Events A, B, C are mutually exclusive events such that P(A)=\;3x+1/3, P(B)=\;1-x/4 and P(C)=\;1-2x/2 The set of possible values of x are in the interval. [AIEEE 2003]

Correct Answer is : [ 13, 12]≤ft[\;1/3, \;1/2][31,21]

Correct Answer is : [ 13, 12]≤ft[\;1/3, \;1/2][31,21]

Test Index

Probability Part 1

Section: Mathematics

Question : 97 of 100

Marks: +1, -0

Events

A, B, C

P(A)=\;\frac{3x+1}{3}, P(B)=\;\frac{1-x}{4}

P(C)=\;\frac{1-2x}{2}

x

Solution:

Given

P(A)=\;\frac{3x+1}{3}, P(B)=\;\frac{1-x}{4}

and

P(C)=\;\frac{1-2x}{2}

We know for any event

X, 0 \le P(X) \le 1

\therefore 0 \le \;\frac{3x+1}{3} \le 1

\Rightarrow -1 \le 3x \le 2

\Rightarrow -\;\frac{1}{3} \le x \le \;\frac{2}{3}

0 \le \;\frac{1-x}{4} \le 1

\Rightarrow -3 \le x \le 1

0 \le \;\frac{1-2x}{2} \le 1

\Rightarrow -1 \le 2x \le 1

\Rightarrow -\;\frac{1}{2} \le x \le \;\frac{1}{2}

In the question given that

A, B

and

C

are mutually exclusive

\;\text{ So }\; P(A \cup B \cup C)=P(A)+P(B)+P(C)

\Rightarrow P(A \cup B \cup C)=\;\frac{3x+1}{3}+\;\frac{1-x}{4}+\;\frac{1-2x}{2}

\therefore 0 \le \;\frac{3x+1}{3}+\;\frac{1-x}{4}+\;\frac{1-2x}{2} \le 1

0 \le 13-3x \le 12

\Rightarrow \;\frac{1}{3} \le x \le \;\frac{13}{3}

From all those relations, we get

\max \{-\;\frac{1}{3}, -3, -\;\frac{1}{2}, \;\frac{1}{3}\} \le x \le \min \{\;\frac{2}{3}, 1, \;\frac{1}{2}, \;\frac{13}{3}\}

So,

\;\frac{1}{3} \le x \le \;\frac{1}{2}

\Rightarrow \Rightarrow x \in \left[\;\frac{1}{3}, \;\frac{1}{2}\right]

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