Probability

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Oct 20, 2009 9:04 pm

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 ?

1/8

1/6

2/9

5/18

1/3

Please try your best to make sense of the answer. I am not a 700 student and the probability of understanding probability, sometimes, can be hard

OA [spoiler]5/18[/spoiler]

Last edited by heshamelaziry on Wed Oct 21, 2009 12:02 am, edited 1 time in total.

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Source: Beat The GMAT — Problem Solving | Tue Oct 20, 2009 9:04 pm

sanjana: Master | Next Rank: 500 Posts; Posts: 182; Joined: Sun Aug 02, 2009 7:19 pm; Thanked: 18 times; GMAT Score:680

by sanjana » Tue Oct 20, 2009 10:49 pm

IMO : D

Total Players to choose from : 9

No of players to choose : 5

Restriction : Team with John and Peter

- - - - -

The order in which the team members are selected doesnt matter.

No of ways of selecting John : 1
No of ways of selecting Peter : 1

No of ways of selecting the remaining 3 : 7c3

Hence P(team with Jon and Peter) = 1*1*7c3/9c5
=5/18

Whats the OA?

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Wed Oct 21, 2009 12:15 am

sanjana wrote:IMO : D

Total Players to choose from : 9

No of players to choose : 5

Restriction : Team with John and Peter

- - - - -

The order in which the team members are selected doesnt matter.

No of ways of selecting John : 1
No of ways of selecting Peter : 1

No of ways of selecting the remaining 3 : 7c3

Hence P(team with Jon and Peter) = 1*1*7c3/9c5
=5/18

Whats the OA?

THANK YOU SANJANA. YOU ARE A NATIONAL TREASURE

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kkpatel1: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Tue Oct 20, 2009 5:59 pm; Location: North Carolina

confused

by kkpatel1 » Wed Oct 21, 2009 1:15 pm

I am a little lost. I do not understand the shortcuts made here.

Such as, "7c3". Where does the 7 and 3 come from and also what does "c" stand for?

and

= 1*1*7c3/9c5, if someone could break down where the 9 and 5 come from.

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briantime: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Sun Oct 18, 2009 3:39 amFollowed by: 1 member; Location: Baden-Baden, Germany; Thanked: 2 times; GMAT Score:690

Re: confused

by briantime » Wed Oct 21, 2009 1:44 pm

kkpatel1 wrote:I am a little lost. I do not understand the shortcuts made here.

Such as, "7c3". Where does the 7 and 3 come from and also what does "c" stand for?

and

= 1*1*7c3/9c5, if someone could break down where the 9 and 5 come from.

9c5: This means that 5 out of 9 people are to be chosen, while the arrangement of the chosen people does not matter.

7c3: We expect that John and Peter are on the team, that means, 7 people are left which can be chosen. Since there are in sum 5 people to pick and John and Peter are already on the team, there are 3 people left to pick, which results in 3 out of 7 = 7c3.

C stands for combination. That means, the order of the chosen group does not matter. This is the fact if the question askes for a team, a group, or anything that doesn't depend on the actual arrangement of the people.

On the other side there are permuations. Permutations invole groups where the arrangement is important. For example, if the number of possible seat arrangement in a cinema is asked.

The rule for combinatorics is:
XcY = X! / (X-Y)!Y!

The rule for permutations is:
XpY = X! / (X-Y)!

If you are not familiar with combinatorics and permutations, you should read this article: https://www.beatthegmat.com/a/2009/10/12 ... asy-method

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Leon1984: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Sun Oct 18, 2009 5:23 pm; Location: Israel; Thanked: 3 times; GMAT Score:640

Hey

by Leon1984 » Wed Oct 21, 2009 5:10 pm

I'll try to put some more light on the great explanations they guys gave you.

Probability is the chance that something happens, and can also be looked as the number of desired outcomes divided by the number of possible outcomes.

The number of possible outcomes is 9C5, which means:
9! 9!
-------- = ----
5!(9!-5!) 5!4!

From here just reduce. This is the denominator. The nominator is broken into 3 stages:
How many ways are there to choose 1 of the guys out of the group? 1 way only, and so is for the other guy. Now, how many ways are there to choose the remaining 3 players? We have only 7 players remaining (we already "took", so we choose 3 out of 7, or 7!/(3!4!)

Do the calculation (just reducing) and you have the nominator. Keep reducing when necessary and get the answer.

Hope it helps, and that I didn't mess up in the process.

Leon

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