# #2 get free 2 SBD now .... jus solve this question

in #contest6 years ago

14/9/2017---to---21/9/2017

# Question

### 3 Management trainees apply for company quarters. There are 3 available company quarters c1,c2,and c3. What is the probability that all three management trainees apply for the same quarter.

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# rules

• This competition will last for 7 days. After that, no more submission will be accepted
• Participants can submit an unlimited number of solutions, however only the latest answer will be considered

# Prizes

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I will guess 3.6%

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(1/3)^3 = ~0.03703703703 = ~3.7037037%

1/27=0.03703703703

6 years ago (edited)

So after reading this... I reviewed what I studied in statistics and I still not sure if I remember it right... But I think all of those who commented before me are wrong.

There are 3 items (x,y.z), each one of them can have one of 3 events (A, B, C). the order is important. So there are 27 outcomes for this problem.

But 3 of the outcomes fit the citera "all three management trainees apply for the same quarter" that is in the three events AAA, BBB, CCC. so we have a probabilty of 3/27 = %11.111.

## My solution is: %11.111

6 years ago (edited)

Ans is 1/9

only 3 quarters can be chosen by then so it will be 3/27 = 1/9

1/9 = 11.1111...%

6 years ago (edited)

It's one-third, or ~33%.

if we make a three by three matrix, we'll see why:

we'll call the management trainees t1, t2 and t3.
and recall that the quarters are represent by c1, c2, c3.

so here is all the possible combinations:
[t1c1][t1c2][t1c3]
times
[t2c1][t2c2][t2c3]
times
[t3c1][t3c2][t3c3]

so that's 3x3x3 = 27 possible combinations

we can see that anytime all three trainees apply for the same job:

t1cx
t2cx
t3cx

happens three possible times.

3/27 = 1/9 =~ 11.11%

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