document.write( "Question 1143539: Suppose that two dice are tossed. For each die, it is equally likely that 1, 2, 3, 4, 5, or 6 dots will turn up. Let S be the sum of the two dice.
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Algebra.Com's Answer #764362 by ikleyn(52787) $\"\"$ $\"About$

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document.write( "(a)  We should check if  P (E1 & E2) = P(E1)*P(E2).\r\n" );
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document.write( "     Calculation P(E1) is easy:  P(E1) = ,  since there are 6 outcomes for the first die, \r\n" );
document.write( "     showing the numbers 1, 2, 3, 4, 5 and 6 with equal probabilities.\r\n" );
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document.write( "     To calculate P(E2), notice that the sum of 6 of two dice can be obtained in 5 (five)  cases\r\n" );
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document.write( "         (1,5), (2,4), (3,3), (4,2) and (5,1),\r\n" );
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document.write( "     So the event E2 has the probability  P(E2) = .\r\n" );
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document.write( "     Now, the event (E1 & E2)  is the intersection of E1 and E2; \r\n" );
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document.write( "     in other words, first die shows a \"3\", while the sum of the faces of the two dice is  6.\r\n" );
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document.write( "     It may happen if and only if BOTH dice face up a \"3\" -- the event with the probability  ,  obviously.\r\n" );
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document.write( "     Thus we have  P(E1) = ;  P(E2) =   and  P(E1 & E2) = .\r\n" );
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document.write( "     The product of probabilities P(E1)*P(E2) = } =  is not equal to  P(E1 & E2) =  --\r\n" );
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document.write( "     hence, the two events E1 and E2 ARE NOT INDEPENDENT.    ANSWER\r\n" );
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document.write( "(b)  If the first die faces up a \"3\", then the sum is even in these  3 (three)  cases\r\n" );
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document.write( "         (3,1), (3,3), (3,5).\r\n" );
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document.write( "     So the probability for S to be an even number is   = .    ANSWER\r\n" );
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\n" ); document.write( "\n" ); document.write( "If you want to learn this subject wider and deeper, look into my lesson\r
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