Question

Suppose that in 10 rolls of a balanced die, the number 6 appeared exactly three times. What is the probability that the first three rolls each yielded the number 6?

Short answer

The probability that the first three rolls each yields the number 6 is 0.001292

Step-by-step solution

Given information

The die is the balanced.

6 appeared exactly three times.

Calculation for the probability value

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