1. ## probability question

Question:1
You select 2 numbers at random from 1,2,3,4,5,6,7,8. What is the probability that the sum of two numbers is 3?

I tried as follows:
1+2=3
p(1)=1/8
p(2)=1/8

1/8 * 1/8= 1/64

==================

Question 2:
You select 3 numbers at random from 1,2,3,4,5,6,7,8,9. What is the probability that the sum of 3 numbers is 6?

I tried as follows:
1+2+3=6
or
2+1+3=6
or
3+2+1=6
or
2+2+2=6

p(1)=1/9
p(2)=1/9
p(3)=1/9

1/9 * 1/9 *1/9 = 1/729

Am I going in wrong path? Please correct me.

2. Originally Posted by Ka9
Question:1
You select 2 numbers at random from 1,2,3,4,5,6,7,8. What is the probability that the sum of two numbers is 3?

I tried as follows:
1+2=3
p(1)=1/8
p(2)=1/8

1/8 * 1/8= 1/64

==================

Question 2:
You select 3 numbers at random from 1,2,3,4,5,6,7,8,9. What is the probability that the sum of 3 numbers is 6?

I tried as follows:
1+2+3=6
or
2+1+3=6
or
3+2+1=6
or
2+2+2=6

p(1)=1/9
p(2)=1/9
p(3)=1/9

1/9 * 1/9 *1/9 = 1/729

Am I going in wrong path? Please correct me.
I am assuming that the same number can be selected more than once.

Question 1
(1,2) and (2,1) give sum 3. So, the probability is 2/64=1/32.

Question 2
The selections (1,2,3), (1,1,4) and (2,2,2) give the sum 6.
(1,2,3) can be permuted in 6 ways.
(1,1,4) can be permuted in 3 ways.
So, the probability is 10/729.

3. Originally Posted by Ka9
Question:1
You select 2 numbers at random from 1,2,3,4,5,6,7,8. What is the probability that the sum of two numbers is 3?
==================
Question 2:
You select 3 numbers at random from 1,2,3,4,5,6,7,8,9. What is the probability that the sum of 3 numbers is 6?
Your work is not correct, but neither is the work in the above reply.
If we select two numbers that an be done in $\dbinom{8}{2}=28$ ways.
Only one of those pairs, $\{1,2\}$ sum to 3. What is that probability?

For question #2. How many subsets of three are there from $\{1,2,3,4,5,6,7,8,9\}~?$
How many of those subsets have elements that sum to three?

4. Hi Plato,
I could not follow the second part. Could you please explain me.
Thanks,

5. Originally Posted by Ka9
Hi Plato,
I could not follow the second part. Could you please explain me.
If that means question #2, then there are $\dbinom{9}{3}=\dfrac{9!}{3!\cdot 6!}$ ways to select a subset of three from $\{1,2,3,4,5,6,7,8,9\}$.
How many of those subsets have elements that sum to six?

6. Hi Plato,
It seems, there are 84 ways to get a sum of 6. Is it what I understood is correct.
Also, is there are video link or or other resources to get help on propbability.

7. No that is completely wrong.
There are eighty-four ways to select three digits from nine different digits.
Here is one of those eighty-four, $\{3,1,2\}$. Their sum is 6.
Can you find a different set of three?

8. Hello Ka9

It seems that you are not familiar with the method Plato was going about. If that is the case, take a look at this:

For number one, 2 numbers are drawn from the first 8 counting numbers. Now, how can we get a sum of 3? Well, the only possible way is to get 1+2. The probability of drawing a 1 is 1/8, and the probability of drawing a 2 is 1/7 (because there are 7 numbers left). Multiply these two together, and we have 1/56. But what if the first number you drew was 2, and the second one was 1? We have to account for this, so we have $\displaystyle \frac{1}{56}+\frac{1}{56}=\frac{1}{28}$

Can you tackle the second problem in a similar manner?

9. Hello rtblue,
Thanks for the reply. It seems I am on the way.

1/9 * 1/8 * 1/7= 1/504

1/504 * (6+1+2)=1/56

1+2+3 = 2+2+2 = 1+1+4

(6+1+2)= 6 ways to write 1,2,3, 1 way to write 2,2,2 and 2 ways to write 114.
Am I understanding correct?

10. Originally Posted by Ka9
(6+1+2)= 6 ways to write 1,2,3, 1 way to write 2,2,2 and 2 ways to write 114. Am I understanding correct?
If we have a set $\left\{ {\begin{array}{*{20}c} {} & 3 \\ 1 & {} \\ {} & 2 \\ \end{array} } \right\}$
The numbers in that set add up to 6. No matter how we arrange those three numbers the sum is still 6

Thus this problem has absolutely nothing to do with order.
There are $84$ ways to select a set of three from $\{1,2,3,4,5,6,7,8,9\}$.
Only one set $\{1,2,3\}$ sums to six.
So the answer to the second problem is $\dfrac{1}{84}$.

11. Hi Plato,
Thanks for the reply. But, left with some confusion.

My understanding is we are taking 1,2,3 to get a sum of 6. So, in the set 1,2,3...9, we are also having 2+2+2 to get 6 and also 4+1+1 to get a sum of 6.
Are we not considering 2,2,2 and 4,1,1 combination too?
in which case the answer differs. Is it?

I am still confused. Please guide me.

12. Well then we are discussing two different problems.
If one says select three numbers from $\{1,2,3,4,5,6,7,8,9\}$ then $[2,2,2]$ does not count.
WHY? Well that is what is called a multi-selection.
Saying I have three numbers is saying I have three different numbers.
It is just the way mathematics is written.

Here is the exact quote from the original question: “Question 2:
You select 3 numbers at random from 1,2,3,4,5,6,7,8,9. What is the probability that the sum of 3 numbers is 6?”
It plainly says select 3 numbers. That mean that $[2,2,2$ does not count.

If the question means to allow that multi-selection it should have said so.
It could have said explicitly that repetition is allowed.
It could have said “select a three digit number using those nine digits”, that would have allowed 114, 141, 411 all to have been considered.

So tell us the exact wording of the original question.
Be sure to include the entire question.

13. Now, it's clear to me. Yes, the question says, select 3 numbers. Thanks though for clear explanation. Sorry! for the confusion.