If `(1+3p)/3,(1-p)/1,(1-2p)/2` are the probabilities of 3 mutually exclusive events then find the set of all values of p.

Explanation

Let P(A) = `(1 + 3p)/(3), P(B) = (1 - p)/(4), P(C ) = (1 - 2p)/(2)`
As A, B and C are three mutually exclusive events.
`therefore P(A) + P(B) + P(C ) le 1`
`implies (1 + 3p)/(3) + (1 - p)/(4) + (1 - 2p)/(2) le 1`
`implies 4 + 12p + 3 - 3p + 6 - 12p le 12`
`implies p ge 1//3 " (1)"`
Also, `0 le P(A) le 1`
`implies 0 le (1 + 3p)/(3) le 1`
`0 le 1 + 3p le 3`
`implies -(1)/(3) le p le (2)/(3)` (2)
`implies 0 le P(B) le 1`
`implies 0 le (1-p)/(4) le 1`
`implies 0 le 1 - p le 4`
`implies -3 le p le 1` (3)
`0 le P(C) le 1`
`implies 0 le (1 - 2p)/(2) le 1`
`implies -(1)/(2) le p le (1)/(2)` (4)
Combining (1), (2), (3), and (4), we get
`(1)/(3) le p le (1)/(2)`