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# GMAT Diagnostic Test Question 36

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GMAT Diagnostic Test Question 36 [#permalink]  06 Jun 2009, 23:54
Expert's post
GMAT Diagnostic Test Question 36
Field: probability
Difficulty: 650

A jar contains B blue balls, 6B + 10 yellow balls and 2B+5 green balls. If there are only blue, yellow and green balls in the jar, what is the probability of taking out a blue or green ball?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$
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Last edited by Bunuel on 06 Oct 2013, 23:37, edited 2 times in total.
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Re: GMAT Diagnostic Test Question 37 [#permalink]  18 Jul 2009, 04:11
1
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P(Green or Yellow) = 1 - P(Blue)

ie P(Green or Yellow) = 1 - ( B/(9B+15)

ie P(Green or Yellow) = (8B + 15)/(9B + 15)...

How can it be solved??? Please correct if I am wrong..
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Re: GMAT Diagnostic Test Question 37 [#permalink]  18 Jul 2009, 04:33
1
KUDOS
Here again,

in the explanation you wrote
Quote:
Probability (Green or Yellow) = (B + 2B + 5) / (9B + 15) = 1/3.

but Green or Yellow would be "(6B + 10 + 2B + 5)" and that would equal (8B + 15) / (9B + 15),
which is what Kyabe got.

Hope to be right this time!
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Re: GMAT Diagnostic Test Question 37 [#permalink]  18 Jul 2009, 06:21
1
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The question stem is incorrect as per your calculation.
I think the question should ask for the probability of taking out a blue or a green ball.

then it would be 1/3.
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Re: GMAT Diagnostic Test Question 37 [#permalink]  19 Sep 2009, 09:40
1
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The question doesn't make any sense. The question states "if there are no other balls?" This means if if all the yellow balls have already been taken out, and what's left in the jar is only blue balls and green balls. This can only mean that the probably of selecting a blue or a green is 100%.
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Re: GMAT Diagnostic Test Question 37 [#permalink]  17 Dec 2009, 22:19
1
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I believe that "no other balls" means that there are only blue, yellow and green balls in the jar. The wording is a bit odd, perhaps an edit in the future?

Way to solve- the total number of balls is B + 6B+10 + 2B+5
To get Blue or green, you would not want Yellow. So find the prob of getting yellow and subratct from 1.

6B+10 / 9B+15 = 2/3 for all positive values of B which can be assumed in this case.

2/3 chance of drawing a Yellow, meaning 1/3 chance of drawing a green or blue
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Re: GMAT Diagnostic Test Question 37 [#permalink]  23 Dec 2009, 03:03
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Hey guys!

I think the key is "blue or green ball". Or means you have to find the probability for each event to occur and then add the probabilities.

The total no of possible outcomes is B+6B+10+2B+5=9B+15

So P(blue occurs)=B/(9b+15)
P(green occurs)=2B+15/9B+15

Add the two fractions and you get 3b+5/3(3B+5)=1/3
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Re: GMAT Diagnostic Test Question 37 [#permalink]  17 Sep 2013, 01:59
1
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Expert's post
obs23 wrote:
A jar contains B blue balls, 6B + 10 yellow balls and 2B+5 green balls. If there are only blue, yellow and green balls in the jar, what is the probability of taking out a blue or green ball?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$

Could someone help - do not we have probability of taking out a blue or green ball = $$P(B) + P(G)$$ or $$P(G) + P(B)$$? And in this case the probability would be $$\frac{2}{3}$$?

We pick one ball.

P=(favorable)/(total)=(B+2B+5)/(B+6B+10+2B+5)=1/3.

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Re: GMAT Diagnostic Test Question 37 [#permalink]  13 Jun 2009, 20:46
Explanation
 Rating:

Total = B + 6B + 10 + 2B + 5 = 9B + 15
Probability of getting a blue or green = $$\frac{(B + 2B + 5)}{(9B + 15)} = \frac{1}{3}$$.

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Re: GMAT Diagnostic Test Question 37 [#permalink]  18 Jul 2009, 13:35
Expert's post
Thank you guys - I will let GT tackle this one.

Please keep the questions and suggestions coming!
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Re: GMAT Diagnostic Test Question 37 [#permalink]  22 Dec 2009, 10:24
It helps to think of it as fractions and plug in

Just plug in a value for B(the number of blue balls)- I chose 2 so that means

there are 2 blue balls, 22 yellow balls (6(2)+10), and 9 green balls (2(2)+5) and 33 total

The answer to the question will be the probability of not selecting a yellow ball

1-22/33= 1/3 since 22/33=2/3 so the answer is D 1/3

I hope this helps!
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Re: GMAT Diagnostic Test Question 37 [#permalink]  23 Dec 2009, 04:51
How would you guys change the wording to avoid the confusion? I suggest this:

Quote:
A jar contains B blue balls, 6B + 10 yellow balls and 2B+5 green balls. If there are only blue, yellow and green balls in the jar, what is the probability of taking out a blue or green ball?

marcos4 wrote:
I'm like you guys, I'm a bit confused with the question.

What should be understood by the statement "if there are no other balls" ?

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Re: GMAT Diagnostic Test Question 37 [#permalink]  08 Feb 2010, 07:57
The question stem has been updated both in the PDF and in this thread. Hope it's better this way.
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Re: GMAT Diagnostic Test Question 37 [#permalink]  22 May 2010, 07:52
Easy to do .

Because B is constant, it does not change, so the number of balls does not change as well.
lets B=1

we have the following balls:
blue -1
yellow - 16
green - 7
total -24

BLUE OR GREEN , MEANS NOT YELLOW , so 1-P(yellow)= 1-16/24=1-2/3=1/3
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Re: GMAT Diagnostic Test Question 37 [#permalink]  01 Nov 2011, 02:53
Total number of balls = 9B+15

Probability of taking a blue ball = B/9B+15

Probability of taking a green ball = 2B+5/9B+15

so probability of taking either a blue ball or a green ball is = (B+(2B+5))/9B+15 = 1/3
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Re: GMAT Diagnostic Test Question 37 [#permalink]  01 Nov 2011, 09:56
Probability of taking Blue or Green ball = Number of blue and Green balls / total balls in the jar
= B + 2B+5/B + 2B+5 + 6B+10 = 3B+5/9B+15 = 1/3
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Re: GMAT Diagnostic Test Question 37 [#permalink]  17 Sep 2013, 01:48
Could someone help - do not we have probability of taking out a blue or green ball = $$P(B) + P(G)$$ or $$P(G) + P(B)$$? And in this case the probability would be $$\frac{2}{3}$$?
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Re: GMAT Diagnostic Test Question 37 [#permalink]  17 Sep 2013, 02:12
Quote:
We pick one ball.
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Re: GMAT Diagnostic Test Question 37 [#permalink]  16 Dec 2013, 01:25
Bunuel wrote:
obs23 wrote:
A jar contains B blue balls, 6B + 10 yellow balls and 2B+5 green balls. If there are only blue, yellow and green balls in the jar, what is the probability of taking out a blue or green ball?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$

Could someone help - do not we have probability of taking out a blue or green ball = $$P(B) + P(G)$$ or $$P(G) + P(B)$$? And in this case the probability would be $$\frac{2}{3}$$?

We pick one ball.

P=(favorable)/(total)=(B+2B+5)/(B+6B+10+2B+5)=1/3.

this is the exact way to solve it like in schoolbook. keep clear&simple.
the only thing makes me happy-I scored only 500 on the test and I can solved it. Probably it is not that hard if you think about it in simple way.
But still I would add info regarding replacement or not.

Thanks for nice practicing to see how differently people think.
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Re: GMAT Diagnostic Test Question 37   [#permalink] 16 Dec 2013, 01:25
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# GMAT Diagnostic Test Question 36

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