Q. 105.0( 1 Vote )

# 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).

Answer :

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

Let

E_{1} = events that John promoted

E_{2} = events that Rita promoted

E_{3} = events that Aslam promoted

E_{4} = events that Gurpreet promoted

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

P(E_{1}) = P(E_{4}) …(i)

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

P(E_{2}) = 2P(E_{1}) …(ii)

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

P(E_{3}) = 4P(E_{1}) …(iii)

Since, sum of all probabilities = 1

⇒ P(E_{1}) + P(E_{2}) + P(E_{3}) + P(E_{4}) = 1

⇒ P(E_{1}) + 2P(E_{1}) + 4P(E_{1}) + P(E_{1}) = 1

⇒ 8P(E_{1}) = 1

…(iv)

(a) P (John promoted) = P(E_{1})

[from (iv)]

P (Rita promoted) = P(E_{2})

= 2P(E_{1}) [from (ii)]

[from (iv)]

P (Aslam promoted) = P(E_{3})

= 4P(E_{1}) [from (iii)]

[from (iv)]

P (Gurpreet promoted) = P(E_{4})

= P(E_{1}) [from (i)]

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

∴, A = E_{1}⋃ E_{4}

P(A) = P(E_{1}⋃ E_{4})

= P(E_{1}) + P(E_{4}) – P(E_{1}⋂ E_{4})

[∵ By general addition rule]

= P(E_{1}) + P(E_{1}) – 0 [from (i)]

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