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egoroff_w [7]
3 years ago
13

Given the following three arithmetic sequences : 1, 3, ... and 1, 4, ... and 1, 5, ... A number is randomly selected from 1 to 1

000, inclusive. What is the probability that the selected number is not a term of any of the given sequences?
Mathematics
1 answer:
ladessa [460]3 years ago
8 0

Answer:

  0.333

Step-by-step explanation:

The first sequence starts at 1 and has a common difference of 2. It will include every odd number in the range.

__

The second sequence starts at 1 and has a common difference of 3. It will include odd numbers and the even numbers 4, 10, 16, .... That is, all even numbers of the form 6n -2 will be included. The last one corresponds to the largest value of n such that ...

  6n -2 ≤ 1000

  6n ≤ 1002

  n ≤ 167

That is, 167 even numbers will also be excluded.

__

The third sequence starts at 1 and has a common difference of 4. Every number in this sequence is also a number in the first sequence.

__

So the numbers in these sequences include all 500 odd numbers and 167 even numbers, for a total of 667 numbers. The probability that a randomly chosen number is not in one of these sequences is ...

  (1000 -667)/(1000) = 333/1000 = 0.333 . . . . p(not a sequence term)

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