## Interview Question

**Country:**India

**Interview Type:**Written Test

Good one, an alternative:

`P(AUBUC) = P(A) + P(B) + P(C) - P(AnB) - P(BnC) - P(CnA) + P(AnBnC).`

Since A, B, C are independent events, the above becomes

```
P(AUBUC) = P(A) + P(B) + P(C) - P(A)*P(B) - P(B)*P(C) - P(C)*P(A) + P(A)*P(B)*P(C), which is
1/3 + 1/4 + 1/2 - 1/12 - 1/8 - 1/6 + 1/24 = 3/4
```

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{not getting post 1} = 2 / 3

- Ehsan August 24, 2013Pr{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