document.write( "Question 1042551: A and B are two independent events. The probability that both occur simultaneously is 1/6 and the probability that neither occurs is 1/3, then P(A) + P(B) = *?
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Algebra.Com's Answer #657525 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
A and B are two independent events.
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document.write( "Therefore the probability that they both occur simultaneously\r\n" );
document.write( "is the product of their respective probabilities.\r\n" );
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document.write( "P(A and B) = P(A)*P(B)\r\n" );
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\n" ); document.write( "The probability that both occur simultaneously is 1/6
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document.write( "P(A)*P(B) = 1/6\r\n" );
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\n" ); document.write( "and the probability that neither occurs is 1/3,
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document.write( "Since A and B are independent, then so are A' and B',\r\n" );
document.write( "the events that they do not occur.  \r\n" );
document.write( "P(A') = 1-P(A)\r\n" );
document.write( "P(B') = 1-P(B)\r\n" );
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document.write( "P(neither occurs) = P(A')P(B') = [1-P(A)}{1-P(B)] = 1/3\r\n" );
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document.write( "Let P(A) = x\r\n" );
document.write( "Let P(B) = y\r\n" );
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document.write( "Then P(A)*P(B) = 1/6 becomes\r\n" );
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document.write( "\"xy+=+1%2F6\"\r\n" );
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document.write( "and  [1-P(A)}{1-P(B)] = 1/3 becomes\r\n" );
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document.write( "\"%281-x%29%281-y%29=1%2F3\"\r\n" );
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document.write( "So we have the system of equations:\r\n" );
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document.write( "\"system%28xy=1%2F6%2C%281-x%29%281-y%29=1%2F3%29\"\r\n" );
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document.write( "Solve the 1st equation for y\r\n" );
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document.write( "\"xy=1%2F6\"\r\n" );
document.write( "\"6xy=1\"\r\n" );
document.write( "\"y=1%2F%286x%29\"\r\n" );
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document.write( "Simplify the 2nd equation:\r\n" );
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document.write( "\"%281-x%29%281-y%29=1%2F3%29\"\r\n" );
document.write( "\"1-y-x%2Bxy=1%2F3\"\r\n" );
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document.write( "Substitute 1/6 for xy\r\n" );
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document.write( "\"1-y-x%2B1%2F6=1%2F3\"\r\n" );
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document.write( "Multiply through by 6\r\n" );
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document.write( "\"6-6y-6x%2B1=2\"\r\n" );
document.write( "\"7-6y-6x=2\"\r\n" );
document.write( "\"-6y-6x=-5\"\r\n" );
document.write( "\"6y%2B6x=5\"\r\n" );
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document.write( "Substitute \"1%2F%286x%29\" for y\r\n" );
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document.write( "\"6%281%2F%286x%29%29%2B6x=5\"\r\n" );
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document.write( "\"1%2Fx%2B6x=5\"\r\n" );
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document.write( "Multiply through by x\r\n" );
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document.write( "\"1%2B6x%5E2=5x\"\r\n" );
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document.write( "\"6x%5E2-5x%2B1=0\"\r\n" );
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document.write( "\"%282x-1%29%283x-1%29=0\"\r\n" );
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document.write( "2x-1 = 0;  3x-1 = 0\r\n" );
document.write( "  2x = 1;    3x = 1\r\n" );
document.write( "   x = 1/2;   x = 1/3\r\n" );
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document.write( "If x = 1/2\r\n" );
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document.write( "\"6y%2B6%281%2F2%29=5\"\r\n" );
document.write( "\"6y%2B3=5\"\r\n" );
document.write( "\"6y=2\"\r\n" );
document.write( "\"y=2%2F6\"\r\n" );
document.write( "\"y=1%2F3\"\r\n" );
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document.write( "and if x = 1/3\r\n" );
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document.write( "\"6y%2B6%281%2F3%29=5\"\r\n" );
document.write( "\"6y%2B2=5\"\r\n" );
document.write( "\"6y=3\"\r\n" );
document.write( "\"y=3%2F6\"\r\n" );
document.write( "\"y=1%2F2\"\r\n" );
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document.write( "So the probabilities are 1/2 and 1/3,\r\n" );
document.write( "regardless of which is P(A) and which \r\n" );
document.write( "is P(B).\r\n" );
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document.write( "So P(A)+P(B) = 1/2+1/3 = 3/6+2/6 = 5/6\r\n" );
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document.write( "Edwin
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