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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- Feb 21st 2013, 02:29 AMkastamonuprobability
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) - Feb 21st 2013, 03:19 AMPlatoRe: probability
- Feb 21st 2013, 03:20 AMMINOANMANRe: probability
Kastamonu

since Steve must be selected ...the favourite outcomes will be the combinations of the remaining 5 applicants by 3 . check it... - Feb 21st 2013, 04:39 AMkastamonuRe: probability
Many Thanks Plato

Many Thanks Mindanman I will check it - Feb 21st 2013, 05:55 AMKmathRe: probability
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. - Feb 21st 2013, 06:02 AMPlatoRe: probability
- Feb 21st 2013, 07:06 AMKmathRe: probability
- Feb 24th 2013, 01:56 AMruss123Re: probability
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)