A bag has 4 red marbles, 5 white marbles, and 6 blue marbles. Three marbles are drawn from the bag (without replacement). What is the probability that they are all the same color?

HelpBot Oct 3, 2020

#2

#3**+2 **

Lol, I never gave the answer.

Here is the answer:

\((\frac {4}{15} \cdot \frac {3}{14} \cdot \frac {2}{13})+(\frac {5}{15}\cdot\frac {4}{14}\cdot\frac {3}{13}) + (\frac {6}{15}\cdot \frac {5}{14}\cdot \frac {4}{13})\)

Compute it using any calculator. Don't expect us to do that as well. (Next time, use the class message board.)

Nacirema
Oct 3, 2020

#4**+2 **

I don't take AoPS classes this is just a problem. I listen to video though.

Also tyty, ill try to get the reasoning?

HelpBot Oct 3, 2020

#5**+3 **

Sorry, had to eat dinner.

... And how did you know I was refering to AoPS. Whatever, irrelevent.

Here is why I did that.

So, if we took the probability that the red is first, then it is 4/15. Because there is no replacement, we have to remove one value from both numerator and denominator: (4-1)/(15-1). One mpre time for the third: (4-2)/(15-2). Then we multiply them all together. Careful: if you subtract 1/1 instead of 1 from the numerator and 1 from the denominator, then you will get a wrong number.

This in mind, tell me how I did the other 2 parts: white and blue. Just reply to this thread.

Nacirema
Oct 3, 2020

#9**+2 **

I cannot fault your logic Naciema. That is because it is correct! Thank you for your answer.

I think your method gives a solution of 34/455

I did it a different way and got ~~a different answer~~ the same answer.

I said that there are 4C3=4 ways to chose 3 red balls

There are 5C3=10 ways to chose 3 white

and 6C3=20 ways to choose 3 blue.

That is a total of 34 favourable outcomes

The sample space is 15C3 = ~~286 ways ~~ 455 ways (error corrected)

~~34/286 = 17/143~~ 34/455

**Just as Nacirema found **

Melody Oct 5, 2020