CGPET 2018 Solved Paper

Since,
(

1+3P

3

),(

1−P

4

) and (

1−2P

2

) are probability of three events.
Hence,
0≤

1+3P

3

≤1,0≤

1−P

4

≤1 and 0≤

1−2P

2

≤1
⇒−1≤3P≤2,−3≤P≤1 and −1≤2P≤1
⇒−

1

3

≤P≤

2

3

,−3≤P≤1 and −

1

2

≤P≤2
⇒

1+3P

3

,

1−P

4

and

1−2P

2

are the probabilities of three mutually exclusive events.
Hence,
0≤

1+3P

3

+

1−P

4

+

1−2P

2

≤1
⇒0≤4+12P+3−3P+6−12P≤12 =

1

3

≤P≤

13

3

Hence, the required value of P.
Maximum {−

1

3

,−3,−

1

2

,

1

3

}≤P≤ Minimum {

2

3

,1,

1

2

,

13

3

}
⇒

1

3

≤P≤

1

2