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Probability

by satish_iitg » Sat Aug 03, 2013 4:02 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?


1/9

1/6

2/9

5/18

1/3
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by Brent@GMATPrepNow » Sat Aug 03, 2013 5:45 am
satish_iitg 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?
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
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by satish_iitg » Sat Aug 03, 2013 9:18 am
Thanks Brent.

I tried to do this using the 1-P(either John or peter) method.

P(john or peter) = [(1 * 7C4)/9C5] + [(1 * 7C4)/9C5]

alternatively, 1 - P(neither john or peter) = 1 - [(7C5)/9C5]

Can you please tell me the mistake(s) in these approaches ?
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by Brent@GMATPrepNow » Sat Aug 03, 2013 11:10 am
satish_iitg wrote:Thanks Brent.

I tried to do this using the 1-P(either John or peter) method.

P(john or peter) = [(1 * 7C4)/9C5] + [(1 * 7C4)/9C5]

alternatively, 1 - P(neither john or peter) = 1 - [(7C5)/9C5]

Can you please tell me the mistake(s) in these approaches ?
I'm not sure why you are using the complement for the part in green.

If you want to use the complement here, you must consider 3 cases:
- neither John nor Peter are on the team
- John on the team, but Peter not on the team
- Peter on the team, but John not on the team

So, P(Peter and John on team) = 1 - P(Peter and John NOT on team)
= 1 - P(neither John nor Peter are on the team OR John on the team, but Peter not on the team OR Peter on the team, but John not on the team)

Use this setup and see how you do.

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by satish_iitg » Sun Aug 04, 2013 10:55 pm
I still dont get the answer to this.

1- [P(john or peter) + P(neither john or peter)]

= 1 - [{ (1 * 7C4)/9C5 } + {(1 * 7C4)/9C5 } + {(7C5)/9C5} ]
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by Brent@GMATPrepNow » Sun Aug 04, 2013 11:21 pm
satish_iitg wrote:I still dont get the answer to this.

1- [P(john or peter) + P(neither john or peter)]

= 1 - [{ (1 * 7C4)/9C5 } + {(1 * 7C4)/9C5 } + {(7C5)/9C5} ]
You're almost there. Keep going.

P(Peter and John on team) = 1 - P(Peter and John NOT on team)
= 1- [P(john or peter) + P(neither john or peter)]
= 1 - [{ (1 * 7C4)/9C5 } + {(1 * 7C4)/9C5 } + {(7C5)/9C5} ]
= 1 - [35/126 + 35/126 + 21/126]
= 1 - [91/126]
= 35/126
= 5/18
= D

Cheers,
Brent
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