document.write( "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
\n" ); document.write( "number \n" ); document.write( "

Algebra.Com's Answer #812636 by ikleyn(52943) $\"\"$ $\"About$

You can put this solution on YOUR website!
.
\n" ); document.write( "Find the probability that a single toss of a die will result in a number less than 4
\n" ); document.write( "if it is given that the toss resulted in an odd number.
\n" ); document.write( "~~~~~~~~~~~~~~\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " It is the CONDITIONAL probability problem.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " There are two major ways to solve it, and I will show you BOTH ways.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "

\r\n" );
document.write( "The full space of events consists of 6 events getting \"1\", '2\", \"3\", \"4\", \"5\" and \"6\".\r\n" );
document.write( "\r\n" );
document.write( "The probability for each event is  .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The problem asks about the conditional probability P(getting less than 4 given that the result is an odd number).\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "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  \r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "      P = P(getting less than 4 AND odd number) / P(the result is an odd number).      (1)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The numerator is   = ,  because there are only 2 positive odd integers less than 4 (they are 1 and 3)\r\n" );
document.write( "of 6 possible outputs.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The denominator is  ,  because there are 3 positive odd integer between 1 and 6 inclusive (they are 1, 3 and 5).\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Thus the probability under the problem's question is  P =  = .    ANSWER\r\n" );
document.write( "

\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So, the first solution is completed.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The second solution uses the \"REDUCED\" space of events.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "

\r\n" );
document.write( "The reduced space of events consists of 3 events getting odd integers between 1 and 6: these events are \"1\", \"3\" and \"5\".\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Of them, favorable are only two, \"1\" and \"3\".\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Therefore, the sough probability is  P =  = ,  giving the same answer.\r\n" );
document.write( "

\r
\n" ); document.write( "\n" ); document.write( "Thus the second solution is completed, too.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "At this point, I stop my teaching.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );