Question

Out of 25 employees of a company, 5 are engineers. Three employees are selected at random for granting leave. What is the probability that (i) all the three are engineers?(ii) none of them is an engineer? (iii) at least one of them is an engineer? 

Answer 1

Solution:

(i) Let A₁ be event that all the three are engineers.

Then

P(A₁)=m/n

m=₅C₃=5!/3!/2!=10,

n=₂₅C₃=25!/3!/22!=2300,

P(A₁)=10/2300=1/230.

(ii) Let A₂ be event that "none of them is an engineer".

Then

P(A₂)=m/n

m=₂₀C₃=20!/3!/17!=1140,

n=₂₅C₃=25!/3!/22!=2300,

P(A₂)=1140/2300=57/115.

(iii) Let A₃ be event that "at least one of them is an engineer".

Then

$A_2\bigcup A_3 =U$,

$A_2 \bigcap A_3 =\varnothing$

P(A₂)+P(A₃)=1

P(A₃)=1-P(A₂)

P(A₃)=1-57/115=58/115.

Answer:

(i)   the probability that all the three are engineers is equal to 1/230;

(ii) the probability that none of them is an engineer is equal to 57/115;

(iii) the probability that at least one of them is an engineer is equal to 58/115.