the probability that a 2 digit number selected at random will be a multiple of 3 and a not multiple of 5
Answers
12,18,21,24,27,33,36,39,42,48,51,54,57,63,66,69,72,78,81,84,87,93,96,99
favorable outcomes = 24
total possible outcomes = 30
p ( event) = 24/30
= 4/5.....
hope it help you mate
Answer:
There are 90 two digit numbers(10-99).
Out of this there are 30 numbers divisible by 3(12,15,............99)
Out of this there are 6 numbers divisible by 15(15, 30, 45, 60, 75, 90), which are also divisible by 5.
Therefore, the favorable cases are 30–6=24.
Hence, the required probability is 24/90 = 4/15.
Explanation:
We need a 2 digit number
So there are 9x10=90 possible 2 digit numbers. (10–99 essentially)
Ok so how many multiples of 3 are there in these 90 numbers.
Well I like to look at this in a simpler way.
How many multiples of 3 are there from 1 to 30?
10! of course… 3X10=30
So there are 30 multiples of 3 from 1 to 90 and including 3 more being 93,96 and 99 we get 33.
Ignore 3,6 and 9 since they are 1 digit numbers… we have 30 2-digit multiples of 3 from 10–99.
What is a multiple of 5?
Well if you notice it’s any number ending in 0 or 5 (you must have seen this before)
So how many of these multiples of 3 end in 0 or 5?
Well the best way is to first take LCM of 3 and 5 which is 15(since both are prime)
So the only multiples of both 3 and 5 are in increments of 15.
so we have 15,30,45,60,75 and 90 which are multiples of both.
So 30–6=24 are multiples of 3 but not of 5
Your answer is 24/90 = 4/15 ≈ 0.2667
Hope this is helpful....