# probability

• Feb 21st 2013, 03:29 AM
kastamonu
probability
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, 04:19 AM
Plato
Re: probability
Quote:

Originally Posted by kastamonu
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)

It would look better to write as $\frac{\binom{5}{3}}{\binom{6}{4}}$
• Feb 21st 2013, 04:20 AM
MINOANMAN
Re: 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, 05:39 AM
kastamonu
Re: probability
Many Thanks Plato
Many Thanks Mindanman I will check it
• Feb 21st 2013, 06:55 AM
Kmath
Re: 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, 07:02 AM
Plato
Re: probability
Quote:

Originally Posted by Kmath
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.

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.
• Feb 21st 2013, 08:06 AM
Kmath
Re: probability
Quote:

Originally Posted by Plato
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.
• Feb 24th 2013, 02:56 AM
russ123
Re: 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 ---
Quote:

Originally Posted by Plato
From a set of six, how many subsets of four are there?
Of those sets of four, how many have Steve as a member?

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)