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

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