4 persons will be employed for a job but 6 persons applies to the job. What is the probability that Steve will be employed?
My solution is C(1,1)STEVE IS SELECTED
C(5,3) 3 WORKERS WILL BE CHOSEN
C(1,1)C(5,3)/C(6,4)
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4 persons will be employed for a job but 6 persons applies to the job. What is the probability that Steve will be employed?
My solution is C(1,1)STEVE IS SELECTED
C(5,3) 3 WORKERS WILL BE CHOSEN
C(1,1)C(5,3)/C(6,4)
Quote:
Originally Posted by kastamonu
It would look better to write as
Kastamonu
since Steve must be selected ...the favourite outcomes will be the combinations of the remaining 5 applicants by 3 . check it...
Many Thanks Plato
Many Thanks Mindanman I will check it
My be I missed something in the question.
All applicants have the same chance to get the job, so the probability that Steve (or any one else) get the job is 1/6.
Quote:
Originally Posted by Kmath
What does that have to do with the question? Nothing.
From a set of six, how many subsets of four are there?
Of those sets of four, how many have Steve as a member?
This is a simple counting question.
So stupid fault! Ok, got it, thanks.Quote:
Originally Posted by Plato
I know I'm probably wrong but I always use to think of problems like this as:
4 out of 6 people will be selected or 4/6 which can be reduced to 2/3 which can be written as the percentage 66.6%, of the people will be selected, therefor he has a 66% percent chance. :s is this even close to something correct or was I completely wrong all along?
--- edit ---
From this I confirmed it's correct. There are 15 subsets of four in a set of six. Ten of the 15 subsets have steve. That works out to 10/15, that is steve will be selected 10 out of 15 times or ten times in 15 difference universes. or 66.6 % of the time.(Cool)Quote:
Originally Posted by Plato