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A bag contains some green balls, some blue balls, and seven red balls. The probability of picking one green ball is \(\frac{1}{7}\) more than that of picking one red ball from the bag, and the probability of picking one blue ball is \(\frac{9}{35}\) more than that of picking one red ball from the bag. Find the total number of blue balls in the bag.

1. 12
2. 14
3. 18
4. 20
5. 16

Correct answer: 16

Solution

Let red probability be r. Then green probability = \(r + \frac{1}{7}\) and blue probability = \(r + \frac{9}{35}\). Since the three probabilities sum to 1, solving gives the red probability and then the total number of balls. Using 7 red balls, the blue count is 16.

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