# Math Help - Probability

1. ## Probability

Hi
I need help with this problem:
10 playing cards are numbered 1 to 10. 3 different cards are randomly selected. Calculate the probability that the sum of the numbers selected are at most 9?
Any help will be greatly appreciated.

2. ## Re: Probability

Originally Posted by mlg
Hi
I need help with this problem:
10 playing cards are numbered 1 to 10. 3 different cards are randomly selected. Calculate the probability that the sum of the numbers selected are at most 9?
Any help will be greatly appreciated.
You need to make a list of sums that meet the criteria.
$123,~124,~125,~126,~234~,\cdots$ ect.

Now $C_3^{10}=\dbinom{10}{3}=~?$

4. ## Re: Probability

Can anyone see anything wrong with the following approach?
10^3 possible outcomes.
84 of them sum to less than or equal to 9.
P = 84/1000 = 21/250

5. ## Re: Probability

Originally Posted by mlg
Can anyone see anything wrong with the following approach?
10^3 possible outcomes.
84 of them sum to less than or equal to 9.
P = 84/1000 = 21/250
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.

6. ## Re: Probability

Thanks again.
Regarding favourable outcomes, for example 123, can you also take the number 132 etc? (or is this repetition?).

7. ## Re: Probability

Originally Posted by mlg
Thanks again.
Regarding favourable outcomes, for example 123, can you also take the number 132 etc? (or is this repetition?).
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 $\le 9~?$

9. ## Re: Probability

Originally Posted by mlg
NO! Order has nothing to do with this question.

So $\dfrac{7}{120}$.

10. ## Re: Probability

Thanks again.
I was taking all the different orders, e.g.. 123, 132, 231 etc.

11. ## Re: Probability

This should be clearly solved using a Tree Diagram,
with branches each of which the First card has $\frac{1}{10}$ probability, Second card $\frac{1}{9}$ and Third card has $\frac{1}{8}$ 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 $4 \times 3!$ outcomes:

$\left (\frac{1}{10}\times \frac{1}{9}\times \frac{1}{8} \right )\times 4\times 3! = \frac{1}{30}$ correct!

But I think it is wrong to consider 7 entries 3 of which are repeated combinations, because they are already considered in the $\times 3!$:

1 1 7
1 2 6
1 3 5
1 4 4
1 5 3
1 6 2
1 7 1

Giving,

$\left (\frac{1}{10}\times \frac{1}{9}\times \frac{1}{8} \right )\times 7\times 3! = \frac{7}{120}$ wrong !?!

12. ## Re: Probability

Again, I do mistakes on Combinations word.
I meant Repeated Arrangements instead of Repeated Combinations above. Sorry.

13. ## Re: Probability

Originally Posted by zikcau25
This should be clearly solved using a Tree Diagram,
with branches each of which the First card has $\frac{1}{10}$ probability, Second card $\frac{1}{9}$ and Third card has $\frac{1}{8}$ 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
@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:
Originally Posted by mlg
10 playing cards are numbered 1 to 10. 3 different cards are randomly selected. Calculate the probability that the sum of the numbers selected are at most 9?
Any help will be greatly appreciated.
There are only ten cards numbered 1 thru 10.
Thus there is no way you can choose three of then and have $144$.

14. ## Re: Probability

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.

15. ## Re: Probability

Originally Posted by zikcau25
Umm, example:

1 4 4

1 on the first card,
4 on the second card,
4 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.
But that is impossible.
There is only one 4 in the whole deck,

Cards dealt out only once. If the first is a 4 the no other card can possibly be a 4.
This is not sampling with replacement. You are confused on that point.

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