Probability 1 Question 14
22. If $\frac{1+3 p}{3}, \frac{1-p}{4}$ and $\frac{1-2 p}{2}$ are the probabilities of three mutually exclusive events, then the set of all values of $p$ is… .
(1986, 2M)
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Answer:
Correct Answer: 22. $\frac{1}{3} \leq p \leq \frac{1}{2}$
Solution:
- Since, $\frac{1+3 p}{3}, \frac{1-p}{4}$ and $\frac{1-2 p}{2}$ are the probability of mutually exclusive events.
$$ \begin{aligned} & \therefore \quad \frac{1+3 p}{3}+\frac{1-p}{4}+\frac{1-2 p}{2} \leq 1 \\ & \Rightarrow \quad 4+12 p+3-3 p+6-12 p \leq 12 \\ & \Rightarrow \quad 13-3 p \leq 12 \\ & \Rightarrow \quad p \geq \frac{1}{3} \\ & \text { and } 0 \leq \frac{1+3 p}{3} \leq 1,0 \leq \frac{1-p}{4} \leq 1,0 \leq \frac{1-2 p}{2} \leq 1 \\ & \Rightarrow \quad 0 \leq 1+3 p \leq 3,0 \leq 1-p \leq 4,0 \leq 1-2 p \leq 2 \\ & \Rightarrow \quad-\frac{1}{3} \leq p \leq \frac{2}{3}, 1 \geq p \geq-3, \frac{1}{2} \geq p \geq-\frac{1}{2} \end{aligned} $$
From Eqs. (i) and (ii), $1 / 3 \leq p \leq 1 / 2$
