Nidhs wrote:Joshua and Jose work att an auto repair center with 4 other workers. For a survey on healthcare insurance, 2 of the 6 workers will be randomly chosen to be interviewd. What is the probability that Joshua and Jose both will be chosen?

1)1/15
2) 1/12
3)1/9
4)1/6
5) 1/3

The correct ans is A and i chose E can someone explain?

We can do this is a probability problem or a combinations problem.

We're choosing 2 out of 6 and we want to know the probability of getting BOTH Josh and Joe, so we want 1 specific pair out of all of the possible pairs.

So, we have 1 desired outome and 6C2 total possible pairs.

6C2 = 6!/2!(6-2)! = 6!/2!4! = 6*5*4!/2*4! = 6*5/2 = 15

Therefore, there's a 1/15 chance of getting the two Js.

We also could have used probability.

Chance that 1st person is one of the two is 2/6.

Once we've chosen one of them, chance that second person is the other J is 1/5.

2/6 * 1/5 = 2/30 = 1/15.

Thank you Stuart.

When there's no additional info in questions like that, do we always assume the choosing will be done 1 by 1? Say not like taking 2 balls out of 20 ball basket at a time?? this really confuses me sometimes even I have done this correctly..

Stuart Kovinsky wrote:

Nidhs wrote:Joshua and Jose work att an auto repair center with 4 other workers. For a survey on healthcare insurance, 2 of the 6 workers will be randomly chosen to be interviewd. What is the probability that Joshua and Jose both will be chosen?

1)1/15
2) 1/12
3)1/9
4)1/6
5) 1/3

The correct ans is A and i chose E can someone explain?

We can do this is a probability problem or a combinations problem.

We're choosing 2 out of 6 and we want to know the probability of getting BOTH Josh and Joe, so we want 1 specific pair out of all of the possible pairs.

So, we have 1 desired outome and 6C2 total possible pairs.

6C2 = 6!/2!(6-2)! = 6!/2!4! = 6*5*4!/2*4! = 6*5/2 = 15

Therefore, there's a 1/15 chance of getting the two Js.

We also could have used probability.

Chance that 1st person is one of the two is 2/6.

Once we've chosen one of them, chance that second person is the other J is 1/5.

2/6 * 1/5 = 2/30 = 1/15.

there are a few ways to solve this problem

1. using combinatorics concept

there are 6!/(2!4!)=15 ways of selecting group of 2 people from this group six workers

next, using the "reducing the pool concept" ... if Jose and Joshua MUST be chosen, then there is only way to do that i.e. 1 * 1 =1

prob = 1/15

2. using slot method

there are 6 * 5 ways of choosing two people individually

and, in order to get joshua and jose....we can choose the first person in 2 ways (b/c it doesnt matter if its joshua or jose) and we can choose the second person in 1 way, i.e. 2* 1 =2

prob = 2/30 or 1/15

3. using probability

you can get Jose-Joshua or Joshua-Jose i.e. there are two ways of getting them

probability Jose-Josha = 1/6 * 1/5 = 1/30
probability Joshua-Jose = 1/6 * 1/5 = 1/30
Since we can have the first one or the second one
we should sum the two
which gets us to 1/15

give kudos if you care to!

san2009 wrote:there are a few ways to solve this problem

1. using combinatorics concept

there are 6!/(2!4!)=15 ways of selecting group of 2 people from this group six workers

next, using the "reducing the pool concept" ... if Jose and Joshua MUST be chosen, then there is only way to do that i.e. 1 * 1 =1

prob = 1/15

2. using slot method

there are 6 * 5 ways of choosing two people individually

and, in order to get joshua and jose....we can choose the first person in 2 ways (b/c it doesnt matter if its joshua or jose) and we can choose the second person in 1 way, i.e. 2* 1 =2

prob = 2/30 or 1/15

3. using probability

you can get Jose-Joshua or Joshua-Jose i.e. there are two ways of getting them

probability Jose-Josha = 1/6 * 1/5 = 1/30
probability Joshua-Jose = 1/6 * 1/5 = 1/30
Since we can have the first one or the second one
we should sum the two
which gets us to 1/15

give kudos if you care to!

Thank you san2009.Your way of approach is good.whatever the way got the same answer.