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CriticalDendrite
 Post subject: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x
PostPosted: Thu, 31 May 2012 06:03:12 UTC 
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The probability of exactly 4 out of 10 people having a red ticket at a charity ball is approximately 0.2508. What is the probability, as a percentage, of a random person having a red ticket? Round your answer to the nearest percent. (Hint: red or not red)


The answer I got for this, using the equation Px=nCx(p)^x(q)^n-x, where n=10, x=4, and Px=0.2508 was a probability of approx. 0.5688 for the value of p(probability of success). I plugged my answer back into the equation where x=4 with a probability of success of approx. 0.5688 and received a Px value of approx. 14%, whereas in the question it is stated that it should be equal to 0.2508. Clearly, I have done something wrong but I can't see it. Any help would be greatly appreciated.
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outermeasure
 Post subject: Re: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x
PostPosted: Thu, 31 May 2012 06:08:33 UTC 
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CriticalDendrite wrote:
The probability of exactly 4 out of 10 people having a red ticket at a charity ball is approximately 0.2508. What is the probability, as a percentage, of a random person having a red ticket? Round your answer to the nearest percent. (Hint: red or not red)


The answer I got for this, using the equation Px=nCx(p)^x(q)^n-x, where n=10, x=4, and Px=0.2508 was a probability of approx. 0.5688 for the value of p(probability of success). I plugged my answer back into the equation where x=4 with a probability of success of approx. 0.5688 and received a Px value of approx. 14%, whereas in the question it is stated that it should be equal to 0.2508. Clearly, I have done something wrong but I can't see it. Any help would be greatly appreciated.


How did you get 0.5688?

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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CriticalDendrite
 Post subject: Re: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x
PostPosted: Thu, 31 May 2012 06:25:09 UTC 
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Here is what I did,

0.2508=10C4(p)^4(1-p)^10-4

0.2508=210(p)^4(1-p)^6

0.2508/210=p^4(1-p)^6

4sqrt(0.2508/210)=1(1-p)2

4sqrt(0.2508/210)=p^2-2p+1

p^2-2p+(1-4sqrt(0.2508/210))=0

plug the values a=1, b=-2, and c=(1-4sqrt(0.2508/210) into the quadratic formula and the roots are approximately 1.431 and 0.5688
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outermeasure
 Post subject: Re: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x
PostPosted: Thu, 31 May 2012 07:28:28 UTC 
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CriticalDendrite wrote:
Here is what I did,

0.2508=10C4(p)^4(1-p)^10-4

0.2508=210(p)^4(1-p)^6

0.2508/210=p^4(1-p)^6

4sqrt(0.2508/210)=1(1-p)2

4sqrt(0.2508/210)=p^2-2p+1

p^2-2p+(1-4sqrt(0.2508/210))=0

plug the values a=1, b=-2, and c=(1-4sqrt(0.2508/210) into the quadratic formula and the roots are approximately 1.431 and 0.5688


How did you get the line in red?

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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CriticalDendrite
 Post subject: Re: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x
PostPosted: Thu, 31 May 2012 23:33:01 UTC 
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That line is supposed to be 4sqrt(0.2508/210)=1(1-p)^2

I took the 4th root of the right side of the equation to get that, but I just looked at it again now and wouldn't that result in p(1-p)^2 rather than 1(1-p)^2 on the right side of the equation?

The reason I did that was because I wasn't sure how to solve the 10th degree polynomial that results from not factoring the right side. Is this wrong?
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CriticalDendrite
 Post subject: Re: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x
PostPosted: Fri, 1 Jun 2012 03:03:28 UTC 
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Thanks for your help, I wanted to try to solve it by factoring by I just ended up putting the 10th degree equation into my graphing calculator to find the zeros. One of which, was approx. 0.4, which is the correct answer.
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outermeasure
 Post subject: Re: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x
PostPosted: Fri, 1 Jun 2012 06:52:38 UTC 
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CriticalDendrite wrote:
That line is supposed to be 4sqrt(0.2508/210)=1(1-p)^2

I took the 4th root of the right side of the equation to get that, but I just looked at it again now and wouldn't that result in p(1-p)^2 rather than 1(1-p)^2 on the right side of the equation?


No.

CriticalDendrite wrote:
Thanks for your help, I wanted to try to solve it by factoring by I just ended up putting the 10th degree equation into my graphing calculator to find the zeros. One of which, was approx. 0.4, which is the correct answer.


There are actually two roots near 0.4

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Mr. Mathturo
 Post subject: Re: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x
PostPosted: Sun, 3 Jun 2012 12:55:58 UTC 
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How much technology are you allowed?

You could also graph the equation (10C4)(p^4)(p^6)-0.2508 =0, and solve for the root (zero).
In that case, I do not get your answer.

Also,
The 4th root of
(x^4) (y^6)
is xy[(y^2)^(1/4)]
or (x)(y^(3/2)).
Thus, I do not agree with your 4th root result.

I also got p=.39999.
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