One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently, the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted} You are told that the chances of John’s promotion is same as that of Gurpreet, Rita’s chances of promotion are twice as likely as Johns. Aslam’s chances are four times that of John.
(a) Determine P (John promoted)
P (Rita promoted)
P (Aslam promoted)
P (Gurpreet promoted)
(b) If A = {John promoted or Gurpreet promoted}, find P (A).
One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently, the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted} You are told that the chances of John’s promotion is same as that of Gurpreet, Rita’s chances of promotion are twice as likely as Johns. Aslam’s chances are four times that of John.
(a) Determine P (John promoted)
P (Rita promoted)
P (Aslam promoted)
P (Gurpreet promoted)
(b) If A = {John promoted or Gurpreet promoted}, find P (A).

Solution:

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

Let’s say
Events that John promoted E_{1}

Events that Rita promoted E_2

Events that Aslam promoted E_3

Events that Gurpreet promoted E_4

Given that chances of John’s promotion is same as that of Gurpreet
P(E_1)=P(E_4) \dots \dots(1)
Given that Rita’s chances of promotion are twice as likely as John
P(E_2)=2P(E_1) \dots \dots(2)
and chances of Aslam’s promotion are four times that of John
P(E_3)=4P(E_1) \dots \dots(3)
As, the sum of all probabilities =1
\Rightarrow P(E_1)+P(E_2)+P(E_3)+P(E_4)=1
\Rightarrow P(E_1)+2P(E_1)+4P(E_1)+P(E_1)=1
\Rightarrow 8P(E_1)=1
\Rightarrow P(E_1)=1/8 \dots \dots 4
(a) P( John promoted )=P\left(E_{1}\right)
=\frac{1}{8}[ from eq.(4) ]
P( Rita promoted )=P\left(E_{2}\right)
From eq. 2 we have
=2 P\left(E_{1}\right)
From eq. 4
=2 \times \frac{1}{8}
=\frac{1}{4}
P( Aslam promoted )=P\left(E_{3}\right)
From eq.3 we have
=4 P\left(E_{1}\right)
From eq.4 it can be written as
=4 \times \frac{1}{8}
=\frac{1}{2}
P( Gurpreet promoted )=P\left(E_{4}\right)
From eq.1
=P\left(E_{1}\right)
=\frac12
\mathrm{P}( Gurpreet promoted )=\mathrm{P}\left(\mathrm{E}_{4}\right)
From 1
=\mathrm{P}\left(\mathrm{E}_{1}\right)
=\frac{1}{8}
(b) It is given that A=( John promoted or Gurpreet promoted)
\begin{array}{l} \therefore, A=E_{1} \cup E_{4} \\ P(A)=P\left(E_{1} \cup E_{4}\right) \end{array}
Using the general rule of addition, we have
\begin{array}{l} =P\left(E_{1}\right)+P\left(E_{4}\right)-P\left(E_{1} \cap E_{4}\right) \\ =P\left(E_{1}\right)+P\left(E_{1}\right)-0[\text { from (i) }] \\ =\frac{1}{8}+\frac{1}{8} \\ =\frac{2}{8} \\ =\frac{1}{4} \end{array}