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)
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?
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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.