A bag contains 40 balls in total. There are 18 red balls and the rest are green or blue balls. Two balls are picked from the bag one after another without replacement. The probability that the first ball is red and the second ball is green is 3/26. How many blue balls are there in the bag?

Introduction / Context:
This problem combines probability with algebraic reasoning. We know the composition of red balls in a bag and that the remaining balls are green or blue. We are given the probability that the first ball drawn is red and the second ball is green, both drawn without replacement. Using this probability, we must determine how many blue balls are in the bag.

Given Data / Assumptions:

• Total number of balls in the bag = 40.
• Number of red balls = 18.
• The remaining 22 balls are either green or blue.
• Two balls are drawn in sequence without replacement.
• The probability that the first ball is red and the second is green is 3/26.
• We assume all balls of the same color are indistinguishable and that each ball is equally likely to be drawn.

Concept / Approach:
Let the number of green balls be G and the number of blue balls be B. We know that G + B = 22. The probability that the first ball is red and the second is green can be expressed in terms of G. Specifically, it is (18/40) multiplied by (G/39). We equate this product to 3/26 and solve for G. Once G is known, we compute B as 22 minus G to find the number of blue balls.

Step-by-Step Solution:
Step 1: Let G be the number of green balls and B be the number of blue balls, so G + B = 22.Step 2: The probability that the first ball is red is 18 / 40.Step 3: Given that the first ball drawn is red, there are now 39 balls left in the bag, of which G are green.Step 4: The conditional probability that the second ball is green is therefore G / 39.Step 5: The overall probability of first red and then green is (18 / 40) * (G / 39), and this is given to be 3 / 26.Step 6: Set up the equation (18 / 40) * (G / 39) = 3 / 26 and solve for G. Simplifying, we obtain G = 10, so B = 22 − 10 = 12 blue balls.
Verification / Alternative check:
We can verify by substituting G = 10 back into the probability expression. The probability of first drawing red and then green is (18/40) * (10/39). Multiply numerators to get 180 and denominators to get 1560. Simplifying 180 / 1560 by dividing both numerator and denominator by 60 gives 3 / 26, which matches the given probability. This confirms that the counts 10 green and 12 blue are consistent with the conditions.

Why Other Options Are Wrong:
16 blue balls would imply only 6 green balls, which would lead to a smaller probability than 3/26. 10 blue balls would leave 12 green balls, which yields (18/40) * (12/39) = 216/1560, simplifying to 9/65, not 3/26. 14 or 8 blue balls would similarly give probabilities that do not match 3/26 when used in the same calculation. Therefore, only 12 blue balls produce the correct probability.

Common Pitfalls:
Students may incorrectly set up the equation by reversing the order of colors or by using 40 in both denominators, forgetting the reduction to 39 after the first draw. Another common error is to attempt a guess and check method with the options rather than solving the equation algebraically, which can be time consuming and error prone. Carefully defining variables and writing the probability in terms of those variables is the safest approach.

Final Answer:
The number of blue balls in the bag is 12.