Question

If (1 + 3p)/3, (1 – p)/4 and (1 – 2p)/2 are the probabilities of three mutually exclusive events, then the set of all values of p is –

• A. $\frac { 1 } { 3 }$ £ p £ $\frac { 1 } { 2 }$
• B. $\frac { 1 } { 3 }$ < p < $\frac { 1 } { 2 }$
• C. $\frac { 1 } { 2 }$ £ p £$\frac { 2 } { 3 }$
• D. $\frac { 1 } { 2 }$ < p < $\frac { 2 } { 3 }$

Answer 1

Answer: (A)

$\frac { 1 } { 3 }$ £ p £ $\frac { 1 } { 2 }$

Answer 2

Since $\frac { ( 1 + 3 p ) } { 3 }$ ,$\left( \frac { 1 - 2 p } { 2 } \right)$are the probabilities of the three events, we must have

0 £ $\frac { 1 + 3 p } { 3 }$ £ 1, 0 £ $\frac { 1 - \mathrm { p } } { 4 }$ £ 1 and 0 £ $\frac { 1 - 2 p } { 2 }$ £ 1

Ž –1 £ 3p £ 2, –3 £ p £ 1 and –1 £ 2p £ 1

Ž –$\frac { 1 } { 2 }$

Also as $\frac { 1 + 3 p } { 3 }$, $\frac { 1 - \mathrm { p } } { 4 }$ and are the probabilities of three mutually exclusive events

0 £ $\frac { 1 + 3 p } { 3 }$ + $\frac { 1 - \mathrm { p } } { 4 }$ + $\frac { 1 - 2 p } { 2 }$ £ 1

Ž 0 £ 4 + 12p + 3 – 3p + 6 – 12p £ 12 Ž $\frac { 1 } { 3 }$ £ p £ $\frac { 13 } { 3 }$

Thus the required value of p are such that

Max. $\left\{ - \frac { 1 } { 3 } , - 3 , - \frac { 1 } { 2 } , \frac { 1 } { 3 } \right\}$£ p min.$\left\{ \frac { 2 } { 3 } , 1 , \frac { 1 } { 2 } , \frac { 13 } { 3 } \right\}$

Ž $\frac { 1 } { 3 }$ £ p £ $\frac { 1 } { 2 }$