Let E1, E2, E3 be three mutually exclusive events such that P ( E 1)=(2+3 p /6), P ( E 2)=(2- p /8) and P ( E 3)=(1- p /2). If the maximum and minimum values of p are p 1 and p 2, then ( p 1+ p 2) is equal to :

Question

Let E1, E2, E3 be three mutually exclusive events such that P ( E 1)=(2+3 p /6), P ( E 2)=(2- p /8) and P ( E 3)=(1- p /2). If the maximum and minimum values of p are p 1 and p 2, then ( p 1+ p 2) is equal to : (A) (2/3) (B) (5/3) (C) (5/4) (D) 1

Tardigrade Admin · Accepted Answer

0 ≤ P ( E i) ≤ 1 text for i =1,2,3 ⇒-2 / 3 ≤ p ≤ 1 E 1 E 2 E 3 are mutually exclusive P ( E 1)+ P ( E 2)+ P ( E 3) ≤ 1 ⇒ 2 / 3 ≤ p ≤ 1 p 1=1, p 2=2 / 3 p 1+ p 2=5 / 3

Q. Let $E_{1}, E_{2}, E_{3}$ be three mutually exclusive events such that $P (E_{1}) = \frac{2 + 3 p}{6}$ , $P (E_{2}) = \frac{2 - p}{8}$ and $P (E_{3}) = \frac{1 - p}{2}$ . If the maximum and minimum values of $p$ are $p_{1}$ and $p_{2}$ , then $(p_{1} + p_{2})$ is equal to :

$\frac{2}{3}$

$\frac{5}{3}$

$\frac{5}{4}$

1

$0 \leq P (E_{i}) \leq 1 for i = 1, 2, 3$
$\Rightarrow - 2/3 \leq p \leq 1$
$E_{1} & E_{2} & E_{3}$ are mutually exclusive
$P (E_{1}) + P (E_{2}) + P (E_{3}) \leq 1$
$\Rightarrow 2/3 \leq p \leq 1$
$p_{1} = 1, p_{2} = 2/3$
$p_{1} + p_{2} = 5/3$

Q. Let E1​,E2​,E3​ be three mutually exclusive events such that P(E1​)=62+3p​, P(E2​)=82−p​ and P(E3​)=21−p​. If the maximum and minimum values of p are p1​ and p2​, then (p1​+p2​) is equal to :

32​

35​

45​

1

0≤P(Ei​)≤1 for i=1,2,3 ⇒−2/3≤p≤1 E1​&E2​&E3​ are mutually exclusive P(E1​)+P(E2​)+P(E3​)≤1 ⇒2/3≤p≤1 p1​=1,p2​=2/3 p1​+p2​=5/3

Q. Let $E_{1}, E_{2}, E_{3}$ be three mutually exclusive events such that $P (E_{1}) = \frac{2 + 3 p}{6}$ , $P (E_{2}) = \frac{2 - p}{8}$ and $P (E_{3}) = \frac{1 - p}{2}$ . If the maximum and minimum values of $p$ are $p_{1}$ and $p_{2}$ , then $(p_{1} + p_{2})$ is equal to :

$\frac{2}{3}$

$\frac{5}{3}$

$\frac{5}{4}$

$0 \leq P (E_{i}) \leq 1 for i = 1, 2, 3$
$\Rightarrow - 2/3 \leq p \leq 1$
$E_{1} & E_{2} & E_{3}$ are mutually exclusive
$P (E_{1}) + P (E_{2}) + P (E_{3}) \leq 1$
$\Rightarrow 2/3 \leq p \leq 1$
$p_{1} = 1, p_{2} = 2/3$
$p_{1} + p_{2} = 5/3$