A probability problem
Ten distinct letters are selected at random from the alphabet $\left \{ A,B,C.......Z \right \}$.Find the probability that the selection does not include two consecutive letters.
"Questions we can't answer are far better than answers we can't question"
Re: A probability problem
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
10টি ভিন্ন অক্ষর নেওয়া যেতে পারে $\binom{26}{10}$ভাবে।
$\frac{\binom{26}{10} - (\binom{24}{8} + \binom{23}{8} + \binom{22}{8} + ... + \binom{8}{8})}{\binom{26}{10}}=\frac{8}{13}$
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10টি ভিন্ন অক্ষর নেওয়া যেতে পারে $\binom{26}{10}$ভাবে।
$\frac{\binom{26}{10} - (\binom{24}{8} + \binom{23}{8} + \binom{22}{8} + ... + \binom{8}{8})}{\binom{26}{10}}=\frac{8}{13}$
সমাধান ভুল হতে পারে।
Re: A probability problem
yeah,your solution is wrong.Try more.The answer is $\frac{\binom{17}{10}}{\binom{26}{10}}$
"Questions we can't answer are far better than answers we can't question"
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Golam Musabbir Joy
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Re: A probability problem
Is that 1/2 * c(13,2)/c(26,2)????
Re: A probability problem
I see my mistake.
$\frac{11 + 3(11\times10) + 11\times10\times9 + 3(\frac{11\times10\times9}{2!}) + \frac{11\times10\times9\times8}{2!} + 2(\frac{11\times10\times9\times8}{3!}) + \frac{11\times10\times9\times8\times7}{2!\,3!} + \frac{11\times10\times9\times8\times7}{4!} + \frac{11\times10\times9\times8\times7\times6}{5!} + \frac{11\times10\times9\times8\times7\times6\times5}{7!}}{\binom{26}{10}}$
$=\frac{8}{2185}$
$\frac{11 + 3(11\times10) + 11\times10\times9 + 3(\frac{11\times10\times9}{2!}) + \frac{11\times10\times9\times8}{2!} + 2(\frac{11\times10\times9\times8}{3!}) + \frac{11\times10\times9\times8\times7}{2!\,3!} + \frac{11\times10\times9\times8\times7}{4!} + \frac{11\times10\times9\times8\times7\times6}{5!} + \frac{11\times10\times9\times8\times7\times6\times5}{7!}}{\binom{26}{10}}$
$=\frac{8}{2185}$