NTA Abhyas JEE Main2020MathematicsProbabilityPractice
The probability distribution of a random variable X is given as X − 5 − 4 − 3 − 2 − 1 0 1 2 3 4 5 P ( X ) p 2 p 3 p 4 p 5 p 7 p 8 p 9 p 10 p 11 p 12 p Then, the value of p is
Options
- A1 72
- B3 73
- C5 72
- D1 74
Correct answer
A. 1 72
Step-by-step solution
Sum of Probabilities = 1 ⇒ p + 2 p + 3 p + 4 p + 5 p + 7 p + 8 p + 9 p + 10 p + 11 p + 12 p = 1 ⇒ 72 p = 1 ⇒ p = 1 72