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anonymous

  • 4 years ago

A positive integer is picked randomly from 91 to 100, inclusive. What is the probability that it is divisible by both 2 and 3?

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  1. anonymous
    • 4 years ago
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    well for the 3 the set is(93,96,99)

  2. anonymous
    • 4 years ago
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    for 2 the set is(92,94,96,98,100)

  3. cristiann
    • 4 years ago
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    Divisible by both 2 and 3 mean divisible by 6 (because 2 and 3 are relative prime numbers: their common divisor is 1) So out of 10 numbers you choose 1 number (which is 96) and the probability is 1/10 (the number of favorable cases over the number of all cases)

  4. anonymous
    • 4 years ago
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    cool.....if it is by both 2 and 4

  5. cristiann
    • 4 years ago
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    Then the above line of reasoning doesn't apply anymore... while a number divisible by 4 is by default also divisible by 2, "both by 2 and 4" means just "by 4"

  6. anonymous
    • 4 years ago
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    ok

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