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Answer :
Number of red counters in the original bag is 3, and the number of green counters is 7.
Let's use algebra to solve the problem. Let U be the number of red counters and G be the number of green counters originally in the bag.
From the first part of the problem, we know that
Probability (selecting a green counter) = G / (U + G) = 3/7
Solving for U in terms of G, we get
U = (7G - 3G) / 3 = 4G/3
So we know that there were 4G/3 red counters and G green counters in the bag originally. But since the number of counters must be a whole number, we can assume that there were 4R red counters and 3G green counters originally, where R and G are both integers and R + G is the total number of counters.
After adding 2 red and 3 green counters, the number of counters in the bag is now R + 2 + G + 3 = R + G + 5.
From the second part of the problem, we know that
P(selecting a green counter) = (G + 3) / (R + G + 5) = 6/13
Solving for R in terms of G, we get
R = (13G - 9G - 15) / 7 = 4G/7 - 15/7
Since R must be an integer, we can try different values of G to see if R is an integer. For example, if G = 7, then R = 3 and the total number of counters is 10.
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The given question is incomplete, the complete question is:
There are only U red counters and G green counters in a bag. A counter is taken at random from the bag. The probability that the counter is green is 3/7. The counter is put back in the bag. 2 more red counters and 3 more green counters are put in the bag. A counter is taken at random from the bag. The probability that the counter is green is 6/13. Find the number of red counters and the number of green counters that were in the bag originally
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