A coach will select the members of a 5-players team from 9 players, including A and B. If the 5 players are chosen at random, What is the probability that the coach chooses a team that includes both A and B?
a. 1/9
b. 1/6
c. 2/9
d. 5/18
e. 1/3
The OA is the option D.
I've got confused with this question. Can any expert give me some help? Thanks in advanced.
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
A coach will select the members of a 5-players team from ...
This topic has expert replies
-
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
Since 5 of the 9 players are included on the team, P(A is included) = 5/9.Vincen wrote:A coach will select the members of a 5-players team from 9 players, including A and B. If the 5 players are chosen at random, What is the probability that the coach chooses a team that includes both A and B?
a. 1/9
b. 1/6
c. 2/9
d. 5/18
e. 1/3
Since 4 of the remaining 8 players are included on the team, P(B is included) = 4/8.
To combine the probabilities, we multiply:
5/9 * 4/8 = 5/18.
The correct answer is D.
Alternate approach:
P = (5-member teams with A and B)/(all possible 5-member teams).
All possible 5-member teams:
From the 9 players, the number of ways to choose 5 = 9C5 = (9*8*7*6*5)/(5*4*3*2*1) = 126.
5-member teams with A and B:
Once A and B have been selected, the coach must select 3 additional players to combine with A and B.
The result will be a 5-member team with A and B.
From the 7 remaining players, the number of ways to choose 3 to combine with A and B = 7C3 = (7*6*5)/(3*2*1) = 35.
Resulting probability:
(5-member teams with A and B)/(all possible 5-member teams) = 35/126 = 5/18.
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(John and Peter both on the team) = (# of teams that include both John and Peter) / (total # of 5-person teams possible)Vincen wrote:A coach will select the members of a 5-players team from 9 players, including A and B. If the 5 players are chosen at random, What is the probability that the coach chooses a team that includes both A and B?
a. 1/9
b. 1/6
c. 2/9
d. 5/18
e. 1/3
a) # of teams that include both John and Peter
- Put John and Peter on the team. This can be accomplished in 1 way
- Select the remaining 3 team-members from the remaining 7 players. Since the order in which we select the 3 players does not matter, we can use combinations. We can select 3 players from 7 players in 7C3 ways (35 ways)
So, the total # of teams that include both John and Peter = (1)(35) = 35
b) total # of 5-person teams
Select 5 team-members from the 9 players. This can be accomplished in 9C5 ways
So, the total # of 5-person teams = 9C5 = 126
Therefore, the probability that the coach chooses a team that includes both John and Pete = 35/126 = 5/18
Answer: D
Aside: If anyone is interested, here's a video on calculating combinations (like 7C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
GMAT/MBA Expert
-
Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7249
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
The number of ways to select the team with no restrictions is 9C5 = 9!/[5!(9-5)!] = 9!/(5!4!) = (9 x 8 x 7 x 6 x 5)/5! =(9 x 8 x 7 x 6 x 5)/(5 x 4 x 3 x 2 x 1) = 126 waysVincen wrote:A coach will select the members of a 5-players team from 9 players, including A and B. If the 5 players are chosen at random, What is the probability that the coach chooses a team that includes both A and B?
a. 1/9
b. 1/6
c. 2/9
d. 5/18
e. 1/3
Since we must have A and B on the team, then there are only 7 players remaining to be chosen for the team, and there will be only 3 spots on the team for them to fill. The number of ways to select the team, which already has A and B selected, is 7C3 = 7!/3![7-3)!] = 7!/(3!4!) = (7 x 6 x 5)/3! = (7 x 6 x 5)/(3 x 2 x 1) = 35 ways
Thus, the the probability that the coach chooses a team that includes both A and B is 35/126 = 5/18.
Answer: D
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
