document.write( "Question 1195025: Suppose that two teams, A and B, play each other five times over a season (not best of five series, but they play all five games). If Team A has a .6 probability of winning any given game complete the probability distribution below (2 points):\r
\n" );
document.write( "\n" );
document.write( "XA P(XA)\r
\n" );
document.write( "\n" );
document.write( "0 .01\r
\n" );
document.write( "\n" );
document.write( "1 .08\r
\n" );
document.write( "\n" );
document.write( "2\r
\n" );
document.write( "\n" );
document.write( "3\r
\n" );
document.write( "\n" );
document.write( "4\r
\n" );
document.write( "\n" );
document.write( "5\r
\n" );
document.write( "\n" );
document.write( "Total 1.0\r
\n" );
document.write( "\n" );
document.write( "What is the probability that Team A sweeps the season series (wins all five matchups)?\r
\n" );
document.write( "\n" );
document.write( "What is the probability that Team A wins at least three games?\r
\n" );
document.write( "\n" );
document.write( "What is the probability that Team A gets swept (does not win a game)? \n" );
document.write( "
Algebra.Com's Answer #827359 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! winning none is 0.4^5=0.01 given \n" ); document.write( "winning one is 0.4^4*0.6*5 ways=0.08 rounded \n" ); document.write( "winning two is 5C2*0.4^3*0.6^2=0.23 \n" ); document.write( "winning three is 5C3 *0.4^2*0.6^3=0.346 \n" ); document.write( "winning four is 5C4*0.4*0.6^4=0.26 \n" ); document.write( "winning all is 0.6^5=0.08 \n" ); document.write( "adds up to 1.006 because of rounding. \n" ); document.write( "- \n" ); document.write( "wins at least 3 is 0.686 or 0.69 \n" ); document.write( "gets swept is 0.01, given above \n" ); document.write( " |