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Question: If \(\frac { 1 + 3 \mathrm { p } } { 3 } , \frac { 1 - \mathrm { p } } { 4 }\) and \(\frac { 1 - 2 ...

If 1+3p3,1−p4\frac { 1 + 3 \mathrm { p } } { 3 } , \frac { 1 - \mathrm { p } } { 4 } and 1−2p2\frac { 1 - 2 \mathrm { p } } { 2 } are the probabilities of three mutually exclusive events, then the set of all values of p is

A

[-3, 13/3]

B

[0, 13/3]

C

[1/3, ½]

D

[1/3, 2/3]

Answer

1/3,½1/3, ½

Explanation

Solution

Let P(1) = 1+3p3,P(B)=1−p4\frac { 1 + 3 \mathrm { p } } { 3 } , \mathrm { P } ( \mathrm { B } ) = \frac { 1 - \mathrm { p } } { 4 } and P(3) =

Since A, B and C are mutually exclusive , then

P(AUBUC) = P(1) + P(2) + P(3) =

Also, 0≤P(A)≤1,0≤P(B)≤1,0≤P(C)≤10 \leq \mathrm { P } ( \mathrm { A } ) \leq 1,0 \leq \mathrm { P } ( \mathrm { B } ) \leq 1,0 \leq \mathrm { P } ( \mathrm { C } ) \leq 1 and 0≤P(AUBUC)≤10 \leq \mathrm { P } ( \mathrm { AUBUC } ) \leq 1Now, 0≤P(A)≤1⇒−13≤p≤230 \leq \mathrm { P } ( \mathrm { A } ) \leq 1 \Rightarrow \frac { - 1 } { 3 } \leq \mathrm { p } \leq \frac { 2 } { 3 }

0≤P(B)≤1⇒−3≤p≤10 \leq \mathrm { P } ( \mathrm { B } ) \leq 1 \quad \Rightarrow \quad - 3 \leq \mathrm { p } \leq 1

0≤P(C)≤1⇒−12≤p≤120 \leq \mathrm { P } ( \mathrm { C } ) \leq 1 \Rightarrow - \frac { 1 } { 2 } \leq \mathrm { p } \leq \frac { 1 } { 2 }

and 0≤P(AUBUC)≤1⇒13≤p≤1330 \leq \mathrm { P } ( \mathrm { AUBUC } ) \leq 1 \Rightarrow \frac { 1 } { 3 } \leq \mathrm { p } \leq \frac { 13 } { 3 }

Hence, common value is 13≤p≤12\frac { 1 } { 3 } \leq \mathrm { p } \leq \frac { 1 } { 2 }