Problem 18

Explanation

We need to calculate the probability for each scenario given in the options.

Total books in each shelf:

• Shelf A = $5 + 3 + 4 = 12$.
• Shelf B = $3 + 3 + 6 = 12$.

The total possible pairs of books taken is $12 \times 12 = 144$.

Analysis of Each Option

We will calculate the probability ($P$) for each event.

1. 2 fiction books

   Taking 1 fiction from Shelf A and 1 fiction from Shelf B.

   $$P(\text{2 Fiction}) = \frac{5}{12} \times \frac{3}{12} = \frac{15}{144}$$

2. 1 fiction book and 1 science book

   There are two possibilities:

   • Fiction from Shelf A and Science from Shelf B: $\frac{5}{12} \times \frac{3}{12} = \frac{15}{144}$
   • Science from Shelf A and Fiction from Shelf B: $\frac{3}{12} \times \frac{3}{12} = \frac{9}{144}$
   $$P(\text{1 Fiction, 1 Science}) = \frac{15}{144} + \frac{9}{144} = \frac{24}{144}$$

3. 1 science book and 1 history book

   There are two possibilities:

   • Science from Shelf A and History from Shelf B: $\frac{3}{12} \times \frac{6}{12} = \frac{18}{144}$
   • History from Shelf A and Science from Shelf B: $\frac{4}{12} \times \frac{3}{12} = \frac{12}{144}$
   $$P(\text{1 Science, 1 History}) = \frac{18}{144} + \frac{12}{144} = \frac{30}{144}$$

4. 2 history books

   Taking 1 history from Shelf A and 1 history from Shelf B.

   $$P(\text{2 History}) = \frac{4}{12} \times \frac{6}{12} = \frac{24}{144}$$

5. 2 science books

   Taking 1 science from Shelf A and 1 science from Shelf B.

   $$P(\text{2 Science}) = \frac{3}{12} \times \frac{3}{12} = \frac{9}{144}$$

Conclusion

Let's compare the probability results (all denominators are 144):

• 2 fiction books: 15
• 1 fiction book and 1 science book: 24
• 1 science book and 1 history book: 30
• 2 history books: 24
• 2 science books: 9

The smallest value is 9.

Thus, the smallest probability is for the child to get 2 science books.