Almost every thing is wrong with it. Did you complete the list I started?
$123,~124,~125,~126,~134,~135,~234$ That is only seven favorable outcomes. Are there any others?
How many total outcomes are there. See here.
What answer do you get?
NO! The adding the numbers in the set $\{1,~2,~3\}$ gives the sum of six no matter how one orders the numbers.
Hence this problem comes-down to counting the number of three element subsets that have a sum $~S\le 9$.
Now $6\le S\le 27$, $1+2+3=6~\&~8+9+10=27$. How many of those subsets have a sum $\le 9~?$
This is a question about content and not about order.
This should be clearly solved using a Tree Diagram,
with branches each of which the First card has probability, Second card and Third card has as he is picking out one card after the other without replacement.
Each of these 4 basic and unique entries sums to 9 and each occurred in 3! = 6 times (ex. 1 2 6, 1 6 2, 2 1 6, 2 6 1, 6 1 2, 6 2 1 etc.) throughout the tree diagrams.
1 1 7
1 2 6
1 3 5
1 4 4
The probability sum of the outcomes:
correct!
But I think it is wrong to consider 7 entries 3 of which are repeated combinations, because they are already considered in the :
1 1 7
1 2 6
1 3 5
1 4 4
1 5 3
1 6 2
1 7 1
Giving,
wrong !?!
@zikcau25;813647], did you bother to read the OP much less the entire theard?
Obviously not, or else you don't understand decks of cards.
Here is the OP: There are only ten cards numbered 1 thru 10.
Thus there is no way you can choose three of then and have $144$.
Umm, example:
1 4 4 (represents 1 arrangement of the 3 numbered cards and that the sum of the 3 numbers on these 3 cards add up to 9 only)
1: the number written on the first card,
4: the number written on the second card,
4: the number written on the third card
Precisely there with spaces in between, I mean 1, 4, 4 taken separately or 1+4+4= 9 (sum) and not 144 as a whole value alone.
Thanks.