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Prove that\begin{array}{1 1}(i)\;P(A)+P(A\cap B)+P(A\cap \bar{B})\\(ii)\;P(A\cup B)=P(A\cap B)+P(A\cap\bar{B})+P(\bar{A}\cap\bar{B})\end{array}

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(p(A(\cap\)B )+\(p(A\cap\)\(\bar{B}\))\(Two\; intersefing \;set\;A\; and\;B\; can \;be \;split\; in\; to\;3\; disjoint\; part\;\) B\(\cap\)\(\bar{A}\).A\(\cap\)B,A\(\cap\)\(\bar{B}\)\(since\; they\; are\; disjoints\; the\; events\; associated \;with \;each\; set \;are\; mutually\; exclulive\;If\; two\; sets\; X\;and\;Y \;are\; mutully\; exclusive\)

From the diagram we can see
p(A)=\(p(A(\cap\)B )+\(p(A\cap\)\(\bar{B}\))


answered Feb 26, 2013 by poojasapani_1
edited Mar 16, 2013 by poojasapani_1

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