TS EAMCET20215 Aug 2021Morning ShiftMathematicsProbabilityActual
The probability distribution of a random variable X is given below. X = x 0 1 2 3 4 5 6 7 P ( x ) 0 . 01 0 . 10 0 . 26 0 . 33 0 . 18 0 . 06 K 0 . 04 Then P ( X ≥ 3 ) - P ( X < 6 ) =
Options
- A0 . 24
- B- 0 . 27
- C0 . 57
- D- 0 . 31
Correct answer
D. - 0 . 31
Step-by-step solution
Since, ∑ x = 0 7 P x = 1 ⇒ 0 . 01 + 0 . 10 + 0 . 26 + 0 . 33 + 0 . 18 + 0 . 06 + K + 0 . 04 = 1 ⇒ K = 1 - 0 . 98 = 0 . 02 Now, P x ≥ 3 - P x < 6 = 0 . 33 + 0 . 18 + 0 . 06 + 0 . 02 + 0 . 04 - 0 . 01 + 0 . 1 + 0 . 26 + 0 . 36 + 0 . 18 + 0 . 66 = - 0 . 31