document.write( "Question 949017: If A and B are independent events, P(A) = 0.35, and P(B) = 0.35, find the probabilities below. (Enter your answers to four decimal places.)
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Algebra.Com's Answer #579630 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
If A and B are independent events, P(A) = 0.35, and P(B) = 0.35, find the probabilities below. (Enter your answers to four decimal places.)
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document.write( "Since they are independent events, by definition of independent\r\n" );
document.write( "events, P(A ᑎ B) = P(A)×P(B) = 0.35×0.35 = 0.1225 \r\n" );
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document.write( "P(A ᑌ B) = P(A) + P(B) - P(A ᑎ B) = 0.35 + 0.35 - 0.1225 = 0.5775\r\n" );
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document.write( "Since they are independent, knowing that B has occurred does not\r\n" );
document.write( "affect the probability of A, so it's the same as P(A) = 0.35.\r\n" );
document.write( "But you can do it by the formula for conditional probability, and\r\n" );
document.write( "get the same thing:\r\n" );
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document.write( "P(A | B) = P(A ᑎ B)/P(B) = 0.1225/0.35 = 0.35 \r\n" );
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document.write( "(d) P(Ac ᑌ Bc) \r\n" );
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document.write( "P(Ac) = 1-P(A) = 1-0.35 = 0.65\r\n" );
document.write( "P(Bc) = 1-P(B) = 1-0.35 = 0.65\r\n" );
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document.write( "If two events are independent, so are their complements*, so\r\n" );
document.write( "P(Ac ᑎ Bc) = P(Ac)×P(Bc) = (0.65)(0.65) = 0.4225\r\n" );
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document.write( "P(Ac ᑌ Bc) = P(Ac) + P(Bc) - P(Ac ᑎ Bc) = 0.65 + 0.65 - 0.4225 = 0.8775.\r\n" );
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document.write( "*If you need to prove that,\r\n" );
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document.write( "P(Ac ᑎ Bc)              <-- use DeMorgan's law:\r\n" );
document.write( "= P[(A ᑌ B)c] \r\n" );
document.write( "= 1 - P(A ᑌ B) \r\n" );
document.write( "= 1 - [P(A) + P(B) - P(A ᑎ B)] \r\n" );
document.write( "= 1 - [P(A) + P(B) - P(A)×P(B)]\r\n" );
document.write( "= 1 - P(A) - P(B) + P(A)×P(B)    <--factor -P(B) out of last two terms \r\n" );
document.write( "= 1 - P(A) - P(B)×[1 - P(A)] \r\n" );
document.write( "= P(Ac) - P(B)×P(Ac)             <--factor P(Ac) out of both terms \r\n" );
document.write( "= P(Ac)×[1-P(B)] \r\n" );
document.write( "= P(Ac)×P(Bc) \r\n" );
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document.write( "Edwin
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