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Answer :
The original bag contained 12 red counters and 9 green counters.
Let’s break down the problem step by step:
1. Initial Scenario:
o Initially, there are r red counters and g green counters in the bag.
o The probability of drawing a green counter is given as 3/7.
2. After Putting the Counter Back:
o We put the counter back in the bag.
o Additionally, 2 more red counters and 3 more green counters are added to the bag.
o Now the total number of counters in the bag is (r + 2) + (g + 3).
3. New Probability:
o We want to find the probability of drawing a green counter from the updated bag.
o This probability is given as 6/13.
Let’s set up the equations based on the given information:
• Initial Probability:[tex][ P(\text{green}) = \frac{g}{r + g} = \frac{3}{7} ][/tex]
• Updated Probability:[tex][ P(\text{green}) = \frac{g + 3}{(r + 2) + (g + 3)} = \frac{6}{13} ][/tex]
Now let’s solve for the original number of red and green counters:
1. Initial Probability Equation:[tex][ \frac{g}{r + g} = \frac{3}{7} ] Solving for (g): [ 7g = 3r + 3g ] [ 4g - 3r = 0 \quad \text{(1)} ][/tex]
2. Updated Probability Equation:[tex][ \frac{g + 3}{(r + 2) + (g + 3)} = \frac{6}{13} ] Solving for (g): [ 13(g + 3) = 6(r + g + 5) ] [ 13g + 39 = 6r + 6g + 30 ] [ 7g - 6r = -9 \quad \text{(2)} ][/tex]
3. Solving the System of Equations: Multiply Equation (1) by 2 and subtract Equation (2): [ 8g - 6r = 0 ] [ 7g - 6r = -9 ] [ 8g - 7g = 9 ] [ g = 9 ]
4. Finding the Number of Red Counters: Substitute (g = 9) into Equation (1): [ 4g - 3r = 0 ] [ 4(9) - 3r = 0 ] [ 36 - 3r = 0 ] [ 3r = 36 ] [ r = 12 ]
Therefore, the original bag contained 12 red counters and 9 green counters.
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