 August 13, 2020, 12:57:10 am

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#### mani.s_

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« on: November 26, 2019, 08:29:27 pm »
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When it's asking for the trends in the probabilities of picking exactly 0, 1, 2, 3 blue bricks, do I talk about how the probability of picking 0 blue brick decreases as the number of trials increases, how the probability of picking 1 blue brick increases as it approaches expected value, then it decreases?

Thanks for ur help #### RuiAce

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« Reply #1 on: November 27, 2019, 10:08:26 am »
+3
Your trends are entirely based on what you feel is appropriate.

Your observation about the probability of picking 0 bricks is one example of this, and obviously judging by the data you've provided, I agree with it. This is just one of who knows how many trends you may discover from your table of values.

how the probability of picking 1 blue brick increases as it approaches expected value, then it decreases?
With this statement though, are you trying to say that the probability of picking 1 more blue brick increases as the total number of blue bricks approaches the expected value? Because otherwise the conclusion doesn't make sense.  #### mani.s_

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« Reply #2 on: November 27, 2019, 04:54:18 pm »
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Your trends are entirely based on what you feel is appropriate.

Your observation about the probability of picking 0 bricks is one example of this, and obviously judging by the data you've provided, I agree with it. This is just one of who knows how many trends you may discover from your table of values.
With this statement though, are you trying to say that the probability of picking 1 more blue brick increases as the total number of blue bricks approaches the expected value? Because otherwise the conclusion doesn't make sense.
Yh that's what I mean for the expected value. Thank you so much for ur help. I had one more question, to explain why the probability decreases as you select exactly 0 blue bricks. Is it because the value of p and n are both small, therefore its a negatively skewed graph, or do I write it decreases simply because their are more combinations/sample space??

Thanks #### RuiAce

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« Reply #3 on: November 28, 2019, 10:25:59 pm »
+1
Yh that's what I mean for the expected value. Thank you so much for ur help. I had one more question, to explain why the probability decreases as you select exactly 0 blue bricks. Is it because the value of p and n are both small, therefore its a negatively skewed graph, or do I write it decreases simply because their are more combinations/sample space??

Thanks If you're asking me, I'd say a bit of both would contribute here. The actual graph of a distribution (both in the discrete and continuous case) can be influenced by more than one thing.  #### RuiAce

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« Reply #4 on: December 05, 2019, 10:43:52 am »
+1
I had another inquiry, for the table of probabilities that I presented I saw another trend that some of the probabilities are the same, and that the pattern continues down. I don't know why this happens. Could someone explain why this happens?? I have attached the probabilities table below highlighting the probabilities that are the same. Thanks for your help!
That looks rigged by the consequence of $p=\frac14$ in your binomial distribution.
$\text{Observe from direct computation that }3\times \binom71 = \binom72.$
$\text{Hence}$ \begin{align*}
\binom72 (0.25)^2 (0.75)^5 &= 3\times \binom71  \times (0.25)^2(0.75)^5\\
&= (3\times 0.25) \times \binom71 (0.25)(0.75)^5\\
&= 0.75 \times \binom71 (0.25) (0.75)^5\\
&= \binom71 (0.25)(0.75)^6.
\end{align*}
Similar algebraic tricks can be done to prove the others. (It may be worth remarking that $\binom72 = \frac34\times \binom82$.) But basically, the trick is just shuffling the terms around.  #### mani.s_

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« Reply #5 on: December 06, 2019, 05:23:26 pm »
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Rui, I had another question. For the column for selecting exactly 1 or 2, we can see that the probability increases and then it decreases. Why does that happen? I was thinking that maybe it increases because it approaches the expected value, but then that's when the number of blue bricks in increases and n is kept constant. So I'm not sure. Could you please help? Thanks

#### RuiAce

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« Reply #6 on: December 06, 2019, 05:42:33 pm »
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You could always compute the value $E(X) = np$ for each value of $n$, and compare how close $E(X)$ is to a fixed $k$ perhaps? If it looks like $E(X)$ is a number close to $k$, then you probably do expect the probability to be higher. If it looks like $E(X)$ has now started going further away from $k$, you'd also expect the probability to be start decreasing.

(This really does sound like what you were trying to say, I hope. There's no problem I can see going down that road if you can justify it well. You just need to have an intuitive idea of what the expected value actually is.)  #### mani.s_

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« Reply #7 on: December 06, 2019, 05:53:51 pm »
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You could always compute the value $E(X) = np$ for each value of $n$, and compare how close $E(X)$ is to a fixed $k$ perhaps? If it looks like $E(X)$ is a number close to $k$, then you probably do expect the probability to be higher. If it looks like $E(X)$ has now started going further away from $k$, you'd also expect the probability to start decreasing.

(This really does sound like what you were trying to say, I hope. There's no problem I can see going down that road if you can justify it well. You just need to have an intuitive idea of what the expected value actually is.)
Hmm I'm really sorry but I'm not sure that I understand. What I was trying to say that for columns 1 and 2 (not rows) we can see that the probability first increases and then decreases. Is there a specific reason why this happens. For example, for the first column (0 blue bricks), we can see that the probability decreases as the trials are increased. For this I can say that since n is higher, there are more combinations/sample space, theirs a higher chance that we get more blue bricks and that's why the probability decreases. Is there a similar reason for selecting exactly 1 or 2 blue bricks?

Thank you so much for your help #### RuiAce

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« Reply #8 on: December 06, 2019, 06:15:02 pm »
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Hmm I'm really sorry but I'm not sure that I understand. What I was trying to say that for columns 1 and 2 (not rows) we can see that the probability first increases and then decreases. Is there a specific reason why this happens. For example, for the first column (0 blue bricks), we can see that the probability decreases as the trials are increased. For this I can say that since n is higher, there are more combinations/sample space, theirs a higher chance that we get more blue bricks and that's why the probability decreases. Is there a similar reason for selecting exactly 1 or 2 blue bricks?

Thank you so much for your help In short, it's because the expected value $E(X)$ is getting close to that value of $k$, but then they start going further and further away from it. (Your values of $k$ are just 1 and 2.)

I can't explain everything to you for an assignment, however.  