If A and B are two events such that P(A) = `(1)/(2)` and P(B) = `(2)/(3)`, then show that (a) `P(Auu B) ge (2)/(3) (b) (1)/(6) le P(A nn B) le (1)/(2)

`P(A)= (1)/(2) and P(B) = (2)/(3)`
`P(A uu B) ge "max {P(A), P(B)} = "(2)/(3)`
and `P(A nn B) le min {P(A), P(B)} = (1)/(2)`
Now, `P(A nn B) = P(A) + P(B) - P(A uu B) ge P(A) + P(B) - 1 = (1)/(6)`
`therefore (1)/(6) le P(A nn B) le (1)/(2) " "(1)`
`P(A nn barB) = P(A) - P(A nn B) = (1)/(2) - P(A nn B)`
From (1), `-(1)/(2) le - P (Ann B) le - (1)/(6)`
`therefore 0 le (1)/(2) - P(A nn B) le (1)/(2) - (1)/(6)`
`therefore P(A nn barB) le (1)/(3)`
`P(barA nn B) = P(B) - P(A nn B) = (2)/(3) - P(A nn B)`
`therefore (2)/(3) - (1)/(2) le (2)/(3) - P(A nn B) le (2)/(3) - (1)/(6)`
`(1)/(6) le P(barA nn B) le (1)/(2)`