Probability

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prernamalhotra: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Wed Apr 02, 2014 2:01 amFollowed by: 1 member; Thanked: 2 times

by prernamalhotra » Mon Jun 30, 2014 4:05 am

A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?

1)1/9

2)1/6

3)2/9

4)5/18

5)1/3

Thank you,
Prerna

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Source: Beat The GMAT — Problem Solving | Mon Jun 30, 2014 4:05 am

GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Mon Jun 30, 2014 4:12 am

A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?

1)1/9

2)1/6

3)2/9

4)5/18

5)1/3

Probability = Favorable outcomes / Total outcomes

Total Outcomes = No. of ways of choosing 5 players out of 9 players randomly = 9C5 = 126

Favorable outcomes = When 7C3 (because John and Peter are already selected) = 35

Probability = 35/126 = [spoiler]5/18 Option D[/spoiler]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Jun 30, 2014 4:39 am

prernamalhotra wrote:A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?

1)1/9

2)1/6

3)2/9

4)5/18

5)1/3

From the 9 players, 5 are to be selected for the team.
P(john is selected) = 5/9.
From the 8 remaining players, 4 are to be selected for the team.
P(Peter is selected) = 4/8.
Multiplying the probabilities, we get:
5/9 * 4/8 = 5/18.

The correct answer is D.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Jun 30, 2014 4:39 am

A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?
A) 1/9
B) 1/6
C) 2/9
D) 5/18
E) 1/3

The combination theory approach:

P(John and Peter both on the team) = (# of teams that include both John and Peter) / (total # of 5-person teams possible)

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 = D

Aside: If anyone is interested, we have a free 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

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