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# Each of the 25 balls in a certain box is either red, blue or

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Intern
Joined: 22 Jan 2010
Posts: 13
Followers: 0

Kudos [?]: 2 [0], given: 5

Each of the 25 balls in a certain box is either red, blue or [#permalink]  06 Mar 2010, 07:42
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Difficulty:

(N/A)

Question Stats:

100% (03:12) correct 0% (00:00) wrong based on 1 sessions
1. Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

(1) The probability that the ball will both be white and an even number painted on it is 0.
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2.
Manager
Joined: 03 Feb 2010
Posts: 68
Followers: 1

Kudos [?]: 32 [0], given: 4

Re: Probability Data Sufficiency Question [#permalink]  06 Mar 2010, 10:51
p(White or Even) = ?

we know, p(White or Even) = P(white) + p(even) - p(white and even).

1. p(white and even) = 0. we donno p(w) or p(e) ...so insuff.
2. p(white) - p(even) = 0.2 . ...insuff.

combining 1&2, we donno p(white), p(even) or p(white) + p(even). so insuff.

ans E.
Director
Joined: 21 Dec 2009
Posts: 592
Concentration: Entrepreneurship, Finance
Followers: 16

Kudos [?]: 354 [0], given: 20

Re: Probability Data Sufficiency Question [#permalink]  06 Mar 2010, 15:39
question stem requires that we calculate:
P(White or Even) = P(White) + P(Even) - P(White and Even)
W=white; E=even

(1) P(W and E) = 0
=> P(W or E) = P(W) + P(E) + 0
P(W); P(E) are both unknowns.........insuff

(2) P(W) - P(E) = 0.2
i.e P(W or E) = P(W) + P(E) - P(W and E)
we don't know P(W), P(E) or P(W and E) .........insuff

combining, we get:
P(W or E) = P(W) + P(E) + 0
no relationship is given to establish either P(W) or P(E) independently
so, OA should be E
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Re: Probability Data Sufficiency Question   [#permalink] 06 Mar 2010, 15:39
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