anonymous
  • anonymous
A positive integer is picked randomly from 91 to 100, inclusive. What is the probability that it is divisible by both 2 and 3?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
well for the 3 the set is(93,96,99)
anonymous
  • anonymous
for 2 the set is(92,94,96,98,100)
cristiann
  • cristiann
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)

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anonymous
  • anonymous
cool.....if it is by both 2 and 4
cristiann
  • cristiann
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"
anonymous
  • anonymous
ok

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