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IMO B

So we need to find the probability of picking a blue marble out of all

.. B/(B+G+Y)

St1: Gives the no for B+G+Y , but we still don't know the value for B.

Hence ST1 is insufficient.

St2: P(Y+G) = 2/5; so P(B) = P (Y+G) -1

So St2 is sufficient.
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IMO B

candies in the jar: B, G, Y
one candy selected at random,
probability of brown candy = \(\frac{(number\ of\ brown\ candies) }{ (total\ candies) }\)

(1) There are 25 candies in the jar.
don't know the number of brown candies in the jar
Not sufficient

(2) The probability that the candy selected will be green or yellow is \(\frac{2}{5}\).
probability it is either green or yellow = \(\frac{2}{5}\)
probability it is neither green nor yellow = \(1 - \frac{2}{5} = \frac{3}{5}\)
since there are only 3 kinds of candies
=> probability of selecting brown candy = probability of selecting neither green nor yellow =\( \frac{3}{5}\)
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Let total number of candies be c.

Probability of selecting brown candy = b/c, b is the number of brown candies.
or
1-probability of selecting a green or yellow candy =1- (g+y)/c, g & y are the numbers of green and yellow candies respectively.

(1) There are 25 candies in the jar.
we know that c=25, but no information about b,g,y.

Insufficient

(2) The probability that the candy selected will be green or yellow is 2/5.
We know the value of (g+y)/c = 2/5
Probability of selecting brown candy = 1 - 2/5 = 3/5

Sufficient.

IMO Option B
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Bunuel
Each of the candies in a jar is brown, green, or yellow. If one candy is to be selected at random from the jar, what is the probability that the candy will be brown?

(1) There are 25 candies in the jar.
(2) The probability that the candy selected will be green or yellow is 2/5.

The probability of Brown \(= \frac{Brown}{Total}\)

(1) \(\frac{Brown}{25}\): We don't know brown. Insufficient.

(2) The probability of \(Green+Yellow=\frac{2}{5}\), So, the probability of Brown \(=1-\frac{2}{5}=\frac{3}{5}\); Sufficient.

The answer is B
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Correct Answer B

Each of the candies in a jar is brown, green, or yellow. If one candy is to be selected at random from the jar, what is the probability that the candy will be brown?

(1) There are 25 candies in the jar.
We do not any information about the number of brown candy
- Not sufficient


(2) The probability that the candy selected will be green or yellow is 2/5.
given
P(green or yellow) =2/5
P(g)+P(y)+P(b)=1
P(b or brown) = 3/5
-sufficient
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Each of the candies in a jar is brown, green, or yellow. If one candy is to be selected at random from the jar, what is the probability that the candy will be brown?

Stat1: There are 25 candies in the jar.
we don't know breakup of brown, green, or yellow candies. Not sufficient.

Stat2: The probability that the candy selected will be green or yellow is 2/5.
P(GUY)= P(G) + P(Y)= 2/5 so, P(B)= 1-P(GUY)= 3/5 ....Sufficient

So, I think B. :)
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Candies in the jar are either brown, green or yellow. No other information is provided in the question other than this. To find the probability that candy will be brown we need total number of candies and no. of brown candies or the probability that selected candy is green or yellow.

Statement 1:
It only provides the total no. of candies in the jar but no information about the number of brown candies. Hence Insufficient

Statment 2 :
This statement provides the probability that the selected candy will green or yellow.
Hence we can find the probability that candy will be brown by subtracting the value from 1.
1-2/5 = 3/5
Sufficient

So the answer will be B
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P(brown) = ?

St1 - b + g + y = 25
Clearly insufficient.

St2 - P(green or yellow) = 2/5
1 - P(green or yellow) = P(brown) = 3/5
Sufficient.

Answer is B.
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