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# The probability for event A to occur is 0.6, and for B to

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Intern
Joined: 23 Dec 2005
Posts: 18

Kudos [?]: 4 [0], given: 0

The probability for event A to occur is 0.6, and for B to [#permalink]

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24 Apr 2006, 03:29
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The probability for event A to occur is 0.6, and for B to occur is 0.5. What is the probability that neither event A nor event B will occur?

Kudos [?]: 4 [0], given: 0

Senior Manager
Joined: 08 Jun 2004
Posts: 494

Kudos [?]: 93 [0], given: 0

Location: Europe

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24 Apr 2006, 03:45
9/10

(1-6/10)+(1-5/10)=9/10.

Kudos [?]: 93 [0], given: 0

Senior Manager
Joined: 09 Mar 2006
Posts: 444

Kudos [?]: 8 [0], given: 0

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24 Apr 2006, 03:57
(1-6/10)*(1-5/10) = 20/100 = 1/5 = 0.2

Kudos [?]: 8 [0], given: 0

Intern
Joined: 28 Nov 2005
Posts: 33

Kudos [?]: [0], given: 0

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24 Apr 2006, 05:26
If we assume that A and B are independant, then we have:
P(N)=1-[P(A)+P(B)-P(AnB)]
=1-[6/10+5/10 - 6/10*5/10]
=1-[11/10- 3/10]
=1-[8/10]
=2/10
=0.2

Is that correct?
(N=Neither A, nor B)

Kudos [?]: [0], given: 0

Intern
Joined: 23 Dec 2005
Posts: 18

Kudos [?]: 4 [0], given: 0

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24 Apr 2006, 08:02
OA is 0.2

Thanks guys!

Kudos [?]: 4 [0], given: 0

Senior Manager
Joined: 08 Jun 2004
Posts: 494

Kudos [?]: 93 [0], given: 0

Location: Europe

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24 Apr 2006, 11:26
Ethan wrote:
OA is 0.2

Thanks guys!

OOOPPPSSS

It should be multiplied not added.

Kudos [?]: 93 [0], given: 0

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5034

Kudos [?]: 438 [0], given: 0

Location: Singapore

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25 Apr 2006, 00:47
Alternatively,

P(!A) = 0.4
P(!B) = 0.5

P(!A AND !B) = 0.4 * 0.5 = 0.2

Kudos [?]: 438 [0], given: 0

25 Apr 2006, 00:47
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# The probability for event A to occur is 0.6, and for B to

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