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by pushkin1982 » Sat Aug 21, 2010 10:15 am
1) Each of the 25 Balls is either red, blue or white and has a number 1 to 10 printed on it. If a ball is selected at random, what is the probability that the ball is either white or has an even number printed on it.

A) Probability that the ball would be both white and with an even number is 0.
B) Probability that ball is white minus probability that the ball has an even no is 0.2.
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Source: — Data Sufficiency |
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by Maciek » Sat Aug 21, 2010 11:51 am
Hi!
[spoiler]
IMO A[/spoiler]

this task is interesting:

A) P(white & even) = 0

P(white or even) = P(white) + P(even) - P(white & even)
there are only 5 white balls with odd number
P(white) = 5/25 = 1/5
we have 5 red balls and 5 blue balls with even number
P(even) = (5 + 5)/25 = 10/25 = 2/5
P(white or even) = 1/5 + 2/5 = 3/5
It is SUFFICIENT

B) P(white) - P(even) = 0,2

if there are more white balls than balls with equal number then some balls with equal number have the same number and the same color

we need more information about order

it is INSUFFICIENT

so I choose A
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by Makushr1 » Sat Aug 21, 2010 12:15 pm
Maciek wrote:Hi!
[spoiler]
IMO A[/spoiler]

this task is interesting:

A) P(white & even) = 0

P(white or even) = P(white) + P(even) - P(white & even)
there are only 5 white balls with odd number
P(white) = 5/25 = 1/5
we have 5 red balls and 5 blue balls with even number
P(even) = (5 + 5)/25 = 10/25 = 2/5
P(white or even) = 1/5 + 2/5 = 3/5
It is SUFFICIENT

B) P(white) - P(even) = 0,2

if there are more white balls than balls with equal number then some balls with equal number have the same number and the same color

we need more information about order

it is INSUFFICIENT

so I choose A
How did you know there are 5 white balls? It never states exactly how many of each ball there are in the question.
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by Maciek » Sat Aug 21, 2010 2:17 pm
you are right Makushr1 that it is not clearly defined.

we have 25 balls
each can be either red, blue or white
each of the balls has a number 1 to 10

red 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
blue 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10
white 1, 2, 3, 4, 5, 6, 7, 8 ,9, 10

if probability(white&even) = 0 we can reduce number of above described balls by 5( 2, 4, 6, 8, 10)

thus we have 25 balls

red 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
blue 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10
white 1, 3, 5, 7, 9

what would you choose and why?
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by pushkin1982 » Sat Aug 21, 2010 9:25 pm
Thanks Everyone! I have got it now.... The correct answer is E
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by adi_800 » Sat Aug 21, 2010 11:36 pm
@Pushkin...
can u pls tell me how much u got in quant when u gave d test??
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by pushkin1982 » Sun Aug 22, 2010 12:38 am
@ Adi I got a 49 in quant and a 38 in English on my first attempt at GMatPrep 1
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by sdotcruz » Tue Aug 31, 2010 12:08 pm
pushkin1982, why is the answer E?
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by pushkin1982 » Tue Aug 31, 2010 12:11 pm
Read The reply by Maciek
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by Gurpinder » Tue Aug 31, 2010 1:54 pm
IMO (E)

25 balls = red, white, blue
each one has a number from 1 to 10

(1) there aren't any that are both white and even. This doesn't tell us how many are white or how many are even. insufficient.

(2) p of white - p of even = 0.2. this doesn't again tell you anything about the # of each. for ex. p of w = .5 and p of even = .3 which subtracted = .2 OR it could be any other combination that gives .2
so... insufficient ....Eliminate B.

Together:
Peven+white = 0. but there are still many posibilities for the other options just like as in statement (2).

Therefore (E).
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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