Question 1156279
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<pre>

The probability to get the sum 6 tossing a pair of dice is  {{{5/36}}}.

    // Simply because there are 5 favorable pairs (1,5), (2,4) (3,3) (4,2) (5,1) of 36 possible outcomes.



Hence, the probability DO NOT get the sum 6 tossing a pair of dice is  the complement  {{{31/36}}}.



Now, the problem asks to find the probability that the first three tosses of the pair of dices 

will do not get a 6, while the 4-th toss will get 6.


This probability is  P = {{{(31/36)^3*(5/36)}}} = {{{(31^3*5)/36^4}}} = 0.088684 (approximately).      <U>ANSWER</U>
</pre>

Solved.


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On tossing a pair of dices, &nbsp;see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Rolling-a-pair-of-fair-dice.lesson>Rolling a pair of fair dice</A> 

in this site. &nbsp;You will find there many other similar solved problems.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic &nbsp;"<U>Solved problems on Probability</U>". 



Save the link to this textbook together with its description


Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson


into your archive and use when it is needed.