I am confused about part (a). I apologise firstly for bombarding you with Q this week. I think the answers are wrong but i am really not sure how they have derived it as three GGG (no blue) should be (1/2)^3, as there are 3 B and 6 G. I am confused.
Probability distributiion
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I am confused about part (a). I apologise firstly for bombarding you with Q this week. I think the answers are wrong but i am really not sure how they have derived it as three GGG (no blue) should be (1/2)^3, as there are 3 B and 6 G. I am confused.
No blue means the same thing as three green. There are six green balls out of nine total.
[MATH]\dfrac{6}{9} * \dfrac{6}{9} * \dfrac{6}{9} = \left ( \dfrac{2}{3} \right )^3 = \dfrac{8}{27}.[/MATH]
Where do you get (1/2)? It is not equally likely to draw a blue ball and green ball.
[MATH]\dfrac{6}{9} * \dfrac{6}{9} * \dfrac{6}{9} = \left ( \dfrac{2}{3} \right )^3 = \dfrac{8}{27}.[/MATH]
Where do you get (1/2)? It is not equally likely to draw a blue ball and green ball.
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pka
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The answers are correct. Look at their sum. It is one as it should be.
They way they have explained it it looks like they used the three options. Sorry I was thinking of something else. I obviously need to take a break and look at this again. Simpler than i thought.
pka
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Is my statement incorrect? If not why are you butting in?
A Because you did not address the actual question, which was why (1/2) was not used.
B Telling someone that probailities are correct because they add up to one is indeed incorrect.
P implies Q does not imply Q implies P as you well know. Did you intend to mislead or were you just careless?
By the way, I posted before you did so you were the one butting in.
True. It is a good check, not a proof.