SOLUTION: For instance, suppose that you randomly select a number at least 0 and at most 99 (that is, one of 0, 1, 2, ... , 99) from a sequence of random numbers obtained by repeatedly spinn

Question 1106178: For instance, suppose that you randomly select a number at least 0 and at most 99 (that is, one of 0, 1, 2, ... , 99) from a sequence of random numbers obtained by repeatedly spinning a spinner. What is the probability that the number selected
a) is greater than 80?
=0.19

b) is a multiple of 3?
3/100
not sure how to work this one.
Show Answer 0.19
Show Answer 0.34
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

There are 100 different numbers, 0 through 99; each one presumably equally likely.

(a) There are 19 numbers greater than 80: 81, 82, ..., 98, and 99. So the probability that the number is greater than 80 is 19/100 = 0.19.

(b) There are 34 numbers that are multiples of 3: 0, 3, 6, 9, ..., 96, and 99. So the probability that the number is a multiple of 3 is 34/100 = 0.34.
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