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Mathematics Question on Probability

Let E1, E2, E3 be three mutually exclusive events such that
P(E1)=2+3p6,P(E2)=2−p8and P(E3)=1−p2.P(E_1)=\frac{2+3p}{6}, P(E_2)=\frac{2−p}{8} and\ P(E_3)=\frac{1−p}{2}.
If the maximum and minimum values of p are p1 and p2, then (p1 + p2) is equal to :

A

23\frac{2}{3}

B

53\frac{5}{3}

C

54\frac{5}{4}

D

1

Answer

53\frac{5}{3}

Explanation

Solution

0≤2+3p6≤10≤\frac{2+3p}{6}≤1
⇒P∈[−23,43]⇒P∈[−\frac{2}{3},\frac{4}{3}]
0≤2−p8≤10≤\frac{2−p}{8}≤1
⇒P∈[−6,2]⇒P∈[−6,2]
0≤1−p2≤10≤\frac{1−p}{2}≤1
⇒P∈[−1,1]⇒P∈[−1,1]
0<P(E1)+P(E2)+P(E3)≤10<P(E_1)+P(E_2)+P(E_3)≤1
0<1312−P8≤10<\frac{13}{12}−\frac{P}{8}≤1
P∈[23,263]P∈[\frac{2}{3},\frac{26}{3}]
Taking intersection of all
P∈[23,1)P∈[\frac{2}{3},1)
P1+P2=53P1+P2=\frac{5}{3}
So, the correct option is (B):53 \frac{5}{3}