Question 1182555
.
Find the probability that a single toss of a die will result in a number less than 4 
if it is given that the toss resulted in an odd number.
~~~~~~~~~~~~~~



            It is the  CONDITIONAL  probability problem.


            There are two major ways to solve it,  and I will show you  BOTH  ways.



<pre>
The full space of events consists of 6 events getting "1", '2", "3", "4", "5" and "6".

The probability for each event is  {{{1/6}}}.


The problem asks about the conditional probability P(getting less than 4 given that the result is an odd number).


So, it is  P(getting less than 4 | the result is an odd number), and by the definition of conditional probability, it is the ratio  


      P = P(getting less than 4 AND odd number) / P(the result is an odd number).      (1)



The numerator is  {{{2/6}}} = {{{1/3}}},  because there are only 2 positive odd integers less than 4 (they are 1 and 3)
of 6 possible outputs.


The denominator is  {{{3/6}}},  because there are 3 positive odd integer between 1 and 6 inclusive (they are 1, 3 and 5).


Thus the probability under the problem's question is  P = {{{((1/3))/((1/2))}}} = {{{2/3}}}.    <U>ANSWER</U>
</pre>


So, &nbsp;the first solution is completed.


The second solution uses the &nbsp;"REDUCED" &nbsp;space of events.



<pre>
The reduced space of events consists of 3 events getting odd integers between 1 and 6: these events are "1", "3" and "5".


Of them, favorable are only two, "1" and "3".


Therefore, the sough probability is  P = {{{favorable/total_in_reduced_space}}} = {{{2/3}}},  giving the same answer.
</pre>

Thus the second solution is completed, &nbsp;too.



At this point, &nbsp;I stop my teaching.