Two identical urns—black and white—each contain 5 blue, 5 red and 10

Firstly the Probability can never be GREATER than 1, probability of 1 itself means that the event is sure to happen..
So choices B and C are flawed....
Of course the choices should be in some order...

Back to the question...
So we are looking at price equal to or more than 25 cents...
Only way it will be less than 25 will be when we get a nickel in each draw, so let us choose easier path with lesser calculations....
So first draw from black a will be 10/20 and the second too will be 10/20..
First draw in white urn will be 10/20 but the second will be 9/19 ..
Probability of not being able to buy... (10/20)(10/20)(10/20)(9/19)=(1/2)(1/2)(1/2)(9/19)=9/(2*2*2*19)=9/152..
So probability of buying is 1-(9/152)=143/152

A
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Are we expected to know values of quarters, dimes and nickles?

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deushyant
Absolutely not . Nickel and Dime are not globally recognized US currencies .By glovally recognised i mean the currencies are not used in global transaction units

The only values you should know are "cents" and "dollars" ... This qustion seems as a self made. Choices B and C are flawed as probab9ility is never more than 1 . Please practice from GMAT trusted sources only. Use the tags for refining your question bank

A lot of words to get through, but not too hard if you take a little time to think about it.

There is only way one that she will not be able to afford a 25 cent chocolate bar: she must draw Green balls FOUR Times: two green from each urn. G-G-G-G ———> 5 + 5 + 5 + 5 = 20 cents

Any other combination of balls will allow her to buy the candy bar. The next lowest amount she could possibly get is: G-G-G-R ——-> 5 + 5 + 5 + 10 = 25 cents, which is enough.

To find the probability that she will be able to afford the candy bar, we can take:

1 - (Prob. of Unfavorable Outcome) = Prob. of Favorable Outcome


Probability of Pulling 2 Green balls from the Black urn, when the 1st ball is replaced is:

(10/20) * (10/20) = (1/2) * (1/2) = 1/4

AND

Probability of pulling 2 Green balls from the White urn when the balls are NOT replaced after taking one:

(10/20) * (9/19) = (1/2) * (9/19) = 9/38


Probability of the Unfavorable Outcome of pulling 2 green from the Black urn and pulling 2 green from the White urn therefore is = (1/4) * (9/38) = 9/152


1 - (9/152) = 143/152

(A)

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We should notice that the only way Jenny will not be able to earn 25 cents after four draws is if she draws a green ball in each of the four draws. If she draws even a single blue ball, she receives a quarter, which is enough to buy the candy bar regardless of the remaining three draws. If she does not draw a blue ball but draws a red ball, then the minimum amount she will earn is 10 + 5 + 5 + 5 = 25 cents. Thus, if we can calculate the probability that she will draw a green ball all four draws, we can simply subtract that number from 1 to calculate the probability that she earns at least 25 cents.

Since the draws from the black urn are made with replacement, the probability that she will draw two green balls from the black urn is 10/20 * 10/20 = 1/4. Since the draws from the white urn are made without replacement, the probability that she will draw two green balls from the white urn is 10/20 * 9/19 = 9/38. Thus, the probability that she draws green balls in all her draws is 1/4 * 9/38 = 9/152. It follows that the probability that she draws at least one non-green ball is 1 - 9/152 = 143/152.

Answer: A
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Why are not considering two ways because in question it is not given he/she will start selection from which colour urn whether black or white. As there colour is different i think we should multiply by 2

To calculate the probability that Jenny will be able to buy a 25-cent candy bar with the proceeds from drawing four balls (2 from each urn), we need to consider all possible combinations of balls that add up to 25 cents and divide that by the total number of combinations of drawing 4 balls (2 from each urn).

For Jenny to buy the candy bar, she needs 25 cents, which can be obtained by either selecting 2 blue balls (25 cents), or 1 blue and 1 red ball (15 cents), or 2 red balls (20 cents), or 1 red and 5 green balls (15 cents), or 1 blue and 10 green balls (15 cents), or 2 green balls (10 cents), or 5 green balls (25 cents).

From the black urn, there are 5 choose 2 = 10 combinations of blue balls, 5 choose 2 = 10 combinations of red balls, and 10 choose 2 = 45 combinations of green balls.

From the white urn, there are 5 choose 2 = 10 combinations of blue balls, 5 choose 2 = 10 combinations of red balls, and 10 choose 2 = 45 combinations of green balls.

Therefore, the number of combinations that result in 25 cents is 10 x 10 + 10 x 10 + 10 x 10 + 10 x 10 + 45 x 45 = 143.

The total number of combinations of drawing 4 balls (2 from each urn) is 20 choose 2 x 20 choose 2 = 19600.

So, the probability that she will be able to buy a 25-cent candy bar is 143/19600 = 143/152.