Question 1153820
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<pre>

Recall that the outcome of rolling a die is any of 6 integer numbers from 1 to 6 inclusive, 

with the uniform probability of {{{1/6}}} for each outcome.


Therefore,  P(getting 3 or 4 on the first toss) = {{{2/6}}} = {{{1/3}}}

and         P(getting 1 or 2 on the second toss) = {{{1/3}}},  again.


The events/outcomes are INDEPENDENT on the first and the second tosses;

therefore, the final probability under the question is the   {{{(1/3)*(1/3)}}} = {{{1/9}}}.    <U>ANSWER</U>
</pre>

Solved.


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If you want to learn this subject and this class of problems &nbsp;BETTER, &nbsp;DEEPER &nbsp;and &nbsp;WIDER, &nbsp;look into 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.