let x be a random variable with the following probability distribution x:-1,0,1 p(x):11/32,1/2,5/32 .find probability function of x^2 and x^2+2
Answers
Answered by
0
Answer:
similar questions ans
Step-by-step explanation:
A random variable X has the probability distribution X: 1,2,3,4,5,6,7,8
P(X):0.15,0.23,0.12,0.10,0.20,0.08,0.07,0.05 . For the events E= {X is a prime number } and F={X<4}, the probability P(E∪F) is:
Correct option is B)
E ={ is a prime number} ={2,3,5,7}
P(E)=P(X=2)+P(X=3)+P(X=5)+P(X=7)
P(E)=0.23+0.12+0.20+0.07=0.62
F={X<4}={1,2,3}
P(F)=P(X=1)=+P(X=2)+P(X=3)
P(F)=0.15+0.23+0.12=0.5
E∩F= {X is prime number as well as <4}={2,3}
P(E∩F)=P(X=2)+P(X=3)
=0.23+0.12
=0.35
Therefore required probability
P(E∪F)=P(E)+P(F)−P(E∩F)
⇒P(E∪F)=0.62+0.5−0.35
⇒P(E∪F)=0.77
most imp questions for jee
Similar questions