AP EAMCET202119 Aug 2021Evening ShiftMathematicsProbabilityActual
Let X be a random variable which takes values 1 , 2 , 3 , 4 such that P ( X = r ) = K r 3 where r = 1 , 2 , 3 , 4 then
Options
- AK = 1 100  and  P 1 2 < X < 5 2 ∣ X > 1 = 8 97
- BK = 1 99  and  P 1 2 < X < 5 2 ∣ X > 1 = 8 99
- CK = 1 100  and  P 1 2 < X < 5 2 ∣ X > 1 = 8 99
- DK = 1 100  and  P 1 2 < X < 5 2 ∣ X > 1 = 10 99
Correct answer
C. K = 1 100  and  P 1 2 < X < 5 2 ∣ X > 1 = 8 99
Step-by-step solution
According to the given condition, we have random variables 11 ,   22 ,   33   and   44 As, P X = r = K r 3   where, r = 1 ,   2 ,   3 ,   4 Hence, P X = 1 = K 1 3 = K , P X = 2 = K 2 3 = 8 K , P X = 3 = K 3 3 = 27 K and P X = 4 = K 4 3 = 64 K As, we know, here ∑ 1 4 P X = r = 1 (sum of probabilities) ⇒ K + 8 K + 27 K + 64 K = 1   ⇒ K = 1 100 Now, P 1 2 < X < 5 2 ∣ X > 1 = P 1 ,   2 ∩   2 ,   3 ,   4 P   2