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Question: If \(( 1 + 3 p ) / 3 , ( 1 - p ) / 4\) and \(( 1 - 2 p ) / 2\) are the probabilities of three mut...

If (1+3p)/3,(1−p)/4( 1 + 3 p ) / 3 , ( 1 - p ) / 4 and (1−2p)/2( 1 - 2 p ) / 2 are the probabilities

of three mutually exclusive events, then the set of all values of p is

A

13≤p≤12\frac { 1 } { 3 } \leq p \leq \frac { 1 } { 2 }

B

13<p<12\frac { 1 } { 3 } < p < \frac { 1 } { 2 }

C

12≤p≤23\frac { 1 } { 2 } \leq p \leq \frac { 2 } { 3 }

D

12<p<23\frac { 1 } { 2 } < p < \frac { 2 } { 3 }

Answer

13≤p≤12\frac { 1 } { 3 } \leq p \leq \frac { 1 } { 2 }

Explanation

Solution

Since (1+3p)3,(1−p)4\frac { ( 1 + 3 p ) } { 3 } , \frac { ( 1 - p ) } { 4 } and (1−2p2)\left( \frac { 1 - 2 p } { 2 } \right) are the probabilities of the three events, we must have

0≤1−2p2≤10 \leq \frac { 1 - 2 p } { 2 } \leq 1

⇒ −1≤3p≤2,−3≤p≤1- 1 \leq 3 p \leq 2 , - 3 \leq p \leq 1 and −1≤2p≤1- 1 \leq 2 p \leq 1

⇒ −13≤p≤23,−3≤p≤1- \frac { 1 } { 3 } \leq p \leq \frac { 2 } { 3 } , - 3 \leq p \leq 1 and −12≤p≤12- \frac { 1 } { 2 } \leq p \leq \frac { 1 } { 2 } .

Also as 1−2p2\frac { 1 - 2 p } { 2 } are the probabilities of three mutually exclusive events,

0≤1+3p3+1−p4+1−2p2≤10 \leq \frac { 1 + 3 p } { 3 } + \frac { 1 - p } { 4 } + \frac { 1 - 2 p } { 2 } \leq 1

⇒ 0≤4+12p+3−3p+6−12p≤120 \leq 4 + 12 p + 3 - 3 p + 6 - 12 p \leq 12 ⇒ 13≤p≤133\frac { 1 } { 3 } \leq p \leq \frac { 13 } { 3 }

Thus the required values of p are such that

max⁡{−13,−3,−12,13}≤p≤min⁡{23,1,12,133}\max \left\{ - \frac { 1 } { 3 } , - 3 , - \frac { 1 } { 2 } , \frac { 1 } { 3 } \right\} \leq p \leq \min \left\{ \frac { 2 } { 3 } , 1 , \frac { 1 } { 2 } , \frac { 13 } { 3 } \right\} ⇒ 13≤p≤12\frac { 1 } { 3 } \leq p \leq \frac { 1 } { 2 }.