anonymous
  • anonymous
Suppose I have 3 dices. I throw them. I want to get a {1,2,5}. So I do this to get the probabilities: \[\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \] In this case, have I considered other combinations like {2, 1, 5}, {5,2,1}, etc?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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KingGeorge
  • KingGeorge
You have not considered those other cases. You're assuming you get a 1, 2, and 5 in that order.
anonymous
  • anonymous
If I want to consider, what should I do?
KingGeorge
  • KingGeorge
You just have to do one more thing. You can just consider all possible orderings. There happen to be \(3!=6\) orderings, and they all have equal probability. Thus, just multiply your original answer by 6.

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More answers

anonymous
  • anonymous
Suppose if there were repeated values in side it: {2, 2, 5}. Then in this case, would I still multiply by 3!?
KingGeorge
  • KingGeorge
long story short, no. I'll explain in a minute.
anonymous
  • anonymous
ok sure. I will wait for your explanation. Thanks! :)
KingGeorge
  • KingGeorge
Hold on, for your first question, are you throwing all three dice at the same time? Or one a a time?
anonymous
  • anonymous
Wouldn't it be the same? They are all iid.
KingGeorge
  • KingGeorge
It's actually a little bit different. If we throw them one at a time, we do the process I explained above. Otherwise, you have to do something different.
anonymous
  • anonymous
I think it is all thrown together.
anonymous
  • anonymous
I don't think this is it. Even if they are all thrown together, the dices are i.i.d. random variables on their own.
KingGeorge
  • KingGeorge
There's a difference between thrown together and thrown at the same time. You get different information about which dice need to which numbers.
anonymous
  • anonymous
hmm... with your second equation, does it count the combinations of it?
KingGeorge
  • KingGeorge
If we were looking at \(\{2, 2, 5\}\) instead, we have to look at a couple different cases. Drawing a tree is a good idea here. Like above, we have a \(2/6\) chance of getting a 2 or a 5. Then, within this probability, there's a \(2/3\) chance we get a 2, and \(1/3\) chance we get a 5. If we get a 5, we need to find the probability of getting 2 2's, and if we get a 2, we find the probability of getting a 2 and a 5.
KingGeorge
  • KingGeorge
Also, you were correct in thinking I was wrong with my other explanation. It doesn't actually matter. I was merely getting confused.
anonymous
  • anonymous
oh I see. Thanks!! :) And just one thing, how do you type the math equations inline? I used the \[ tags but it goes to the next line.
KingGeorge
  • KingGeorge
use \( instead of \[
anonymous
  • anonymous
ohh...haha...thanks for your help and explanation! :)
KingGeorge
  • KingGeorge
I should note that doing it the other way I suggested is also a valid way of doing it, but you get \[{3 \over 6} \cdot{2 \over 6}\cdot{1 \over 6}={3! \over 6^3}\]Not what I originally got.
anonymous
  • anonymous
Thanks! :)

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