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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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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%.

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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"Any school that meets you and still lets you in is not a good enough school to go to" - my mom upon hearing i got in Thanks mom.

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" ?

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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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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There are times when I do not mind kudos...I do enjoy giving some for help

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 \(\frac{2}{3}\)?

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 \(\frac{2}{3}\)?

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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Good things come to those who wait… greater things come to those who get off their ass and do anything to make it happen...