Question

Probability question

Answer 1

We randomly choose a non-empty subset with at most three elements of the set
A = {2, 4; 6; 8; 10}. Determine the probability that the sum of the elements chosen subset to be a perfect square.

Answer 2

16 will be your only perfect square, because 4 is too small, and 36 is already too big.

5C3 combinations overall, (choosing 3 numbers out of 5) which gives 10 combinations.

2+6+8 = 16
2+4+10 = 16

the rest are smaller or greater than 16

Answer 3

Holfd on with at most 3, or you can choose only 3, and not a sum of 3 numbers, because what I did was for only choosing sums of 3 numbers.

Answer 4

@SolomonZelman at most means<=3

Answer 5

so a sum of 2 numbers or just 1 number is also an option, right ?

Answer 6

yes

Answer 7

Then,
5C3 for all possible sums of 3 (which =10)
5C2 for all possible sums of 2 (which=10)
and 5 possible choices of 1 element. So add all up, to get 25 options altogether.

The working ones are:



for sums of 3,

2+6+8 = 16
2+4+10 = 16



for sums of 2,

10 + 6 = 16 (because the rest are 4<x<16 with one option of 18 that is not a perfect square either)

for one number,
element "4"

Answer 8

So you have 2 working options of 3 elements, 1 option of 2 elements and 1 option of 1 element.
This gives you that 4 options out of 25 are perfect square.

Answer 9

sorry you cannot reach 25

Answer 10

yes, and the perfect square of 4, I did mention.

Answer 11

I found 10 subsets am i wrong?

Answer 12

10 subsets? Can you show how, please ?

Answer 13

{a,b,c} {x,y} {z}
2,4,10 2,2 4
2,10,4 6,10
10,2,4 10,6
10,4,2
4,10,2
4,2,10

Answer 14

I thought that when it comes to choosing 2 players for the basketball team (for example) you can't say that choosing ` Sam and John ` isn't the same thing as choosing `John and Sam`.

The same people are playing, right?

I would say that

2 + 4 + 10

and all possible permuttions of the sum of `2` `4` and `10` are one option, which is just sum of 2, 4 and 10. No matter in what order is the sum written.

Answer 15

all possible `permutations`, (not permuttions) TYPO

Answer 16

you mean the subset {2,4,10} is the same with the subset {10,4,2}

Answer 17

No, I am just referring to the sums. But yeh, in your case, it would be nPr (not nCr) sorry -:(

Answer 18

So you get 6 + 6 subsets for {x,y,z}

2+6+8 and 2+4+10

Answer 19

for example {2,2} can be a subset?

Answer 20

for {x,y{} you only get 6+10 and 10+6, since you can't use the element 2 twice, you only have one "2" in the set.

Answer 21

okay now to clear it:
{a,b,c} {x,y} {z}
2,4,10 10,6 4
but {8,2,6 } looks good as well can i count him also?

Answer 22

yes, 8,2,6 is just 3P3

Answer 23

3 P 3 = 3! / (3 - 3)! = 3! / 0! = 3! / 1 = 3! = 1 × 2 × 3 = 6

Answer 24

Permutation of 3 taken by 3 for {8,2,6}=6
Permutation of 3 taken by 3 for {2,4,10}=6}=>6+6=12

Answer 25

Just giving you all possible permutations of ALL POSSIBLE SUBSETS

5 P 3 = 5! / (5 - 3)! = 5! / 2! = (5 × 4 × 3 × 2!) / 2! = 5 × 4 × 3 = 60
+
5 P 3 = 5! / (5 - 2)! = 5! / 3! = (5 × 4 × 3!) / 3! = 5 × 4 = 20
+
(5 option of all 1 number subsets, even non-perfect-squares) 5
──
85

Answer 26

So 85 is all possible options, and we found the prefect square ones:

for {a,b,c} = 12 subsets (as 2 × 3P3 )
for {x,y} = 2 subsets ( `10+6`, or `6+10` )
for {z} = 1 subset (number 4)

Answer 27

So overall, 15 subsets out of 81 are perfect squares. Now, reduce the fraction
••••

Answer 28

and the P=15/85(85 NOT 81)

Answer 29

Yes, I already corrected that error :)

Answer 30

well i think
I got it

Answer 31

I re-wrote it wrong again... good that you noticed it.

Answer 32

So 15/85
(÷5)

Answer 33

I must recap the permutation and combinatorics thing

Answer 34

What do you mean to recap the permutation and combination thing? To RVW ?

Answer 35

*review sry, I'm not proficiency in english

Answer 36

The difference between COMBINATIONS (written as `nCr` where you choose `r` numbers out of `n` number) and PERMUTATIONS (written as `nPr` where you choose `r` numbers out of `n` number), is that in PERMUTATIONS the order matter and in Combinations makes no difference.

For example, saying 1,2 and 2,1 is the same in COMBINATIONS, but for permutations is 2 different things...

Answer 37

If you are familiar with factorial (written as ` n! = 1 × 2 × 3 × ... × ( n-1 ) × n ` . ) I can relate the 2,
ComB. and PerM.

Answer 38

okay thanks for the tip;
I got too deep into the calculus so I've kind of forgotten this things;

Answer 39

Well, anyway, just one more MATH sentence.

` n C r × r ! = n P r `

Answer 40

BTW- if you need help typing any symbol in on the keyboard? Will help with anything that I know:)

Answer 41

thank you so much I guess yo're an ace when it comes to ASCII code

Answer 42

`✂` No.... I am not so good at computers as others think.. tnx though:) `✂`

Answer 43

\(\huge\color{ blue }{\huge {\bbox[5pt, cyan ,border:2px solid purple ]{ \frac{x}{dx} }}}\)

Answer 44

Have a good one... !