if digits are not repeated,
- (a)(i)if digits are not repeated,[2]
- (a)(ii)if digits can be repeated and the number formed is odd.[3]
- (b)(i)Find the number of different selections that include the 2 oatmeal biscuits.[2]
- (b)(ii)Find the probability that fewer than 3 chocolate biscuits are selected.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [M1] First digit in 2 ways. 2 × 4 × 3 × 2 or 2 × 4P3 OE as final answer
- [A1] Total = 48 ways
- [M1] 2×5×5×3 Seeing 5² mult; this mark is for correctly considering the middle two digits with replacement
- [M1] Mult by 6; this mark is for correctly considering the first and last digits
- [A1] = 150 ways
- [M1] OO**** in 18C4 ways 18Cₓ or the sum of five 2-factor products with n = 14 and 4, may be x by 2C2: 4C0 × 14C4 + 4C1 × 14C3 + 4C2 × 14C2 + 4C3 × 14C1 + 4C4 (× 14C0)
- [A1] = 3060
- [B1] Choc Not Choc: 0 6 = 1 × 16C6 = 8008 0.2066, 1 5 = 4C1 × 16C5 = 17472 0.4508, 2 4 = 4C2 x 16C4 = 10920 0.2817 OR Choc Oats Ginger: 0 0 6, 0 1 5, 0 2 4, 1 0 5, 1 1 4, 1 2 3, 2 0 4, 2 1 3, 2 2 2 The correct number of ways with one of 0, 1 or 2 chocs, unsimplified or any three correct number of ways of combining choc/oat/ginger, unsimplified
- [M1] Total = 36400 ways sum the number of ways with 0, 1 and 2 chocs and two must be totally correct, unsimplified OR sum the nine combinations of choc, ginger, oats, six must be totally correct, unsimplified
- [M1] Probability = 36400/ 20C6 dividing by 20C6 (38760) oe
- [A1] = 0.939 (910/969)
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