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Question

Events A,B,C are mutually exclusive events such that P(A)=3x+13, and P(B)=1x4, and P(C)=12x2. The set of possible values of x is in the interval

A
[13,23]
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B
[13,133]
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C
[0,1]
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D
[13,12]
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Solution

The correct option is D [13,12]
P(A)=3x+13,P(B)=1x4,P(C)=12x20P103x+13103x+1313x23similarly01x414x103x1similarly012x2122x1012x12Also0P(ABC)10P(A)+P(B)+P(C)P(AB)P(BC)P(CA)+P(ABC)10P(A)+P(B)+P(C)1
Since A , B and C are mutually exhaustive hence P(AB)=P(BC)=P(CA)=P(ABC)=0
03x+13+1x4+12x210133x121123x13013x133
comparing all the inequalities of x and taking the intersection we get
13x12orx[13,12]

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