If all those little (let's call them fubbies) were allowed to stand wherever they wanted to, but only in one of the positions they're occupying right now, what would be the probability that they would get the wave perfect? Assume that all the fubbies have their own timing at which they'll put their hand up, unaffected by their position. Also assume that there are only \(7\) of them, and ignore the ones behind. They make it too hard.
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Omkar Kulkarni
·
2 years, 1 month ago

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@Omkar Kulkarni
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Only in the current position (as shown in gif) the wave is perfect? All other arrangements, its not?
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Arpan Banerjee
·
2 years, 1 month ago

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TopNewestGiven that \(H\) and \(W\) are real numbers and

\[\large{\theta=\pi\left(10^{\huge{\frac{\sin^{-1}\left(\log\left(\int_{0}^{\infty}x^{\left(W\cdot0 \cdot 0 \cdot H \cdot 0 \cdot 0\right)!}e^{-x}dx\right)\right)}{\cos\pi+i\sin\pi}+1}}\right)}\]

\[+1+2+3+4+6+7+8+9+10+11+12+13,\]

find the value of \(\tan\theta^{\circ}\). – Victor Loh · 2 years, 1 month ago

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– Megh Choksi · 2 years, 1 month ago

\( - \cot 1^{\circ}\)Log in to reply

@Llewellyn Sterling – Victor Loh · 2 years, 1 month ago

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– Llewellyn Sterling · 2 years, 1 month ago

Just...wow.Log in to reply

If all those little (let's call them fubbies) were allowed to stand wherever they wanted to, but only in one of the positions they're occupying right now, what would be the probability that they would get the wave perfect? Assume that all the fubbies have their own timing at which they'll put their hand up, unaffected by their position. Also assume that there are only \(7\) of them, and ignore the ones behind. They make it too hard. – Omkar Kulkarni · 2 years, 1 month ago

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– Arpan Banerjee · 2 years, 1 month ago

Only in the current position (as shown in gif) the wave is perfect? All other arrangements, its not?Log in to reply

– Omkar Kulkarni · 2 years, 1 month ago

Yup.Log in to reply

clapping– Llewellyn Sterling · 2 years, 1 month agoLog in to reply

– Omkar Kulkarni · 2 years, 1 month ago

Hard, eh? :PLog in to reply

– Llewellyn Sterling · 2 years, 1 month ago

I understand it, and still trying to answer that question. xDLog in to reply