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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?

 Oct 3, 2020
 #1
avatar+246 
+2

There are three cases, all are red, all are white and all are blue. 

 

The total marbles (NOT PROBABILITY) is 15. Next time, look at the transcript or actually pay attention in class.

 Oct 3, 2020
edited by Nacirema  Oct 3, 2020
 #2
avatar+399 
+3

I did but I still needed help. And btw your answer is wrong

 Oct 3, 2020
 #3
avatar+246 
+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
avatar+399 
+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?

 Oct 3, 2020
 #5
avatar+246 
+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
 #6
avatar+399 
+5

It's fine. Thanks I think I get it now. Brb I'm going to try to solve it

 Oct 3, 2020
edited by HelpBot  Oct 3, 2020
 #7
avatar+246 
+1

Message me if you need any more help.👌

Nacirema  Oct 4, 2020
 #8
avatar+399 
+3

I will, tyty

HelpBot  Oct 4, 2020
 #9
avatar+111124 
+2

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

 

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  

 Oct 5, 2020
edited by Melody  Oct 5, 2020
 #10
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+2

15C3 = 455

Guest Oct 5, 2020
 #11
avatar+111124 
+1

Ahr, thank you guest, now our answers are the same   laugh

Melody  Oct 5, 2020
 #12
avatar+246 
+2

Nice to hear I ACTUALLY GOT A PROBABILITY/COUNTING QUESTION RIGHT!!!!

 

I didn't bother with the computations since that is a bit nasty and I am FAR too lazy to go onto mathway or the forum calculator. cheeky

Nacirema  Oct 5, 2020
 #13
avatar+399 
+2

Tytyty very much! :D I'm not too hot on probability but i get this thx :D

 Oct 5, 2020
edited by HelpBot  Oct 5, 2020

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