Question 1161036


yes, it is B.

For the first die, the probability that the number is less than {{{3}}} is {{{2/6}}} or {{{1/3}}} because there are only {{{2}}} numbers ({{{1}}} and {{{2}}}) which are less than {{{3}}}. 

The second die needs to land on a number greater than {{{3}}} (and it does not include {{{3}}}). This means that there are {{{3}}} numbers that can meet the requirements, {{{4}}}, {{{5}}}, and {{{6}}}, which makes the chances {{{3/6}}} or {{{1/2}}}.

Multiply the two fractions together:

{{{(1/3 )(1/2 )= 1/6=0.16666666666666666=0.167}}}