu^2 = 2. Then ab\dfrac{a}{b} is equal to:
    1. (1)30
    2. (2)3
    3. (3)60
    4. (4)12
  2. MathsJEE Main 22 Jan 2026, Shift 1·Q4·MCQ·Easy–moderate
    Open

    Classical probability with divisibility

    Two distinct numbers aa and bb are selected at random from 1,2,3,…,501, 2, 3, \ldots, 50. The probability, that their product abab is divisible by 3, is
    1. (1)5611225\dfrac{561}{1225}
    2. (2)6641225\dfrac{664}{1225}
    3. (3)2721225\dfrac{272}{1225}
    4. (4)825\dfrac{8}{25}
  3. MathsJEE Main 22 Jan 2026, Shift 1·Q7·MCQ·Easy–moderate
    Open

    Probability distribution of random variable

    If a random variable xx has the probability distribution
    xx01234567
    p(x)p(x)02k2kkk3k3k2k22k^22k2kk2+kk^2+k7k27k^2
    then P(3<x≤6)P(3 < x \le 6) is equal to
    1. (1)0.34
    2. (2)0.22
    3. (3)0.64
    4. (4)0.33
  4. MathsJEE Main 23 Jan 2026, Shift 1·Q24·Numerical·Moderate
    Open

    Classical probability counting

    From the first 100 natural numbers, two numbers first aa and then bb are selected randomly without replacement. If the probability that a−b≥10a - b \ge 10 is mn\dfrac{m}{n}, gcd⁡(m,n)=1\gcd(m, n) = 1, then m+nm + n is equal to .
  5. MathsJEE Main 23 Jan 2026, Shift 2·Q11·MCQ·Easy–moderate
    Open

    Total probability theorem

    Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is pq\dfrac{p}{q}, gcd⁡(p,q)=1\gcd(p, q) = 1, then p+qp + q is equal to
    1. (1)22
    2. (2)23
    3. (3)24
    4. (4)21
  6. MathsJEE Main 24 Jan 2026, Shift 1·Q17·MCQ·Easy–moderate
    Open

    Binomial distribution

    From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is :-
    1. (1)7107\dfrac{7}{10^7}
    2. (2)81108\dfrac{81}{10^8}
    3. (3)67108\dfrac{67}{10^8}
    4. (4)73108\dfrac{73}{10^8}
  7. MathsJEE Main 24 Jan 2026, Shift 2·Q24·Numerical·Moderate
    Open

    Classical probability with subsets

    Let SS be a set of 5 elements and P(S)P(S) denote the power set of SS. Let EE be an event of choosing an ordered pair (A,B)(A, B) from the set P(S)×P(S)P(S)\times P(S) such that A∩B=∅A \cap B = \varnothing. If the probability of the event EE is 3p2q\dfrac{3^p}{2^q}, where p,q∈Np, q \in \mathbb{N}, then p+qp + q is equal to .
  8. MathsJEE Main 28 Jan 2026, Shift 1·Q9·MCQ·Moderate
    Open

    Bayes theorem

    A bag contains 10 balls out of which kk are red and (10−k)(10 - k) are black, where 0≤k≤100 \le k \le 10. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:
    1. (1)711\dfrac{7}{11}
    2. (2)755\dfrac{7}{55}
    3. (3)7110\dfrac{7}{110}
    4. (4)1455\dfrac{14}{55}
    We corrected a printing error in the original paper

    Source printed “the bag contains l red and 9 black balls” (letter l for the digit 1).

  9. MathsJEE Main 28 Jan 2026, Shift 2·Q4·MCQ·Easy–moderate
    Open

    Expectation of a random variable

    The probability distribution of a random variable XX is given below :
    XX4k4k307k\dfrac{30}{7}k327k\dfrac{32}{7}k347k\dfrac{34}{7}k367k\dfrac{36}{7}k387k\dfrac{38}{7}k407k\dfrac{40}{7}k6k6k
    P(X)P(X)215\dfrac{2}{15}115\dfrac{1}{15}215\dfrac{2}{15}15\dfrac{1}{5}115\dfrac{1}{15}215\dfrac{2}{15}15\dfrac{1}{5}115\dfrac{1}{15}
    If E(X)=26315E(X) = \dfrac{263}{15}, then P(X<20)P(X < 20) is equal to :
    1. (1)35\dfrac35
    2. (2)815\dfrac{8}{15}
    3. (3)1115\dfrac{11}{15}
    4. (4)1415\dfrac{14}{15}
  10. MathsJEE Main 2 Apr 2026, Shift 1·Q23·Numerical·Moderate
    Open

    Classical probability counting

    Let a,b,c∈{1,2,3,4}a, b, c \in \{1, 2, 3, 4\}. If the probability, that ax2+22 bx+c>0ax^2 + 2\sqrt2\,bx + c > 0 for all x∈Rx \in \mathbb{R}, is mn\dfrac{m}{n}, gcd⁡(m,n)=1\gcd(m, n) = 1, then m+nm + n is equal to .
  11. MathsJEE Main 2 Apr 2026, Shift 2·Q8·MCQ·Moderate
    Open

    Probability of reaching a score

    A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is mn\dfrac{m}{n}, gcd⁡(m,n)=1\gcd(m, n) = 1, then m+nm + n is equal to:
    1. (1)53
    2. (2)55
    3. (3)107
    4. (4)105
  12. MathsJEE Main 4 Apr 2026, Shift 1·Q21·Numerical·Moderate
    Open

    Independent trials counting outcomes

    A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is pp, then 96p96p is equal to .
  13. MathsJEE Main 4 Apr 2026, Shift 2·Q21·Numerical·Moderate
    Open

    Classical probability

    From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to ab\dfrac{a}{b}, where a,b∈Na, b \in \mathbb{N} and gcd⁡(a,b)=1\gcd(a, b) = 1, then a+ba + b is equal to .
  14. MathsJEE Main 5 Apr 2026, Shift 1·Q7·MCQ·Moderate
    Open

    Bayes theorem

    A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:
    1. (1)710\dfrac{7}{10}
    2. (2)1017\dfrac{10}{17}
    3. (3)1219\dfrac{12}{19}
    4. (4)719\dfrac{7}{19}
    We corrected a printing error in the original paper

    Source printed “On the the envelope” (doubled word).

  15. MathsJEE Main 5 Apr 2026, Shift 2·Q8·MCQ·Easy–moderate
    Open

    Total probability theorem

    The probabilities that players AA and BB of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If AA is selected the captain, the probability that the team wins the tournament is 0.8 and if BB is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is:
    1. (1)0.74
    2. (2)0.76
    3. (3)0.72
    4. (4)0.78
  16. MathsJEE Main 6 Apr 2026, Shift 1·Q10·MCQ·Easy–moderate
    Open

    Total probability theorem

    A bag contains (N+1)(N + 1) coins – NN fair coins, and one coin with ‘Head' on both sides. A coin is selected at random and tossed. If the probability of getting ‘Head' is 916\dfrac{9}{16}, then NN is equal to:
    1. (1)5
    2. (2)7
    3. (3)8
    4. (4)9
  17. MathsJEE Main 6 Apr 2026, Shift 2·Q9·MCQ·Moderate
    Open

    Classical probability of pairings

    A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is:
    1. (1)63925\dfrac{63}{925}
    2. (2)17231\dfrac{17}{231}
    3. (3)16231\dfrac{16}{231}
    4. (4)64925\dfrac{64}{925}
  18. MathsJEE Main 8 Apr 2026, Shift 2·Q6·MCQ·Easy–moderate
    Open

    Bayes theorem

    A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are 25\dfrac25, 15\dfrac15 and 25\dfrac25. The probabilities that the candidate reaches late at the examination centre are 15\dfrac15, 13\dfrac13 and 14\dfrac14 if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is:
    1. (1)1137\dfrac{11}{37}
    2. (2)1237\dfrac{12}{37}
    3. (3)1337\dfrac{13}{37}
    4. (4)1437\dfrac{14}{37}