Some of the books on a certain shelf are in English, and the rest of the books are in Spanish....

Understanding the Question

We need to find the probability that at least one of 2 randomly chosen books will be English.

Let's clarify what we know:

• Books on shelf are either English or Spanish (no other languages)
• We're selecting 2 books without replacement
• We want P(at least one English book)

Key probability insight: \(\mathrm{P(at\,least\,one\,English)} = 1 - \mathrm{P(both\,Spanish)}\)

Here's the crucial point: To find a unique probability value, we need the exact number of Spanish and English books. A ratio alone won't suffice because the same ratio with different absolute numbers yields different probabilities when selecting without replacement.

Analyzing Statement 1

Statement 1 tells us: The ratio of Spanish to English books is 3:1.

This means for every 3 Spanish books, there's 1 English book. Our possibilities include:

• 3 Spanish, 1 English (4 total)
• 6 Spanish, 2 English (8 total)
• 9 Spanish, 3 English (12 total)
• And so on...

Let's test two scenarios to demonstrate they yield different probabilities:

Scenario 1: 3 Spanish, 1 English (4 books total)

• P(both Spanish) = \(\frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}\)
• P(at least one English) = \(1 - \frac{1}{2} = \)\(\frac{1}{2}\)

Scenario 2: 6 Spanish, 2 English (8 books total)

• P(both Spanish) = \(\frac{6}{8} \times \frac{5}{7} = \frac{30}{56} = \frac{15}{28}\)
• P(at least one English) = \(1 - \frac{15}{28} = \frac{13}{28} \approx \)0.464

Since \(\frac{1}{2} \neq \frac{13}{28}\), we've proven that different scenarios with the same 3:1 ratio produce different probabilities.

[STOP - Not Sufficient!] Statement 1 is NOT sufficient.

This eliminates answer choices A and D.

Analyzing Statement 2

Important: We now forget Statement 1 completely and analyze Statement 2 independently.

Statement 2 tells us: There are fewer than 20 books on the shelf.

This constraint gives us an upper bound but reveals nothing about the Spanish-English distribution. Consider these possibilities:

• 19 books: all Spanish, 0 English → P(at least one English) = \(0\)
• 10 books: 5 Spanish, 5 English → P(at least one English) = \(\frac{3}{4}\)
• 15 books: 14 Spanish, 1 English → P(at least one English) = \(\frac{2}{15}\)

Without knowing the distribution, we cannot determine a unique probability.

[STOP - Not Sufficient!] Statement 2 is NOT sufficient.

This eliminates answer choices B and D (already eliminated).

Combining Both Statements

Now we use both pieces of information:

• Spanish:English ratio is 3:1 (Statement 1)
• Total books < 20 (Statement 2)

With a 3:1 ratio, the total number of books must be a multiple of 4 (since 3 + 1 = 4). Given the constraint of fewer than 20 books, our possible totals are: 4, 8, 12, or 16.

This gives us exactly four possible scenarios:

• 4 books total: 3 Spanish, 1 English → P(at least one English) = \(\frac{1}{2}\)
• 8 books total: 6 Spanish, 2 English → P(at least one English) = \(\frac{13}{28}\)
• 12 books total: 9 Spanish, 3 English
• 16 books total: 12 Spanish, 4 English

We've already demonstrated that the 4-book and 8-book scenarios yield different probabilities (\(\frac{1}{2}\) vs \(\frac{13}{28}\)).

Why do different totals give different probabilities? In "without replacement" problems, removing a book has a proportionally larger impact on smaller totals. Think of it this way: removing 1 book from 4 changes the pool by 25%, but removing 1 book from 16 changes it by only 6.25%.

Since we have multiple valid scenarios that produce different probability values, the statements together are NOT sufficient.

[STOP - Not Sufficient!] Even combined, we cannot determine a unique answer.

This eliminates answer choices A, B, C, and D.

The Answer: E

Even with both pieces of information, we cannot determine a unique probability because different valid book totals (4, 8, 12, or 16) yield different probability values.

Answer Choice E: "The statements together are not sufficient."