One of the four persons John, Rita, Aslam

Solution:

Given Sample Space, S = John promoted, Rita promoted, Aslam promoted, Gurpreet promoted

Let E1 = events that John promoted

E2 = events that Rita promoted

E3 = events that Aslam promoted

E4 = events that Gurpreet promoted

It is given that chances of John’s promotion is same as that of Gurpreet

P (E1) = P (E4) ……1

It is given that Rita’s chances of promotion are twice as likely as John

P (E2) = 2P (E1) ……2

and Aslam’s chances of promotion are four times that of John

P (E3) = 4 P (E1) ……3

Since, sum of all probabilities = 1

⇒ P (E1) + P (E2) + P (E3) + P (E4) = 1

⇒ P (E1) + 2P (E1) + 4 P (E1) + P (E1) = 1

⇒ 8P (E1) = 1

⇒ P (E1) = 1/8 ….4

(a) $P$ (John promoted) $=P\left(E_{1}\right)$

$=\frac{1}{8}$ [from (iv)]

$P$ (Rita promoted) $=P\left(E_{2}\right)$

From 2 we have

$=2 P\left(E_{1}\right)$

From 4

$=2 \times \frac{1}{8}$

$=\frac{1}{4}$

$P$ (Aslam promoted) $=P\left(E_{3}\right)$

From 3 we have

$=4 \mathrm{P}\left(\mathrm{E}_{1}\right)$

From 4 we can write as

$=4 \times \frac{1}{8}$

$=\frac{1}{2}$

$P$ (Gurpreet promoted) $=P\left(E_{4}\right)$

 

From 1

$=P\left(E_{1}\right)$

$=1 / 2$

$P$ (Gurpreet promoted) $=P\left(E_{4}\right)$

 

From 1

$=P\left(E_{1}\right)$

$=\frac{1}{8}$

(b) Given $A=$ (John promoted or Gurpreet promoted)

$\therefore, A=E_{1} \cup E_{4}$

$P(A)=P\left(E_{1} \cup E_{4}\right)$

By general rule of addition, we have

$=P\left(E_{1}\right)+P\left(E_{1}\right)-0[$ from (i) $]$

$=\frac{1}{8}+\frac{1}{8}$

$=\frac{2}{8}$

$=\frac{1}{4}$