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joemama142000
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another probabiliy ps 18 Jan 2006, 13:01
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The valet parking attendent parked 4 cars that belonged to 4 corresponding people. Each car had 1 key. If the attendent gave the 4 keys to the 4 people at random, what is the probablility that only 1 person received the correct key?



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another probabiliy ps 18 Jan 2006, 14:02
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Probability of first guy gets correct key = 1/4
Probability of second guy gets incorrect key = 2/3
Probability of second guy gets incorrect key = 1/2
Probability of second guy gets incorrect key = 1

Probability off all this 4 happning = 1/4 * 2/3 * 1/2 * 1 = 1/12

Probability of Same thing happning to all 4 guys one at a time = 4/12 = 1/3

So answer should be 1/3...

Though feels I am missing something! as 1/3 is high probability..
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chiragr
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another probabiliy ps 18 Jan 2006, 14:27
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Nop it says 'Only ONE' so not more than 1! o/w 4/4! is correct..!
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another probabiliy ps 18 Jan 2006, 16:54
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Total number of ways to distribute keys = 4! = 24

Lets give correct key to first person and incorrect key to other three.

Person P1= Correct key i.e key #1
Person P2 = either K3 or K4
Person P3 = either K4 or K2
Person P4 = either K2 or K3

P1-P2-P3-P4
K1-K3-K4-K2
K1-K4-K2-K3

So total cases = 2

Same way we can give correct key to P2, P3 and P4
So toal cases = 4*2 = 8

Prob = 8/24 = 1/3
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Aiyer
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another probabiliy ps 20 Jan 2006, 15:43
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joemama142000
The valet parking attendent parked 4 cars that belonged to 4 corresponding people. Each car had 1 key. If the attendent gave the 4 keys to the 4 people at random, what is the probablility that only 1 person received the correct key?
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I believe it is 11/24.

The question says exactly one person receiving correct key.

So we cannot consider the probability of the 4th person getting the correct key. Because if that were the case then someone else also got the correct key.

so,
probability of 1st person getting the correct key = (1/4)*(2/3)*(1/2)*(1)
probability of 2nd person getting the correct key = (3/4)*(1/3)*(1/2)*(1)
probability of 3rd person getting the correct key = (3/4)*(2/3)*(1/2)*(1)

Adding 1/12+1/8+1/4 =11/24.

what is the OA??
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another probabiliy ps 21 Jan 2006, 22:20
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RWWW + WRWW+WWRW+WWWR

P(RIght Key)= 1/4
P(Wrong Key)=1-1/4=3/4

1/4*(3/4)^3 * 4= 27/64
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giddi77
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another probabiliy ps 22 Jan 2006, 02:26
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Also got 1/3.

Similar approach as ps_dahiya's.
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joemama142000
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another probabiliy ps 22 Jan 2006, 03:15
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thanks for your effort. however, I dont have an Oa for this. sorry

here are my thoughts

toatl number of ways 4!

toatl number of ways 1 key for correct person =8

8/24
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another probabiliy ps 23 Jan 2006, 18:08
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I agree with chiragr. I got 1/3 with same reasoning.
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thegmatguy
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another probabiliy ps 23 Jan 2006, 19:05
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joemama142000
The valet parking attendent parked 4 cars that belonged to 4 corresponding people. Each car had 1 key. If the attendent gave the 4 keys to the 4 people at random, what is the probablility that only 1 person received the correct key?
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My vote to 1/3.

Aiyer is partly correct only if it was mentioned that the attendent returned the keys to those persons in a specific order. I said partly because Aiyer didnot consider the 4th person



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