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

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.

# Answer submission

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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**##### You have to upvote and must resteemit this post in order to be eligible for the competition

# Prizes

Name of prize | Max no. of prizes | Prizes |
---|---|---|

First prize | 1 | 2 SBD |

Second prize | 1 | 0.5 SBD |

Third prize | 1 | 0.1 SBD |

all paticipent | all | follow and up voted by me |

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The source of questions is this book

answer is displayed on 21/9/2017

Or results also

knowledge-seeker (58)6 years agoI will guess 3.6%

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napkin (50)6 years ago(1/3)^3 = ~0.03703703703 = ~3.7037037%

khalifandro (52)6 years ago1/27=0.03703703703

ahmadmanga (65)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

pps (44)6 years ago (edited)Ans is 1/9

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

anyway (47)6 years ago1/9 = 11.1111...%

bluabaleno (55)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%