Question

Let p(z)=0.54, p(y)=0.22, and p(z u y)= 0.62. Find each probability.

Answer 1

Answer : p(only z)=0.40 and p(only u)=0.08

Explanation :

We have given that ,

p(u)=0.22,

p(z∪u)=0.62,

p(z)=0.54

By using probability rule of union of 2 events, we have,

p(z)+p(u)-p(z∩u)=p(z∪u)

⇒0.54+0.22-p(z∩u)=0.62

⇒0.76-p(z∩u)=0.62

⇒(-)p(z∩u)=0.62-0.76

⇒(-)p(z∩u)= (-)0.14

⇒p(z∩u)=0.14

Now,

p(only z)=p(z)-p(z∩u)

=0.54-0.14

=0.40

and,

p(only u)=p(u)-p(z∩u)

=0.22-0.14

=0.08

∴ Each probability will be

p(only z)=0.40

p(only u)=0.08