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# Probabilty Problem

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Joined: 01 Nov 2010
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Location: NJ
Schools: Rutgers MQF
WE 1: 2 years IT
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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.

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

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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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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: Probabilty Problem   [#permalink] 08 Feb 2011, 14:11
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