Question

If $\frac{1+4p}{4},\frac{1-p}{4},\frac{1-2p}{4}$ are probabilities of three mutually exclusive and exclusive and exhaustive events, then the possible value of p belong to the set

• A. $(0,\frac{2}{3})$
• B. $[0,\frac{1}{2}]$
• C. $[-\frac{1}{4},\frac{1}{2}]$
• D. $[-\frac{2}{3},\frac{2}{3}]$

Answer 1

The correct option is C $[-\frac{1}{4},\frac{1}{2}]$

Since the probabilities are greater than 0 or less than (=)1.

Also, the probabilities are mutually excessive

and exhaustive.

$0<\frac{1+4p}{4}<1$ or $0<1+4p<4...\left(1\right)$

$0<\frac{1-p}{4}<1$

$0<1-p<4...\left(2\right)$

$0<1\frac{1-2p}{4}<1$

$0<1-2p<4...\left(3\right)$

from (1), (2), (3), we get

$-\frac{1}{4}<p<\frac{3}{4}$

$1>p>-3$ or $-3<p<1$

$-\frac{3}{2}<p<\frac{1}{2}$

$\therefore$ interval of p can be $[\frac{-1}{4},\frac{1}{2}]$

Option C