If `((1-3p))/2,((1+4p))/3,((1+p))/6` are the probabilities of three mutually excusing and exhaustive events, then the set of all values of `p` is a. (0,1) b. (-1/4,1/3) c. (0,1/3) d.

Options

A. (0, 1)
B. `[-1//4, 1//3]`
C. `(0, 1//3)`
D. `(0, prop)`

Answer

B

Explanation

Since `(1-3p)/(2),(1+4p)/(3),(1+p)/(6)` are the probabilities of three mutually exculsive and exhaustive events. Therefore,
`(1-3p)/(2) ge 0, (1+4p)/(3) ge 0, (1+p)/(6) ge 0`
and, `(1-3p)/(2)+(1+4p)/(3)+(1+p)/(6)=1`
`implies -(1)/(4) le p le (1)/(3) implies p in [-1//4, 1//3]`