facing a problem... Need solution.
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A bacteria has 8 flazela. 3 of them are green, 2 of them are blue, 2 of them are white & the last one is pink. What is the probability of blue flazela not getting placed beside the white ones.......
- nafistiham
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Re: facing a problem... Need solution.
Well, it seems like a 'pearls in the necklace' type problem. But, bacteria is not a 2d object that has flagella just in a circle. it has them all around it. so, how can you define being 'placed beside white ones' ?Shahriar tanvir wrote:A bacteria has 8 flazela. 3 of them are green, 2 of them are blue, 2 of them are white & the last one is pink. What is the probability of blue flazela not getting placed beside the white ones.......
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: facing a problem... Need solution.
I think in this case,we have to consider a $2D$ bacteria,or else a bacteria drawn on a sheet of paper.nafistiham wrote:Well, it seems like a 'pearls in the necklace' type problem. But, bacteria is not a 2d object that has flagella just in a circle. it has them all around it. so, how can you define being 'placed beside white ones' ?Shahriar tanvir wrote:A bacteria has 8 flazela. 3 of them are green, 2 of them are blue, 2 of them are white & the last one is pink. What is the probability of blue flazela not getting placed beside the white ones.......
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
- nafistiham
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Re: facing a problem... Need solution.
well, so I am describing my solution defining ' blue flazela not getting placed beside the white ones'
as, if any blue and white flagella is staying alongside, we discount it
suppose there is no condition, so, number of all the available permutation is
\[\frac{7!}{3!\cdot 2!\cdot2! }\]
and we can discount
\[\frac{2!\cdot 6!}{3!}\]
so the probability will be
\[\frac{\frac{7!}{3!\cdot 2!\cdot2! }-\frac{2!\cdot 6!}{3!}}{\frac{7!}{3!\cdot 2!\cdot2! }}\]
as, if any blue and white flagella is staying alongside, we discount it
suppose there is no condition, so, number of all the available permutation is
\[\frac{7!}{3!\cdot 2!\cdot2! }\]
and we can discount
\[\frac{2!\cdot 6!}{3!}\]
so the probability will be
\[\frac{\frac{7!}{3!\cdot 2!\cdot2! }-\frac{2!\cdot 6!}{3!}}{\frac{7!}{3!\cdot 2!\cdot2! }}\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.