AktiraAktira JEE
MathsJEE Main 28 Jan 2026, Shift 2·Q4·MCQ·Easy–moderate

Probability · 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}

Video solution coming soon

We're animating every PYQ step by step. Subscribe to get them first.

Subscribe

See the full 28 Jan 2026, Shift 2 paper