GingerAle

You should know, Mr. BB, that using “brovos” for a woman is often considered mockery. The correct form is “brave” and “brava” is the singular form.

Happy New Year, JB. My biggest fan. 

I originally wrote that post with three scenes. I deleted the funniest scene because the Shanghai’d Post Production Supervisor or Shanghai’d Technical Consultant would have factored it to invisibility.  

I’ll send you a copy of the deleted scene.

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This post is one in a set that I’ve periodically monitored for view counts.    

There is more to this “rude dude” than bad manners. 

The solution(s) presented in the prior posts have one or more errors; most notably an incorrect value for (p) –the probability of a single (successful) event

To solve: Find (p), the probability of a single event.

Statement:  If I shoot 5 pieces of paper at the recycling bin, at least one of them will make it inside the recycling bin with probability \(\frac {211}{243}\)

Equivalent statistical statement:

\(\Pr(X\geq 1)= 1- \left(\sum_ \limits {i=0}^{1} \binom{5}{i} p^{i}(1-p)^{5-i}\right)=\frac {211}{243}\)

\(\text {Equivalent statement: $Pr(X\geq 1)= \Pr(X > 0) = (\frac {211}{243})$  |  Binomial events are discreet integers. }\)

\(\text {Complementary statement $Pr(X < 1) = \Pr(X = 0) =  \frac {32}{243}$ | Probability of zero (0) successes in five attempts. } \)

\(\Pr(X = 0) =  \binom{5}{0} p^{5}(1-p)^{0}=\frac {32}{243} \)

\(\rightarrow p^5=\frac {32}{243}  \text { |  Solve complementary statement for (p)} \)

\( p = \sqrt [\leftroot{-1}\uproot{1}5]{\frac {32}{243}}\rightarrow p = \frac {2}{3} \text { |  Probability of failure on a single throw}\\ \)

\(1-\frac {2}{3} =\frac {1}{3}  \text { |  Complement of failure = success }\\ \)

\(\text {Use probability of success on a single throw  $(\frac {1}{3})$  to solve:} \\ \) 

If I shoot 6 pieces of paper at the recycling bin, what's the probability at least two of them make it inside the recycling bin

Equivalent statistical statement:

\(\Pr(X\geq 2)= 1- \left(\sum_ \limits {i=0}^{1} \binom{6}{i}(\frac {1}{3})^{i}(1-\frac {1}{3})^{6-i}\right)=64.88\% \)

\({}\)

GA

Solution:

One method for visualizing a solution is to view the cards in a 4 X 4 matrix, where the columns represent the players and the rows represent the cards in the players' hands. Valid conditions for success occur when the players have single “A” cards located in any one of the four positions of their respective hands.

\(\left[ {\begin{array}{cccc} A & A & A & A \\ X & X & X & X \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {First arrangement of “A} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{cccc} A & A & A & X \\ X & X & X & A \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{cccc} A & A & X & A \\ X & X & A & X \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {Fifth arrangement} \\ \left[ {\begin{array}{cccc} X & X & X & X \\ X & X & X & X \\ X & X & X & X \\ A & A & A & A \\ \end{array} } \right] \text {Last arrangement} \)

From this, it’s easy to see there are (4^4) = 256 arrangements of success.

Dividing the number of successes by the total number of combinations

(4⁴) / (nCr(16,4)), gives (64/445) ≈ 0.1406593.

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Here are two other solution methods for this question:

Solution:

This is a voltage divider circuit. The switch places a third resistor in parallel with (R1) or (R2)

\(\text {One method of analysis for this circuit is to calculate the resistance values needed to divide 10 volts into a 6v and 4v proportion. }\\ \text {The Voltage Divider Formula resolves these values. }\\ R2_{vd} = Vs * \frac {R2}{(R1+R2)} | \text {where vd is the voltage drop and Vs is the source voltage.} \\ R1_{vd} = Vs * \frac{R1}{(R1+R2)} \\ \text {With a 10 volt source and a 6v drop on R2 and R2 set to 5k ohms, }\\ \text {Then } 6_{vd} = 10_{vs} * \frac {5000}{(R1 +5000) }\\ 6 * (R1 + 5000) =50000 \; \;| \text { divide out voltage and multiply by resistance.}\\ 6*R1 = 20000 \ \rightarrow R1= 3333.33 \;\; | \text { in ohms.}\\ \text {If (R2) is 5k$\Omega$ then (R1) is also 5k$\Omega$ because (R1 + R2) = 10k$\Omega$. }\\ \text {The switch connects a parallel resistor (R3) to (R1) lowering the resistance to 3333.33 ohms. }\\ \)

\(\\ \text {Use the Parallel Resistance Formula to find the needed resistance of R3.}\\ Rp = \frac {(R1 + R3)} {(R1 * R3)} \;\; |\text {Parallel resistance }\\ R3 = \frac { (-Rp * R1)}{(Rp - R1)} \;\;\;\;\; |\text {Solve for R3 }\\ R3 = \frac {(-3333.33 * 5000)}{(3333.33 - 5000)}\\ R3 = \frac {(-16666650)} {(-1666.67)} \rightarrow R3 = 10000 \Omega \\ \text {summary}\\ R1 = 5k \Omega \\ R2 = 5k \Omega \\ R3 = 10k \Omega \\ \)

GA

I’m probably the only person on this forum that has a clue about what you are referencing. 

The reason I have a clue is I’m a genetically enhanced chimp with advanced mind reading skills.

This skill is a great advantage for interpreting babble and bullshit questions that lack a contextual preface.

You can use a variable to store the constant of the Golden ratio. I’ve use the Golden ratio and many other constants in the forum’s calculator –sometimes several constants simultaneously. They are easy to change on the fly, too.  I’ve posted at least three examples. You can peruse through my profile and find them.

Oh, and those “physic thingies that don't make any sense,” are very useful when needed. There are over 300 of them in the constants list. Your values have to be limited because if there were 30,000 of them, then it would take you a long time to find the one you need –you may as well just type it in. 

There is one constant that should be on the list: The constant for Quantum Dumbness at a distance. It works well for dumbness infectious via the net. 

GA

JESUS CHRIST!! Are you having a bad day or are you just fucking stupid?

Read my goddamn post! I clearly answer your question! The answer is in boldface.

I have questions for you:

How did you ever figure out how to post on this forum? 

How is it that you are still alive?

Darwin’s natural selection takes too long!