ITPEC FE Morning October 2023 Question 3

1/15 — the three empty spaces are an unordered choice of 3 out of 10, and only 8 of those choices form one adjacent block.

• •All equally likely choices of the empty spaces: C(10, 3) = 10 × 9 × 8 ÷ 6 = 120
• •A block of three adjacent spaces can start at space 1, 2, …, 8: 10 − 3 + 1 = 8 blocks
• •Probability: 8 ÷ 120 = 1 ÷ 15

Answer (c)

Why not others:

• •(a) 1/90 — divides the 8 blocks by ordered choices, 10 × 9 × 8 = 720, while counting the favourable cases unordered: 8 ÷ 720 = 1 ÷ 90
• •(b) 1/72 — no consistent count gives it; with total and favourable counted the same way the ratio is always 8 ÷ 120
• •(d) 1/12 — that is 10 ÷ 120, counting 10 block positions as if the row wrapped round in a circle; in a straight row a block cannot start at space 9 or 10

Key rule: Count the total and the favourable outcomes the same way (both as combinations or both in order); k adjacent slots in a row of n fit in n − k + 1 positions.

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