A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously...

Understanding the Question

We need to find the exact value of n, where n is the number of defective bulbs in a box of 10 bulbs.

Given Information

• Total bulbs in box: 10
• Fewer than half are defective (so \(\mathrm{n < 5}\))
• This means n can only be: 0, 1, 2, 3, or 4
• Two bulbs will be drawn simultaneously

What We Need to Determine

For sufficiency, we need information that narrows down n to exactly one value from {0, 1, 2, 3, 4}.

Key Insight

Since we're dealing with just 5 possible values, we can test each one directly against any probability constraint. This approach is often faster than setting up and solving equations.

Analyzing Statement 1

Statement 1 tells us: The probability that both bulbs drawn will be defective is \(\frac{1}{15}\).

Let's test which value of n gives us this probability. When drawing 2 bulbs from 10, there are \(\mathrm{C(10,2) = 45}\) total ways.

Testing Different Values

• If n = 0 or 1: Can't draw 2 defective bulbs → Probability = 0 (not \(\frac{1}{15}\))
• If n = 2: Only 1 way to draw both defective → Probability = \(\frac{1}{45}\) (not \(\frac{1}{15}\))
• If n = 3: \(\mathrm{C(3,2) = 3}\) ways to draw 2 defective → Probability = \(\frac{3}{45} = \frac{1}{15}\) ✓
• If n = 4: \(\mathrm{C(4,2) = 6}\) ways to draw 2 defective → Probability = \(\frac{6}{45} = \frac{2}{15}\) (not \(\frac{1}{15}\))

Conclusion

Only n = 3 produces the probability of \(\frac{1}{15}\). [STOP - Sufficient!]

Statement 1 is sufficient.

This eliminates choices B, C, and E.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2 tells us: The probability that one bulb will be defective and the other non-defective is \(\frac{7}{15}\).

Again, with 45 total ways to draw 2 bulbs, let's test our possible values.

Testing Different Values

For each n, the number of ways to draw 1 defective and 1 non-defective = \(\mathrm{n × (10-n)}\)

• If n = 1: \(\mathrm{1 × 9 = 9}\) ways → Probability = \(\frac{9}{45} = \frac{1}{5}\) (not \(\frac{7}{15}\))
• If n = 2: \(\mathrm{2 × 8 = 16}\) ways → Probability = \(\frac{16}{45}\) (not \(\frac{7}{15}\))
• If n = 3: \(\mathrm{3 × 7 = 21}\) ways → Probability = \(\frac{21}{45} = \frac{7}{15}\) ✓
• If n = 4: \(\mathrm{4 × 6 = 24}\) ways → Probability = \(\frac{24}{45} = \frac{8}{15}\) (not \(\frac{7}{15}\))

Conclusion

Only n = 3 produces the probability of \(\frac{7}{15}\). [STOP - Sufficient!]

Statement 2 is sufficient.

The Answer: D

Since each statement alone is sufficient to determine that n = 3, the answer is D.

Answer Choice D: "Each statement alone is sufficient."