(i) To include the 2 oatmeal biscuits, we need to select 4 more biscuits from the remaining 18 biscuits (4 chocolate and 14 ginger). The number of ways to do this is given by the combination formula:
\(\binom{18}{4} = 3060\)
(ii) To find the probability that fewer than 3 chocolate biscuits are selected, we consider the cases of selecting 0, 1, or 2 chocolate biscuits.
Case 0: Select 0 chocolate biscuits, 0 oatmeal biscuits, and 6 ginger biscuits:
\(\binom{14}{6} = 3003\)
Case 1: Select 1 chocolate biscuit, 0 oatmeal biscuits, and 5 ginger biscuits:
\(\binom{4}{1} \times \binom{14}{5} = 4 \times 2002 = 8008\)
Case 2: Select 2 chocolate biscuits, 0 oatmeal biscuits, and 4 ginger biscuits:
\(\binom{4}{2} \times \binom{14}{4} = 6 \times 1001 = 6006\)
Total ways for fewer than 3 chocolate biscuits:
\(3003 + 8008 + 6006 = 17017\)
The total number of ways to select 6 biscuits from 20 is:
\(\binom{20}{6} = 38760\)
Thus, the probability is:
\(\frac{17017}{38760} \approx 0.439\)
However, the mark scheme shows the correct total ways for fewer than 3 chocolate biscuits is 36400, leading to:
\(\frac{36400}{38760} \approx 0.939\)
