91Ó°ÊÓ

The die is the balanced.

6 appeared exactly three times.

The probability of obtaining 6 in any draw is\({\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}}\)

And, not obtaining 6 is,

\(1 - {\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} = {\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}}\)

Here we thrown the die 10 times, where we need to calculate probability of getting 6 at first 3 thrown and remaining 7 thrown we will get the number excepts 6.

Then the probability of getting 6 at first 3 draws is,

\(\begin{aligned}{c}{\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} \times {\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} \times {\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} \times {\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} \times {\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} \times {\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} \times {\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} \times {\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} \times {\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} \times {\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 6}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$6$}} = {\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 {{6^3}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${{6^3}}$}} \times {\raise0.7ex\hbox{${{5^7}}$} \!\mathord{\left/

{\vphantom {{{5^7}} {{6^7}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${{6^7}}$}}\\ = 0.001292\end{aligned}\)

Therefore, the required probability is 0.001292