Probability

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Tame the CAT: Senior | Next Rank: 100 Posts; Posts: 46; Joined: Sun Mar 04, 2007 6:27 am

by Tame the CAT » Fri Apr 27, 2007 5:16 am

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Source: Beat The GMAT — Problem Solving | Fri Apr 27, 2007 5:16 am

Neo2000: Legendary Member; Posts: 519; Joined: Sat Jan 27, 2007 7:56 am; Location: India; Thanked: 31 times

by Neo2000 » Fri Apr 27, 2007 7:18 am

Number of ways of picking 5 people from 9 = 9C5
Number of ways of picking 5 such that John and Peter are already there = 7C3

Probability of picking this team = 7C3/9C5 = 5/18

Hence Option D

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Cybermusings: Legendary Member; Posts: 559; Joined: Tue Mar 27, 2007 1:29 amFollowed by: 2 members; Thanked: 5 times

by Cybermusings » Sat Apr 28, 2007 12:54 am

John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

A 1/9
B 1/6
C 2/9
D 5/18
E 1/3

Number of ways of making a free selection of a team of 5 players from 9 players = 9C5 = 9*8*7*6/4*3*2*1 = 9*7*2 = 63*2 = 126
Number of ways of selecting 3 players from amongst 7 players when John and Peter have already been selected = 2C2 * 7C3 = 1 * [7*6*5 / 3*2*1] = 35
Thus probability = 35/126 = 5/18
Hence D

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8154; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Mon May 20, 2019 6:17 pm

Tame the CAT wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

A 1/9
B 1/6
C 2/9
D 5/18
E 1/3

The number of ways of choosing 5 players from 9 is 9C5 = 9!/[5!(9 - 5)!] = 9!/(5!4!) = (9 x 8 x 7 x 6 )/( 4 x 3 x 2) = 3 x 7 x 6 = 126.

If John and Peter are already chosen for the team, then only 3 additional players can be chosen for the five-player team. The number of ways of choosing the 3 additional players from the remaining 7 players is 7C3 = 7!/[3!(7 - 3)!] = 7!/(3!4!) = (7 x 6 x 5)/(3 x 2) = 7 x 5 = 35.

Thus, the probability that John and Peter will be chosen for the team is 35/126 = 5/18.

Answer: 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 » Tue May 21, 2019 6:12 am

Tame the CAT wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

A 1/9
B 1/6
C 2/9
D 5/18
E 1/3

We want:
a) # of teams that include both John and Peter
b) 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 (this can be accomplished in 7C3 ways)
So, the total # of teams that include both John and Peter = (1)(7C3) = 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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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