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Events $A,B,C$ are mutually exclusive events such that $P(A)=\frac{3x+1}{3},\; \; \; P(B)=\frac{1-x}{4}\; \; \; \; and\; \; P(C)=\frac{1-2x}{2}.$ Then set of possible values of $x$ are in the interval

Option 1)
$\left [ \frac{1}{3},\frac{2}{3} \right ]\;$

Option 2)
$\; \; \left [ \frac{1}{3},\frac{13}{3} \right ]\;$

Option 3)
$\; \left [ 0,1 \right ]$

Option 4)
$\; \left [ \frac{1}{3},\frac{1}{2} \right ]\;$

Answers (1)

best_answer

As se learnt in

Probability of occurrence of an event -

Let S be the sample space then the probability of occurrence of an event E is denoted by P(E) and it is defined as

$P\left ( E \right )=\frac{n\left ( E \right )}{n\left ( S \right )}$

$P\left ( E \right )\leq 1$

$P(E)=\lim_{n\rightarrow\infty}\left(\frac{r}{n} \right )$

- wherein

Where n repeated experiment and E occurs r times.

$\begin{matrix} \\ 0\leq P(A)\leq 1 \\ 0\leq P(B)\leq 1 \\ 0\leq P(C)\leq 1 \\ P(A)+P(B)+P(C)\leq 1 \end{matrix}$ $\left \{ Conditions\ on\ probability\ value \right.$

Thus $0\leqslant \frac{3x+1}{3}\leqslant 1..............(1)$

$\Rightarrow -\frac{1}{3}\leqslant x \leqslant\frac{2}{3}..............1a$

$0 \leq \frac{1-x}{4}\leq 1.....................(2)$

$\Rightarrow -3 \leqslant x\leqslant 1...................2a$

$0\leqslant \frac{1-2x}{2}\leqslant 1...................(3)$

$\Rightarrow \frac{-1}{2}\leqslant x \leqslant \frac{1}{2}...................3a$

$P(A)+P(B)+P(C)\leqslant 1$

$\Rightarrow \frac{3x+1}{3} + \frac{1-x}{4}+ \frac{1-2x}{2}\leq 1$

$\Rightarrow 4(3x+1) +3(1-x)+6(1-2x)\leqslant 1$

$\Rightarrow -3x+0\leqslant -1$

$\Rightarrow x\geq \frac{1}{3}....................(4)$

From 1a,2a,3a and 4 we have

$\frac{1}{3}\leqslant x \leqslant \frac{1}{2}$

Option 1)

$\left [ \frac{1}{3},\frac{2}{3} \right ]\;$

Incorrect

Option 2)

$\; \; \left [ \frac{1}{3},\frac{13}{3} \right ]\;$

Incorrect

Option 3)

$\; \left [ 0,1 \right ]$

Incorrect

Option 4)

$\; \left [ \frac{1}{3},\frac{1}{2} \right ]\;$

Correct

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