ishitadas1994@gmail.com wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14
(B) 2/7
(C) 6/35
(D) 6/29
(E) 6/42
We and let r, g, and b be the number of balls that have red color only, green color only, and both colors, respectively. We can create the equations:
r + g + b = 58,
b = (2/7)(r + b)
And
b = (3/7)(g + b)
Let’s simplify the second equation by multiplying it by 7:
7b = 2r + 2b
5b = 2r
r = (5/2)b
Likewise, let’s simplify the third equation by multiplying it by 7:
7b = 3g + 3b
4b = 3g
g = (4/3)b
So plug (5/2)b for r and (4/3)b for g into the first equation, we have:
(5/2)b + (4/3)b + b = 58
15b + 8b + 6b = 58 x 6
29b = 58 x 6
b = 2 x 6 = 12
Therefore, the probability of picking a ball with both colors is 12/58 = 6/29.
Answer: D
How did you arrive at the two equations Sir ?
Can we solve this using an easier alternate method ?
Response:First, notice that r denotes the number of balls that have only red color, g denotes the number of balls that have only green color and b denotes the number of balls that have both red and green colors. The total number of balls is red only + green only + both; therefore, the total number of balls is r + g + b. The first equation is obtained by simply equating this quantity to 58.
For the equation b = (2/7)(r + b), we use the sentence “2/7 of the balls that have red color also have green color.” Notice that the balls that have red color is r + b (i.e. only red + both) and if a ball that contains the red color also contains the green color, then it contains both colors. Thus, 2/7 of r + b should equal b and that’s how we obtain the equation (2/7)(r + b) = b. The equation b = (3/7)(r + b) is obtained similarly.
Alternate Solution:As an alternate solution, let the number of balls that have both the colors be 6n. We choose the coefficient 6 because LCM(2, 3) = 6. Since 2/7 of the balls that have red color also have green color, there are 6n / (2/7) = 7 * 3n = 21n balls with red color (including the ones which also have green color). Similarly, since 3/7 of the balls that have green color also have red color, there are 6n / (3/7) = 7 * 2n = 14n balls with green color (including the ones which also have red color). We will use the following formula:
Total = #(red) + #(green) - #(both)
58 = 21n + 14n - 6n
58 = 29n
n = 2
Thus, there are 6n = 6*2 = 12 balls containing both colors. Since the total number of balls is 58, the probability that a randomly selected ball contains both colors is 12/58 = 6/29.
Answer: D _________________
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