Question

X and Y are two independent events. The probability that X and Y occur is 1/12, and the probability that neither occur is 1/2, the probability of occurrence of X can be:

Answer 1

P(XΠY)=1/12
P(XUY)'=1/2 ⇒P(XUY)=1/2
so
P(XUY)=P(X) + P(Y) - P(XΠY)   ....    1
and
P(XΠY)=P(X).P(Y)   ⇒P(Y)=1/(12×P(X))   ......2
so from 1 and 2
hence
1/2=t +1/12t - 1/12          (p(X)=t let)
so t =1/4 or 1/3

Answer 2

Answer:

The probability of occurrence of X can be is 1/4 or 1/3

Step-by-step explanation:

We are given that The probability that X and Y occur is 1/12

So, P(X∩Y)=$\frac{1}{12}$

Now we are given that the probability that neither occur is 1/2

P(XUY)'=$\frac{1}{2}$

 ⇒P(XUY)=$\frac{1}{2}$

We know that

P(XUY)=P(X) + P(Y) - P(X∩Y) ----- A

Since we are given that X and Y are two independent events.

So, P(X∩Y)=P(X).P(Y)   

⇒P(Y)=$\frac{1}{12 P(X)}$ ----B

Substitute values in B

$\frac{1}{2}=P(X)+\frac{1}{12 P(X)}-\frac{1}{12}$

Substitute P(X)=x

$\frac{1}{2}=x+\frac{1}{12x}-\frac{1}{12}$

$x=\frac{1}{4} , \frac{1}{3}$

Hence  the probability of occurrence of X can be is 1/4 or 1/3