A positive integer is picked randomly from 91 to 100, inclusive.

What is the probability that it is divisible by both 2 and 3?

Question

A positive integer is picked randomly from 91 to 100, inclusive.

What is the probability that it is divisible by both 2 and 3?

anonymous · Answer

well for the 3 the set is(93,96,99)

anonymous · Answer

for 2 the set is(92,94,96,98,100)

cristiann · Answer

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)

anonymous · Answer

cool.....if it is by both 2 and 4

cristiann · Answer

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 · Answer

ok