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A state auto-inspection station has two teams. Team 1 is lenient and

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A state auto-inspection station has two teams. Team 1 is lenient and [#permalink]

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08 Feb 2011, 13:55
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Hey guys, this is a probability problem that you probably wont find on the GMAT. It is out of my book for my probability class that I'm in right now. Either way, it's a quality problem and pertains to probability/combinatorics and is useful practice.

A state auto-inspection station has two teams. Team 1 is lenient and passes all automobiles of a recent vintage; team 2 rejects all autos on first inspection because their "headlights are not properly adjusted." Four unsuspecting drivers take their autos to the station for inspection on four different days and randomly select one of the two teams.

A) If all four cars are new and in excellent condition, what is the probability that three of the four will be rejected?

B) What is the probability that all four will pass?

[Reveal] Spoiler:
A) 4(1/2)^4 = .25
B) (1/2)^4 = .0625

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Re: A state auto-inspection station has two teams. Team 1 is lenient and [#permalink]

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08 Feb 2011, 14:10
PennState08 wrote:
Hey guys, this is a probability problem that you probably wont find on the GMAT. It is out of my book for my probability class that I'm in right now. Either way, it's a quality problem and pertains to probability/combinatorics and is useful practice.

Q: A state auto-inspection station has two teams. Team 1 is lenient and passes all automobiles of a recent vintage; team 2 rejects all autos on first inspection because their "headlights are not properly adjusted." Four unsuspecting drivers take their autos to the station for inspection on four different days and randomly select one of the two teams.
A) If all four cars are new and in excellent condition, what is the probability that three of the four will be rejected?
B) What is the probability that all four will pass?

[Reveal] Spoiler:
A) 4(1/2)^4 = .25
B) (1/2)^4 = .0625

A. We need the probability of the event that three of the four cars will be rejected (R) and the one will pass (P): $$P(RRRP)=C^3_4*(\frac{1}{2})^3*\frac{1}{2}=\frac{1}{4}$$ - $$C^3_4$$ choosing which three of the four cars will be rejected, $$(\frac{1}{2})^3$$ - the probability of these three being rejected and $$\frac{1}{2}$$ - the probility of the fourth one besing passed;

B. The same way the probability that all four will pass will be: $$P(SSSS)=(\frac{1}{2})^4$$.
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Re: A state auto-inspection station has two teams. Team 1 is lenient and [#permalink]

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08 Feb 2011, 14:11
Probability of rejection and passing is equally likely and is 1/2.

P(Rejection) = 1/2
P(Passing) = 1/2

Let's denote Rejection as R
and Passing as P

1. Three rejections out of the four days can happen in following four ways:

PRRR - Car on the second, third and fourth day are rejected
RPRR
RRPR
RRRP

So;
P(3 Rejections) = 4(1/2*1/2*1/2*1/2) = 4/16 = 1/4 = 25%

2. All pass
Can happen in only one way
PPPP

P(4 Passing) = 1/2*1/2*1/2*1/2 = 1/16 = 6.25%
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Re: A state auto-inspection station has two teams. Team 1 is lenient and [#permalink]

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27 Oct 2016, 22:25
Hi,

The second part was easy. The first part I did as following:

We have 4 cars who have to choose 2 possible inspections, so total: 4C2
To fail, 3 of the following cars had to choose the second inspection. Hence: 3C1
To pass, 1 of them had to choose the 1st inspection. Hence: 1C1

So, probability in the 1st case (a) is given by: ((3C1*1C1)/4C2) = 1/4

Is this the best approach?
Re: A state auto-inspection station has two teams. Team 1 is lenient and   [#permalink] 27 Oct 2016, 22:25
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