document.write( "Question 824318: Using the values, X = 0, 1, 2, 4, 6 and respectively P(X) = .1, .2, ?, .3, .3 find the probability that X is greater than 1.
\n" ); document.write( "
\n" ); document.write( " A) .6
\n" ); document.write( " B) .1
\n" ); document.write( " C) .7
\n" ); document.write( " D) .3
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #496298 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "X    P(X)\r\n" );
document.write( "--------- \r\n" );
document.write( "0    .1\r\n" );
document.write( "1    .2 \r\n" );
document.write( "2     ?\r\n" );
document.write( "4    .3\r\n" );
document.write( "6    .3\r\n" );
document.write( "\r\n" );
document.write( "For this to be a probability distribution, the sum\r\n" );
document.write( "of the probabilities must = 1.  So since the probability\r\n" );
document.write( "of the other four values of X have sum\r\n" );
document.write( ".1+.2+.3+.3 = .9, then P(2) = 1.0-.9 = .1 and the\r\n" );
document.write( "distribution is:\r\n" );
document.write( "\r\n" );
document.write( "X    P(X)\r\n" );
document.write( "--------- \r\n" );
document.write( "0    .1\r\n" );
document.write( "1    .2 \r\n" );
document.write( "2    .1\r\n" );
document.write( "4    .3\r\n" );
document.write( "6    .3\r\n" );
document.write( "--------\r\n" );
document.write( "Sum = 1\r\n" );
document.write( "\r\n" );
document.write( "So P(X > 1) = P(X=2)+P(X=4)+P(X=6) = .1+.3+.3 = .7,\r\n" );
document.write( "\r\n" );
document.write( "Or what would have been the same, \r\n" );
document.write( "\r\n" );
document.write( "P(X > 1) = 1 - P(X < 2) = 1-P(X=0)-P(X=1) = 1-.1-.2 = .3\r\n" );
document.write( "\r\n" );
document.write( "Edwin

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