If (1+4p)4,(1−p)2,and (1−2p)2 are the probabilities of three

Question

\frac{(1 + 4 p)}{4}, \frac{(1 - p)}{2}, and \frac{(1 - 2 p)}{2}

are the probabilities of three mutually exclusive events , then the value of p is

Options:

A . 12

B . 13

C . 14

D . 15

Answer: Option A
:
A

\frac{1 + 4 p}{4}, \frac{1 - p}{2}, \frac{1 - 2 p}{2}

are the probabilities of three mutually exclusive events.

∴ 0 \leq \frac{1 + 4 p}{p} \leq 1, 0 \leq \frac{1 - p}{2} \leq 1

0 \leq \frac{1 - 2 p}{2} \leq 1 a n d 0 \leq (\frac{1 + 4 p}{4}) + (\frac{1 - p}{2}) + \frac{1 - 2 p}{2} \leq 1

∴ \frac{- 1}{4} \leq p \leq \frac{3}{4} 1 - \leq p \leq 1, \frac{1}{2} \leq p \leq \frac{1}{2} a n d p \leq \frac{5}{2}

\Rightarrow p = \frac{1}{2}

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