document.write( "Question 912044: A loaded coin is tossed 5 times , find the
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Algebra.Com's Answer #553516 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "  P(H)+P(T)=1\r\n" );
document.write( "       P(H)=3P(T)\r\n" );
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document.write( " 3P(T)+P(T)=1\r\n" );
document.write( "      4P(T)=1\r\n" );
document.write( "       P(T)=1/4\r\n" );
document.write( "       P(H)=3P(T)\r\n" );
document.write( "P(H)+3(1/4)=3/4\r\n" );
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document.write( "This is a binomial distribution:\r\n" );
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document.write( "Let x be the number of heads:\r\n" );
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document.write( "x                                     P(x)\r\n" );
document.write( "0    C(5,0)(3/4)^0(1/4)^5 =  1*3^0/4^5 =    1*1/1024 =   1/1024                     \r\n" );
document.write( "1    C(5,1)(3/4)^1(1/4)^4 =  5*3^1/4^5 =    5*3/1024 =  15/1024\r\n" );
document.write( "2    C(5,2)(3/4)^2(1/4)^3 = 10*3^2/4^5 =   10*9/1024 =  90/1024\r\n" );
document.write( "3    C(5,3)(3/4)^3(1/4)^2 = 10*3^3/4^5 =  10*27/1024 = 270/1024\r\n" );
document.write( "4    C(5,4)(3/4)^4(1/4)^1 =  5*3^4/4^5 =   5*81/1024 = 405/1024\r\n" );
document.write( "5    C(5,5)(3/4)^5(1/4)^0 =  1*3^5/4^5 =  1*243/1024 = 243/1024   \r\n" );
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document.write( "mean = n*p = 5*(3/4) = 15/4 = 3.75\r\n" );
document.write( "variance = n*p*(1-p) = 5*(1/4)(3/4) = 15/16 = 0.9375\r\n" );
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document.write( "Edwin

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