CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image