(i) To complete the table, we use the given information:

• There are 12 biscuits wrapped in gold foil, 7 of which are chocolate-covered. Therefore, 5 are not chocolate-covered.
• There are 17 chocolate-covered biscuits in total, so 10 must be unwrapped.
• There are 30 biscuits in total, so 13 are not chocolate-covered. Of these, 5 are wrapped, so 8 are unwrapped.

(ii) The probability that a biscuit is wrapped in gold foil is given by:

\(\frac{12}{30} = 0.4\)

(iii) The probability that an unwrapped biscuit is chocolate-covered is:

\(\frac{10}{18} = \frac{5}{9} \approx 0.556\)

(iv) The probability that a biscuit is unwrapped, given that it is chocolate-covered, is:

\(\frac{10}{17} \approx 0.588\)

(v) The probability that Nasir takes exactly 2 wrapped biscuits is calculated using combinations:

\(\frac{\binom{12}{2} \times \binom{18}{2}}{\binom{30}{4}}\)

Calculating each part:

• \(\binom{12}{2} = 66\)
• \(\binom{18}{2} = 153\)
• \(\binom{30}{4} = 27,405\)

Thus, the probability is:

\(\frac{66 \times 153}{27,405} = 0.368\)