Thread: Red marbles, white marbles, in a bag.

1. Red marbles, white marbles, in a bag.

So,
I have a bag containing 10 blue marbles and 5 red marbles. If I draw 8 marbles without replacement, what is the probability of me getting 1 red marbles out of 8?
2 out 8; 3; 4;5?
How do you calculate that?

Additionally, how would you calculate the probability of my getting at least 1 red marbles out of 8? At least 2; 3; 4; 5?
And how do you calculate that?

Thanks

2. Re: Red marbles, white marbles, in a bag.

Hello, Sabre!'

Here's the first half of your question . . .

I have a bag containing 10 blue marbles and 5 red marbles.
If I draw 8 marbles without replacement, what is the probability of getting:

(a) 1 red marble and 7 blue marbles?
There are: . ${15\choose8} \,=\,6,\!435$ possible outcomes.

To get 1 red and 7 blue: . ${5\choose1}{10\choose7} \:=\:5\cdot 120 \:=\:600$ ways.

Therefore: . $P(\text{1 red}) \:=\:\frac{600}{6,\!435} \:=\:\frac{40}{429}$

(b) 2 red marbles and 6 blue marbles?
To get 2 red and 6 blue: . ${5\choose2}{10\choose6} \:=\:10\cdot210 \:=\:2,\!100$ ways.

Therefore: . $P(\text{2 red}) \:=\:\frac{2,\!100}{6,\!435} \:=\:\frac{140}{429}$

(c) 3 red marbles and 5 blue marbles?
To get 3 red and 5 blue: . ${5\choose3}{10\choose5} \:=\:10\cdot252 \:=\:2,\!520$ ways.

Therefore: . $P(\text{3 red}) \:=\:\frac{2,\!520}{6,\!435} \:=\:\frac{56}{143}$

(d) 4 red marbles and 4 blue marbles?
To get 4 red and 4 blue: . ${5\choose4}{10\choose4} \:=\:5\cdot210 \:=\:1,\!050$ ways.

Therefore: . $P(\text{4 red}) \:=\:\frac{1,\!050}{6,\!435} \:=\:\frac{70}{429}$

(e) 5 red marbles and 3 blue marbles?
To get 5 red and 3 blue: . ${5\choose5}{10\choose3} \:=\:1\cdot120 \:=\:120$ ways.

Therefore: . $P(\text{5 red}) \:=\:\frac{120}{6,\! 435} \:=\:\frac{8}{429}$

3. Re: Red marbles, white marbles, in a bag.

Thank you Soroban,

Questions
Originally Posted by Soroban
There are: . ${15\choose8} \,=\,6,\!435$ possible outcomes.
Above, when you calculate the outcome, what you are calculating is the number of ways, right? A.k.a “number of combinations of n things taken x at a time” with
, yes?

But does it take into account that there are marbles of two different colors here?
I'm really quite at loss, so sorry if I'm totally wrong and stupid.

4. Re: Red marbles, white marbles, in a bag.

Originally Posted by Sabre
Above, when you calculate the outcome, what you are calculating is the number of ways, right? A.k.a “number of combinations of n things taken x at a time” with , yes?
But does it take into account that there are marbles of two different colors here?
I'm really quite at loss, so sorry if I'm totally wrong and stupid.
Suppose that $r$ is the number of red balls in the sample, $0\le r\le 5$.
Then $8-r$ is the of blue balls in the sample. Right?

$\mathcal{P}(r)=\dfrac{\dbinom{5}{r}\dbinom{10}{8-r}}{\dbinom{15}{8}}$

5. Re: Red marbles, white marbles, in a bag.

"Additionally, how would you calculate the probability of my getting at least 1 red marbles out of 8? At least 2; 3; 4; 5?
And how do you calculate that

The simplest way to do that is to look at the opposite. If you don't get "at least one red marble", you must have gotten all 8 blue. There are a total of 15 marbles, 10 of which are blue. The probabability the first marble is blue is 10/15= 2/3. There are now 14 marbles, 9 of them blue so the probability the second marble is also blue is 9/14. Continuing in that way, the probability all 8 are blue is (10/15)(9/14)(8/14)...(2/7)(1/6) which we could write as (10!)(15- 10)!/(15!)= (10!)(5!)/(15!). The probability of "at least one red marble" is 1 minus that:
1- (10!)(5!)/(15!).

6. Re: Red marbles, white marbles, in a bag.

I think I understood. I mean I can place the logic in my head now. Let's see if I got this right.

In a bag with 10 white marbles and 3 red marbles I take 7 without replacement:
Finding the probabilities of getting 1 red marble means comparing the number of ways I can obtain exactly 1 red marble and 6 white marbles (desirable outcome) when drawing 7 marbles to all possible outcomes when drawing 7 marbles.

A. First I calculate the number of all possible combinations when taking 7 out of 13.

${13\choose7} \,=$ (13!)/(7!)(13-7)!= 1,716 possible combinations

Then I calculate the number of ways I can obtain my desirable outcome of 1 red and 6 white, which is the number of combinations for obtaining 1 red out of 3 multiplied by the number of combinations for 6 out of 10 white.

${3\choose1}{10\choose6} \:=$((3!)/(1!)(3-1)!)*((10!)/(6!)(10-6)!) = (3)(210) = 630 ways of obtaining the desirable outcome

Then I calculate P which is

$\mathcal{P}(1 red)=\:\frac{630}{1716}=\:\frac{3}{8}=\:\0,37$
__________________________________________________ __________________________________________________ ________________

B. To find the probability of 2 reds marbles we have:

$\mathcal{P}(2 red)=\dfrac{\dbinom{3}{2}\dbinom{10}{5}}{\dbinom{1 3}{7}}=\:\frac{252}{1716}=\:\0,15$

Yeah, thanks again everybody, I was confused about the notion of 'number of ways' which is actually only the number of combinations for a particular outcome.

7. Re: Red marbles, white marbles, in a bag.

Originally Posted by Sabre
I think I understood. I mean I can place the logic in my head now. Let's see if I got this right.
In a bag with 10 white marbles and 3 red marbles I take 7 without replacement:
Finding the probabilities of getting 1 red marble means comparing the number of ways I can obtain exactly 1 red marble and 6 white marbles (desirable outcome) when drawing 7 marbles to all possible outcomes when drawing 7 marbles.
A. First I calculate the number of all possible combinations when taking 7 out of 13.
[texto ]{13\choose7} \,=[/tex] (13!)/(7!)(13-7)!= 1,716 possible combinations
Then I calculate the number of ways I can obtain my desirable outcome of 1 red and 6 white, which is the number of combinations for obtaining 1 red out of 3 multiplied by the number of combinations for 6 out of 10 white.
${3\choose1}{10\choose6} \:=$((3!)/(1!)(3-1)!)*((10!)/(6!)(10-6)!) = (3)(210) = 630 ways of obtaining the desirable outcome
Then I calculate P which is
$\mathcal{P}(1 red)=\:\frac{630}{1716}=\:\frac{3}{8}=\:\0,37$
__________________________________________________ __________________________________________________ ________________
B. To find the probability of 2 reds marbles we have:
$\mathcal{P}(2 red)=\dfrac{\dbinom{3}{2}\dbinom{10}{5}}{\dbinom{1 3}{7}}=\:\frac{252}{1716}=\:\0,15$
Yeah, thanks again everybody, I was confused about the notion of 'number of ways' which is actually only the number of combinations for a particular outcome.
Why the h_ll did you change the question in midstream?
It seems to someone who has taught this for many years, this is nonsense.
You have no idea what any of this means!

8. Re: Red marbles, white marbles, in a bag.

Did I get the answers right?
What do you mean I changed the question midstream?
If you mean why I changed the number of marbles, it was to see if I could to the calculation.
Apparently not?

Please could you make your reply more useful. I really don't have any idea what you mean.