Question

# If $$\dfrac{1+3P}{3},\dfrac{1-P}{4}$$ and $$\displaystyle \frac{1-2P}{2}$$ are the probabilities of three mutually exclusive events, then $$P\in$$

A
[0,1]
B
[0,12]
C
[13,1]
D
[13,12]

Solution

## The correct option is D $$\left[\displaystyle \frac{1}{3},\frac{1}{2} \right]$$Given, $$P(E_{1})=\displaystyle \frac { 1+3P }{ 3 } , P(E_{2})=\displaystyle \frac { 1-P }{ 4 },P(E_{3})=\displaystyle \frac { 1-2P }{ 2 }$$Using $$0 \leq P(E_{i}) \leq 1$$, we get $$\displaystyle 0\le \frac { 1+3P }{ 3 } \le 1\\ \displaystyle \Rightarrow -\frac { 1 }{ 3 } \le P\le \frac { 2 }{ 3 } ....(1)$$$$\displaystyle 0\le \frac { 1-P }{ 4 } \le 1\\ \displaystyle \Rightarrow -3\le P\le 1.....(2)$$$$\displaystyle 0\le \frac { 1-2P }{ 2 } \le 1\\ \displaystyle \Rightarrow -\frac { 1 }{ 2 } \le P\le \frac { 1 }{ 2 } ....(3)$$Also, $$E_{1},E_{2},E_{3}$$ are three mutually exclusive events.So, $$P(E_{1})+P(E_{2})+P(E_{3})=\displaystyle \frac { 1+3P }{ 3 } +\frac { 1-P }{ 4 } +\frac { 1-2P }{ 2 } \displaystyle =\frac{13-3P}{12}$$We know $$\displaystyle 0\le \frac{13-3P}{12}\le 1$$$$\Rightarrow \displaystyle 0\le 13-3P \le 12$$$$\Rightarrow \displaystyle -13 \le -3P \le -1$$$$\Rightarrow \displaystyle \frac{1}{3} \le P \le \frac{13}{3}....(4)$$Hence, the range of  values of $$P$$ common to $$(1), (2), (3), (4)$$ are between $$\displaystyle \left[ \frac { 1 }{ 3 } ,\frac { 1 }{ 2 } \right]$$.Maths

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