1.

If `((1-3p))/2,((1+4p))/3,((1+p))/6`are theprobabilities of three mutually excusing and exhaustive events, then the setof all values of `p`isa. (0,1)b. (-1/4,1/3) c.(0,1/3) d.A. (0, 1)B. `[-1//4, 1//3]`C. `(0, 1//3)`D. `(0, prop)`

Answer» Correct Answer - B
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]`


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