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There are 58 balls in a jar. Each ball is painted with at le
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07 May 2019, 02:17
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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 \(\frac{2}{7}\) of the balls that have red color also have green color, while \(\frac{3}{7}\) 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?
Re: There are 58 balls in a jar. Each ball is painted with at le
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19 Oct 2019, 02:10
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The solution to this problem might be easier than you think. Looking at the provided solutions, indeed, we see that only one has a denominator that is a divisor of 58.
Re: There are 58 balls in a jar. Each ball is painted with at le
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14 Sep 2020, 14:41
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sal60 wrote:
The solution to this problem might be easier than you think. Looking at the provided solutions, indeed, we see that only one has a denominator that is a divisor of 58.
Hence, without any calcultation, the answer is D.
Could anyone provide the longer solution to this problem -- assuming the reader missed this easy solution?
Re: There are 58 balls in a jar. Each ball is painted with at le
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15 Sep 2020, 05:41
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Carcass 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 \(\frac{2}{7}\) of the balls that have red color also have green color, while \(\frac{3}{7}\) 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) \(\frac{6}{14}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{6}{35}\)
(D) \(\frac{6}{29}\)
(E) \(\frac{6}{42}\)
One approach is to use the Double Matrix Method. This technique can be used for questions featuring a population in which each member has two characteristics associated with it (aka overlapping sets questions). Here, we have a population of balls, and the two characteristics are: - has red or doesn't have red - has green or doesn't have green
Aside: We can also use Venn diagrams and formulae to solve overlapping sets questions. However, as difficulty levels increase, it becomes harder to apply those other approaches, whereas the Double Matrix Method works every time.
So our diagram will look something like this:
Let x = the number of balls with red on them There are 58 balls all together, 58 - x = the number of balls WITHOUT red on them Likewise, if we let y = the number of balls with green on them Then 58 - y = the number of balls WITHOUT green on them Finally, since each ball is painted at least one color, we know that there are ZERO balls with no colors on them. So will place a 0 in the bottom-right box.
[m]2/7 of the balls that have red color also have green color Since x = the number of balls with red on them, we know that (2/7)x = the number of balls with red AND green This also tells us that (5/7)x = the number of balls that have red but NOT green We get:
3/7 of the balls that have green color also have red color Since y = the number of balls with green on them, we know that (3/7)y = the number of balls with green AND red We get:
From here we can create two equations. First, since 2x/7 and 3y/7 both represent the number of balls with red AND green, it must be the case that: 2x/7 = 3y/7 Multiply both sides of the equation by 7 to get: 2x = 3y We can rewrite this as: 2x - 3y = 0
Second, we must recognize that the two boxes in the right-hand column must add to 58 - y So we can write: 5x/7 + 0 = 58 - y In other words: 5x/7 = 58 - y Multiply both sides of the equation by 7 to get: 5x = 406 - 7y Add 7y to both sides of the equation: 5x + 7y = 406
We now have the following equations: 2x - 3y = 0 5x + 7y = 406
Take the top equation and multiply both sides by 5 to get the equivalent equation: 10x - 15y = 0 Take the bottom equation and multiply both sides by 2 to get the equivalent equation: 10x + 14y = 812 Subtract the bottom equation from the top equation to get: -29y = -812 Solve: y = 28
We already know that 3y/7 represents the number of balls with red AND green Plug in y = 28 to get: 3y/7 = 3(28)/7 = 12
So there are 12 balls that are both red and green P(the selected ball is red and green) = 12/58 = 6/29
Re: There are 58 balls in a jar. Each ball is painted with at le
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07 Aug 2021, 00:41
if we solve this algebraically the equations can be tricky, Instead, the best approach can be choosing numbers, balls will be in integer, and since fraction is 2/7 choose multiples of 7 for red color (7,14,21,28,35,42,49,56) we can quickly see that 42 works from there fill the Double matrix accordingly, we get 12 balls for both red and green painted.
There are 58 balls in a jar. Each ball is painted with at le
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12 Aug 2021, 15:25
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Let's say, probability of green ball= P(G) and red ball = P(R) Also, the conditional probability P(G|R) = P(G intersect R) / P(R) = 2/7 and P(R|G) = P(G intersect R) / P(G) = 3/7 And P(R union G) = P(G) + P(R) - P(G intersect R) = 1
If you solve these equations, you will get P(G intersect R)= 6/29
Re: There are 58 balls in a jar. Each ball is painted with at le
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24 Oct 2024, 06:21
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Re: There are 58 balls in a jar. Each ball is painted with at le [#permalink]