What we see: a probability Data Sufficiency question. A box holds 8 balls, red and blue only. We need the number of red balls.
Step 1. Set up the unknown.
What we see: 8 balls in total.
What it means: if \(r\) balls are red, \(8 - r\) are blue, and \(r\) is a whole number from 0 to 8.
What we do and why: each statement gives a probability. We turn it into an equation in \(r\) and count how many whole-number answers it has.
Step 2. Statement (1): both balls red has probability \(\frac{5}{14}\).
What we see: two draws without putting the first ball back.
What it means: \(\frac{r}{8} \times \frac{r - 1}{7} = \frac{5}{14}\), so \(r(r - 1) = \frac{5}{14} \times 56 = 20\).
What we do and why: \(r(r - 1) = 20\) gives \(r = 5\) or \(r = -4\). A count cannot be negative, so \(r = 5\). One value, so statement (1) is enough.
Step 3. Statement (2): red first, then blue, has probability \(\frac{15}{56}\).
What we see: a red draw followed by a blue draw.
What it means: \(\frac{r}{8} \times \frac{8 - r}{7} = \frac{15}{56}\), so \(r(8 - r) = 15\).
What we do and why: \(r = 3\) gives \(3 \times 5 = 15\), and \(r = 5\) gives \(5 \times 3 = 15\). Both work, because swapping red and blue gives the same product. Two values, so statement (2) is not enough.
The trap: solving statement (2), finding \(r = 5\) first, and stopping. The product \(r(8 - r)\) is the same for \(r\) and \(8 - r\), so it always has a matching second answer unless \(r = 4\).
Faster way: for (1), "two numbers in a row that multiply to 20" is 4 and 5 at a glance. For (2), notice the equation does not change when you swap red and blue, which means two answers.
Takeaway: When an equation treats two groups the same way (red times blue), expect two answers. Always check whether the second root is also a valid count.
Answer: A
