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Question Number 18464 by chernoaguero@gmail.com last updated on 22/Jul/17

3 numbers are chosen from 1 to 30. The probability that they are not consecutive is

$$\mathrm{3}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{from}\:\mathrm{1}\:\mathrm{to}\:\mathrm{30}.\:\mathrm{The} \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{they}\:\mathrm{are}\:\mathrm{not}\:\mathrm{consecutive} \\ $$$$\mathrm{is} \\ $$

Answered by Tinkutara last updated on 22/Jul/17

Total number of ways in which a triplet of 3 numbers can be chosen is^(30) C_3 . Now we have 28 pairs of 3 consecutive integers, i.e. (1, 2, 3), (2, 3, 4), (3, 4, 5), ..., (28, 29, 30). Probability of these consecutive integers = ((28)/(^(30) C_3 )) = ((28)/(4060)) = (1/(145)) ∴ Probability that they are not consecutive integers is = ((144)/(145))

$$\mathrm{Total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{in}\:\mathrm{which}\:\mathrm{a}\:\mathrm{triplet} \\ $$$$\mathrm{of}\:\mathrm{3}\:\mathrm{numbers}\:\mathrm{can}\:\mathrm{be}\:\mathrm{chosen}\:\mathrm{is}\:^{\mathrm{30}} {C}_{\mathrm{3}} . \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{28}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{3}\:\mathrm{consecutive} \\ $$$$\mathrm{integers},\:\mathrm{i}.\mathrm{e}.\:\left(\mathrm{1},\:\mathrm{2},\:\mathrm{3}\right),\:\left(\mathrm{2},\:\mathrm{3},\:\mathrm{4}\right),\:\left(\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right), \\ $$$$...,\:\left(\mathrm{28},\:\mathrm{29},\:\mathrm{30}\right). \\ $$$$\mathrm{Probability}\:\mathrm{of}\:\mathrm{these}\:\mathrm{consecutive}\:\mathrm{integers} \\ $$$$=\:\frac{\mathrm{28}}{\:^{\mathrm{30}} {C}_{\mathrm{3}} }\:=\:\frac{\mathrm{28}}{\mathrm{4060}}\:=\:\frac{\mathrm{1}}{\mathrm{145}} \\ $$$$\therefore\:\mathrm{Probability}\:\mathrm{that}\:\mathrm{they}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{consecutive}\:\mathrm{integers}\:\mathrm{is}\:=\:\frac{\mathrm{144}}{\mathrm{145}} \\ $$

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