In a bag, there are three types of balls colored red, white, and blue. The probability of drawing a red ball from the total balls is $\frac{1}{2}$, and the probability of drawing a blue ball is $\frac{2}{7}$. The number of white balls in the bag is 6. All the balls are numbered sequentially starting from 1, 2, 3, and so on. What is the probability of selecting a ball whose number is a multiple of 3 or 7 from the total balls?

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Accepted Answer

**Solution:**

1. Let the number of red balls be *R*, blue balls be *B*, and white balls are given as 6.

2. Given probability of selecting a red ball is:
   $$\frac{R}{R + B + 6} = \frac{1}{2}$$
   From this, we get:
   $$2R = R + B + 6 \implies R = B + 6 \quad 	ext{(Equation 1)}$$

3. Given probability of selecting a blue ball is:
   $$\frac{B}{R + B + 6} = \frac{2}{7}$$
   From this, we get:
   $$7B = 2(R + B + 6) = 2R + 2B + 12$$
   Simplifying:
   $$7B - 2B = 2R + 12 \implies 5B = 2R + 12 \quad 	ext{(Equation 2)}$$

4. Substitute *R* from Equation 1 into Equation 2:
   $$5B = 2(B + 6) + 12 = 2B + 12 + 12 = 2B + 24$$
   $$5B - 2B = 24 \implies 3B = 24 \implies B = 8$$

5. Using *B = 8* in Equation 1:
   $$R = 8 + 6 = 14$$

6. Total number of balls:
   $$R + B + 6 = 14 + 8 + 6 = 28$$

7. The balls are numbered from 1 to 28. We need to find the probability of selecting a ball numbered with a multiple of 3 or 7.

8. Multiples of 3 up to 28 are:
   $$3, 6, 9, 12, 15, 18, 21, 24, 27$$ (9 numbers)

9. Multiples of 7 up to 28 are:
   $$7, 14, 21, 28$$ (4 numbers)

10. Note that 21 is counted twice (multiple of both 3 and 7), so total favorable numbers:
    $$9 + 4 - 1 = 12$$

11. Therefore, the required probability is:
    $$\frac{12}{28} = \frac{3}{7}$$

**Answer:** $\frac{3}{7}$

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