Q) A bag contains some red and some blue marbles.How many red marbles are in the bag?
(1) If two marbles are selected from the bag, the probability of selecting one red and one blue marble is 1/4
(2) There are a total of 8 marbles in the bag.
Marble Menace
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- way2ashish
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Statement 1: Say the total number of marbles in the bag is N and there are R red marbles. Hence, there are (N - R) blue marbles.way2ashish wrote:Q) A bag contains some red and some blue marbles.How many red marbles are in the bag?
(1) If two marbles are selected from the bag, the probability of selecting one red and one blue marble is 1/4
(2) There are a total of 8 marbles in the bag.
Number of ways to select two marbles from N marbles = NC2 = N(N - 1)/2
Number of ways to select one red marble from R marbles and one blue marbles from (N - R) marbles = (RC1)*((N - R)C1) = R*(N - R)
Hence, the probability of selecting one red and one blue marble = [R*(N - R)]/[N(N - 1)/2]
Hence, [R*(N - R)]/[N(N - 1)/2] = 1/4
As we have two unknowns N and R we cannot find the value of them from only one equation.
Statement 2: Clearly this statement is not sufficient
1 & 2 Together: N = 8
Now we have only one unknown R and we can solve the equation we obtained from statement 1 and find the value of R.
Sufficient
The correct answer is C.
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Statement 2: The total number of marbles = 8.way2ashish wrote:Q) A bag contains some red and some blue marbles. How many red marbles are in the bag?
(1) If two marbles are selected from the bag, the probability of selecting one red and one blue marble is 1/4
(2) There are a total of 8 marbles in the bag.
Statement 1: P(one red and one blue) = 1/4.
The following cases satisfy both statements:
Case 1: B=7, R=1, for a total of 8 marbles.
P(BR) = 7/8 * 1/7 = 1/8.
P(RB) = 1/8 * 7/7 = 1/8.
Since either ordering (BR or RB) will yield one red marble and one blue marble, we add the fractions:
P(one red and one blue) = 1/8 + 1/8 = 1/4.
In this case, R=1.
Case 2: B=1, R=7, for a total of 8 marbles.
P(BR) =1/8 * 7/7 = 1/8.
P(RB) = 7/8 * 1/7 = 1/8.
Since either ordering (BR or RB) will yield one red marble and one blue marble, we add the fractions:
P(one red and one blue) = 1/8 + 1/8 = 1/4.
In the case, R=7.
Since the number of red marbles can be different values, the two statements combined are INSUFFICIENT.
The correct answer is E.
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- mariofelixpasku
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Mitch Hunt is right.
1.) R/(R+B) * B/(R+B-1) = 1/8 NS
2.) R/8 * B/7 = 1/8 NS
1.) + 2.) not sufficient
1.) R/(R+B) * B/(R+B-1) = 1/8 NS
2.) R/8 * B/7 = 1/8 NS
1.) + 2.) not sufficient