Four identical balls are placed in a box. The balls are numbered 2; 3; 4 and 5. Two balls are picked from the box at random without replacement and the total score is recorded. Find the probability that (i) the total score is 7, (ii) the product of the scores is 6.

The given expression is $1 \times 2^4 + 1 \times 2^2 + 1 \times 2^1 + 1$.
This can be written as $1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0$.
The coefficients of the powers of 2, from highest to lowest, form the base 2 number.
The coefficients are 1, 0, 1, 1, 1.
So, the number in base 2 is $\boxed{10111_2}$.

Part (a)(ii)

The number $103_x$ in base 10 is given by:

$$1 \times x^2 + 0 \times x^1 + 3 \times x^0$$ $$x^2 + 0 + 3$$ $$x^2 + 3$$

We are given that $103_x = 67_{10}$.

$$x^2 + 3 = 67$$ $$x^2 = 67 - 3$$ $$x^2 = 64$$ $$x = \sqrt{64}$$

Since $x$ represents a base, it must be a positive integer.

$$x = \boxed{8}$$

Part (b)
The balls are numbered 2, 3, 4, and 5. Two balls are picked without replacement.

The total number of ways to pick 2 balls from 4 without replacement is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The possible pairs and their scores are:

• (2, 3): Sum = 5, Product = 6
• (2, 4): Sum = 6, Product = 8
• (2, 5): Sum = 7, Product = 10
• (3, 4): Sum = 7, Product = 12
• (3, 5): Sum = 8, Product = 15
• (4, 5): Sum = 9, Product = 20

Part (b)(i)

From the list above, the pairs with a total score of 7 are (2, 5) and (3, 4). There are 2 such pairs.

$$\text{Probability (total score is 7)} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$ $$\text{Probability (total score is 7)} = \frac{2}{6} = \boxed{\frac{1}{3}}$$

Part (b)(ii)

From the list above, the pair with a product of 6 is (2, 3). There is 1 such pair.

$$\text{Probability (product of scores is 6)} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$ $$\text{Probability (product of scores is 6)} = \boxed{\frac{1}{6}}$$