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# If $$(1+3p)/3,(1-p)/4$$ and $$(1-2p)/2$$ are the probabilities of three mutually exclusive events, then the set of all values of p is.............

Solution

## Let,$$P(A)=\dfrac{1+3p}{3},P(B)=\dfrac{1-p}{4},P(C)=\dfrac{1-2p}{2}$$A,B and C are three mutually exclusive events.$$\therefore P(A)+P(B)+P(C)\leq 1$$$$\Rightarrow \dfrac{1+3p}{3}+\dfrac{1-p}{4}+\dfrac{1-2p}{2}\leq 1 \leq 1$$$$\Rightarrow 4+12p+3-3p+6-12p \leq 12$$$$\Rightarrow p \leq 1/3$$................(1)Also,$$0 \leq P(A) \leq 1 \Rightarrow 0 \leq \dfrac{1+3p}{3}\leq 1$$$$0 \leq 1+3p \leq 3$$$$\Rightarrow -\dfrac{1}{3}\leq p \leq \dfrac{2}{3}$$............(2)$$0\leq P(B)\leq 1\Rightarrow 0 \leq \dfrac{1-p}{4}\leq 1$$$$\Rightarrow 0 \leq 1-p\leq 4$$$$\Rightarrow -3 \leq p \leq 1$$.................(3)$$0 \leq P(C)\leq 1\Rightarrow 0 \leq \dfrac{1-2p}{2}\leq 1$$$$\Rightarrow -\dfrac{1}{2}\leq p \leq \dfrac{1}{2}$$.................(4)Combining eq.(1),(2),(3) and (4) ,we get$$\dfrac{1}{3}\leq p \leq \dfrac{1}{2}$$Mathematics

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