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
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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