Joshua and Jose work at an auto repair center with 4 other

Question

Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?

A. 1/15
B. 1/12
C. 1/9
D. 1/6
E. 1/3

A. 1/15 
B. 1/12 
C. 1/9 
D. 1/6 
E. 1/3

jaynayak · Accepted Answer

Two Methods

1) Probability of chosing Josh first = 1/6
 Probability of chosing Jose second = 1/5
 total = 1/30
 Probability of chosing Jose first = 1/6
 Probability of chosing Josh second = 1/5
 Total = 1/30
Final = 1/30 + 1/30 = 1/15

2) Number of ways two persons can be chosen 6C2 = 15
 Number of ways Josh and Jose are the two persons = 1
 Total = 1/15

BrentGMATPrepNow · Answer

P(Joshua and Jose both chosen) = P(one of the two men is chosen first   AND   the other man is chosen second)
= P(one of the two men is chosen first)   x   P(the other man is chosen second)
= 2/6   x    1/5
= 1/15

Answer: A

Cheers,
Brent

ezhilkumarank · Answer

We should consider Joshua and Jose as one entity and then proceed for the probability.

Total ways of choosing 2 people (Joshua and Jose) out of 6 people -- 
 (6!)/(4! * 2!) 
 =15

Favorable outcomes - 1 (Since Joshua and Josh should be picked up)

Probability -  1/15

Any other thoughts.....

mainhoon · Answer

Interestingly in such questions if one notices the denominator, we know it has to be 15, as 6C2=15, so that rules out B C D. Leaves us with A and B. A says 1/15, B says 5/15. of course there is one choice and that is 1/15, but even if you did not get it the right way, this elimination should help.

trunksli · Answer

I reworded this problem to have Josh and Jose be two guys, and the other 4 are women. What are the chances of picking only guys?

There are two slots, first slot has (2/6) chance of picking a guy, the second slot has (1/5) chance of picking the other guy.

Picture them in a group. The first time you randomly pick someone, there are six people, and two guys. Now, assuming you have picked a guy, you are left with a group of 5 people standing there (1 guy, 4 girls). The chance of picking a guy then is 1/5. Multiply the two probabilities together, and you have 2/30, or 1/15

pankajattri · Answer

Probability of an event = (number of positive outcomes)/(total number of outcomes)

Here number of positive outcomes = choose Josh and Jose when two people are picked at random = 2C2

Total Outcomes = choose two people from a group of 6 people = 6C2

Probability = 2C2 / 6C2 = 1/15

ingoditrust · Answer

total ways to select 2 people out of 6 = 6c2 = 15
the number of ways to select J & J = 2c2 = 1

= 1/ 15 (answer)

email2vm · Answer

I am not able to solve this questions with below method:

probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5

probability required = 1-2/5 = 3/5

Where am I doing wrong?

Regards,
Ravi

Bunuel · Answer

You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.

email2vm · Answer

Hmmm...silly again.

Only Joshua but not jose probability = 4/15 
only jose but not joshua = 4/15

2/5 +4/15+4/15= 14/15

so actual probabilty required is 1-14/15 = 1/15

Cheers Bunuel!!

nyoebic · Answer

Bunuel

I went for a simpler approach and although I get the right answer, I'm not sure if it is correct. Here is how I went about it:

Since the question is only concerned about Josh and Jose, the probability that either one of them is chosen first is: 2/6
After the first is chosen, there are 5 people left, out if which one is either Josh or Jose. Probability of getting them is then: 1/5

So, (2/6)*(1/5)= 1/15

Is this a right approach for this question? To me this seems so much more straightforward...Thoughts? Suggestions?

Bunuel · Answer

Yes, that's correct.

jabhatta2 · Answer

Hi - why doesnt the following work

(1/6) * (1/5) = 1/30

just wondering why this didnt work ...

Bunuel · Answer

What is the logic behind this?

If you say that 1/6 is the probability of choosing Joshua first and 1/5 is the probability of choosing Jose second, then you are missing a case when you are choosing Jose first and Joshua second, so 1/6*1/5 + 1/6*1/5 = 1/15.

ManoramaS · Answer

Bunuel  Please help me why 1-4/6*3/5 is wrong ?

Bunuel · Answer

Hi, Your doubt is already addressed here: https://gmatclub.com/forum/joshua-and-j ... l#p1407022 Hope it helps.

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