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If ((1-3p))/2,((1+4p))/3,((1+p))/6 are t...

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.

Text Solution

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Denote the events by `E_1,E_2 " and " E_3`
`P(E_1)=(1-3p)/2,P(E_2)=(1+4p)/3,P(E_3)=(1+p)/6`
Since events `E_1, E_2, E_3` are mutually exclusive
`P(E_1cupE_2cupE_3)=P(E_1)+P(E_2)+P(E_3)`
`=(1-3p)/2+(1+4p)/3+(1+p)/6=(3-9p+2+8p+1+p)/6=1`
Thus `E_1, E_2, E_3` are exhaustive as given in the problem.
Now `0lt=P(E_i)lt=1` for i=1, 2, 3 gives
`0lt=P(E_1)lt=1`
`implies 0lt=(1-3p)/2lt=1 implies 0 lt=1-3p tl= 2`
`implies -1 tl= -3p lt= 1 implies -1 lt= 3p lt= 1`
`implies -(1)/(3) lt= p lt= 1/3 " " ....(i)`
Also `0 lt= P(E_2) lt= 1`
`implies 0 lt= (1+4p)/(3) lt= 1 implies 0 lt= 1=4plt=3`
`implies -1 lt= 4p lt= 2 implies -(1)/(4) lt= p lt= 1/2 " " ....(ii)`
and `0 lt= P(E_3) lt= 1`
`implies 0 lt= (1+p)/6 lt= 1`
`implies 0 lt= 1+p lt= 6`
`implies -1 lt= p lt=5 " "....(iii)`
The intersection of (i), (ii) and (iii) gives
`-(1)/(4) lt p lt 1/3`
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