Question

If a number is randomly chosen from the following list, what is the probability that the number is a multiple of 5?

[40, 30, 40, 45, 25, 40, 34, 3, 28, 50, 55, 50]

Answer 1

i see 9 of them that are mults of 5

Answer 2

9/12

Answer 3

9 /12
3/ 4

Answer 4

and 12 total gives us:

9/12 = 3/4

Answer 5

Here's a fun question. In the set of natural numbers from 1 to 19, if 3 numbers are chosen at random and added together, what is the probability that the sum will be a multiple of 3?

Answer 6

with or without repetition?

Answer 7

Without.

Answer 8

20 total in the pool

19+18+17 = 30+15+9 = 40+13 = 53 is the largest sum

Answer 9

well...54 lol

Answer 10

which accounts for multiples of 3 up to 18*3

Answer 11

but since we have 3 chances to pull that pool becomes 20*19*18 in total right?

Answer 12

6840 ways .... but that doesnt account for duplicates does it lol

Answer 13

I think you're missing some options.

Answer 14

im sure I am ;)

Answer 15

Firstly there's only 19 numbers, not 20.

Answer 16

1 through 19 inclusive.

Answer 17

Now you pick 3 of them to sum.

The number of possible ways you can pick those 3 is \[{19 \choose 3} = 969\]

Answer 18

But the real question is, how many will sum to be a multiple of 3?

Answer 19

To answer this, we have to look at remainders.

Each of the numbers can be expressed as:

\[n = 3q + r, (q \in \mathbb{N}, r \in \{0,1,2\})\]

Answer 20

1 for example is 3(0) + 1

Answer 21

19 is 3(6) + 1

Answer 22

14 is 3(4)+2

Answer 23

So we can see that the sum of 3 numbers will be a multiple of 3 if and only if the r's for the three numbers add to a multiple of 3.

Answer 24

\[n_1 + n_2 + n_3 = 3(q_1 + q_2 + q_3) + r_1 + r_2 + r_3\]

Answer 25

The possible options for that are:
\[r_1 + r_2 + r_3 = 0\]\[r_1 + r_2 + r_3 = 3\]\[r_1 + r_2 + r_3 = 6\]
So we can have 0+0+0, 1+2+0, 2+2+2, and 1+1+1

Answer 26

So we have to go through and enumerate all the remainders in our set.

You can make a table:
n r
1 1
2 2
3 0
4 1
5 2
6 0
7 1
8 2
9 0
10 1
11 2
12 0
13 1
14 2
15 0
16 1
17 2
18 0
19 1

# of 0's: 6
# of 1's: 7
# of 2's: 6

Answer 27

lol .... i thought you said it be fun ;)

Answer 28

it is fun! =)

Answer 29

So the number of ways we can have (without repetition)

0+0+0 is 6*5*4 = 120
1+1+1 is 7*6*5 = 210
2+2+2 is 6*5*4 = 120
1+2+3 is 6*7*6 = 252

Adding all those possible ways we have 702 options for summing to a multiple of 3.
Dividing by the total number of possible ways we can pick 3 numbers we get

\[\frac{702}{969} = 72.44\%\]