CareerCup

P{not getting post 1} = 2 / 3
Pr{not getting post 2} = 3 / 4
Pr{not getting post 3} = 1 / 2
Pr{not getting any posts} = 2 / 3 * 3 / 4 * 1 / 2 = 1 / 4
Pr{Getting at least one post} = 1 - Pr{not getting any posts} = 3 / 4

11/24

1/4
3/24+2/24+6/24+1/24+2/24+3/24+1/24 = 18/24

The odds will change if other candidates are also interviewing for multiple posts.

This question is a joke.
You cannot answer it without assuming very unrealistic things (which are not mentioned).
1) Independence between interviews of any of the people
2) Uniform distribution (all candidates look the same, act the same, have the same skills, and interviewer also felt the same and asked the same questions ... everything is the same for all interviews)

Now calculate the probability that he goes home unemployed:
2*3*1
-------- = 1/4
3*4*2

Use product rule on numerator and denominator (in numerator, leave him out of every possible choice, so the numbers are 1 less).

So his probability for getting a job is 3/4

{Note, probability distribution are not required as this is a uniform distribution... simply counting is enough}

P(at least one offer) = 1 - p(no offer) = 1 - p(no offer for post 1)*p(no offer for post 2)*p(no offer for post 3) by assuming 3 interviews are independent and each person has equal opportunity for each interview.

p = 1 - 2/3*3/4*1/2 = 3/4

Stupid question.