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
OA is a
why is E the only best option? which probability approach did you use to get your answer?[/spoiler]
Probability
This topic has expert replies
-
- Moderator
- Posts: 772
- Joined: Wed Aug 30, 2017 6:29 pm
- Followed by:6 members
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Approach 1:Roland2rule wrote: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
From the 6 workers, the number of combinations of 2 that can be formed = 6C2 = (6*5)/(2*1) = 15.
Of these 15 pairs, only 1 is good: Joshua and Jose.
Thus:
P(Joshua and Jose are both chosen) = 1/15.
Approach 2:
Joshua and Jose constitute 2 of the 6 workers.
Thus:
P(Joshua or Jose is chosen on the 1st pick) = 2/6.
Since Joshua or Jose is chosen on the 1st pick, only 1 of the 5 remaining workers will be Joshua or Jose.
Thus:
P(Joshua or Jose is chosen on the 2nd pick) = 1/5.
Since we want both events to happen, we multiply the probabilities:
P(Joshua and Jose are both chosen) = 2/6 * 1/5 = 1/15.
The correct answer is A.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
P(Joshua and Jose both chosen) = P(one of the two men is chosen first AND the other man is chosen second)Roland2rule wrote: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
[/spoiler]
= 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
-
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
We could also do All Groups - Groups With Neither Guy - Groups With Only One of the Guys:
All Groups = 6 choose 2 = 6!/(4!2!) = 15
Neither Guy = 4 choose 2 = 4!/(2!2!) = 6
Only One Guy = (4 choose 1) * (2 choose 1) = 8
If Only One Guy doesn't make sense, I'm choosing one of the other four people first, then choosing one of our two guys to go with them.
That leaves us with 15 - 6 - 8 = 1 group with both guys. 1 group out of 15 is 1/15, and we're done.
All Groups = 6 choose 2 = 6!/(4!2!) = 15
Neither Guy = 4 choose 2 = 4!/(2!2!) = 6
Only One Guy = (4 choose 1) * (2 choose 1) = 8
If Only One Guy doesn't make sense, I'm choosing one of the other four people first, then choosing one of our two guys to go with them.
That leaves us with 15 - 6 - 8 = 1 group with both guys. 1 group out of 15 is 1/15, and we're done.
-
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
It seems like this is easy conceptually too. We want to pick both guys, so there's only one way to do that: 2 choose 2. If we're just picking two random guys, there are 6 choose 2, or 15 ways to do that, as shown in my previous post.
From there:
Our target = the one group
Our total = 6 choose 2 = 15 groups
Probability = target/total = 1/15
From there:
Our target = the one group
Our total = 6 choose 2 = 15 groups
Probability = target/total = 1/15
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7245
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
The probability that Joshua and Jose are chosen is 1/6C2 = 1/(6 x 5)/2 = 1/15BTGmoderatorRO wrote: ↑Sat Sep 16, 2017 7:26 pmJoshua 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
OA is a
why is E the only best option? which probability approach did you use to get your answer?[/spoiler]
Answer: A
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews