BTGmoderatorLU wrote:Source: Veritas Prep
A container contains 4 red marbles and 8 blue marbles. A second container contains 6 red marbles and x blue marbles. One marble is drawn from each of the two containers. If the probability of drawing a pair of marbles of the same color is 1/2, what is the value of x?
A. 0
B. 4
C. 6
D. 10
E. 12
We can PLUG IN THE ANSWERS, which represent the number of blue marbles in the second container.
When the correct answer is plugged in, the probability of selecting 2 red marbles OR 2 blue marbles = 1/2.
B: 4 blue marbles, implying a total of 10 marbles in the second container
Case1: 2 red marbles
P(red marble from the 1st container) = 4/12
P(red marble from the 2nd container) = 6/10.
To combine these probabilities, we multiply:
4/12 * 6/10 =
1/5.
Case 2: 2 blue marbles
P(blue marble from the 1st container) = 8/12.
P(blue marble from the 2nd container) = 4/10.
To combine these probabilities, we multiply:
8/12 * 4/10 =
4/15.
Since a favorable outcome will be yielded by Case 1 OR by Case 2, we ADD the fractions in blue:
1/5 + 4/15 = 3/15 + 4/15 = 7/15.
Here, the resulting probability is too low.
Eliminate B.
D: 10 blue marbles, implying a total of 16 marbles in the second container
Case1: 2 red marbles
P(red marble from the 1st container) = 4/12.
P(red marble from the 2nd container) = 6/16.
To combine these probabilities, we multiply:
4/12 * 6/16 =
1/8.
Case 2: 2 blue marbles
P(blue marble from the 1st container) = 8/12.
P(blue marble from the 2nd container) = 10/16.
To combine these probabilities, we multiply:
8/12 * 10/16 =
5/12.
Since a favorable outcome will be yielded by Case 1 OR by Case 2, we ADD the fractions in blue:
1/8 + 5/12 = 3/24 + 10/24 = 13/24.
Here, the resulting probability is too high.
Eliminate D.
Since B yields a probability that is TOO LOW, while D yields a probability that is TOO HIGH, the correct answer must be BETWEEN B AND D.
The correct answer is
C.
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