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)

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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