document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1194140>Question 1194140</A>: <I><SPAN STYLE=\"word-break: break-all;\"> X and Y are independent, X ~ P(λ1), Y ~ P(λ2). Given probabilities P(X1 > 0) = 0.58, P(X2 > 0) = 0.43. Find the probability of P(X+Y=1) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848530 by </B><A HREF=/tutors/aboutme.mpl?userid=parmen><B>parmen(42)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=parmen><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=848530\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> **1. Determine λ1 and λ2:**\r<BR>\n" );
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document.write( "* We know that for a Poisson distribution, P(X > 0) = 1 - P(X = 0) <BR>\n" );
document.write( "* Since X ~ P(λ1), P(X = 0) = e^(-λ1)<BR>\n" );
document.write( "* Therefore, P(X > 0) = 1 - e^(-λ1) <BR>\n" );
document.write( "* Given P(X > 0) = 0.58, we can solve for λ1:<BR>\n" );
document.write( "   0.58 = 1 - e^(-λ1) <BR>\n" );
document.write( "   e^(-λ1) = 0.42 <BR>\n" );
document.write( "   λ1 = -ln(0.42) ≈ 0.8675\r<BR>\n" );
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document.write( "* Similarly, for Y ~ P(λ2):<BR>\n" );
document.write( "   P(Y > 0) = 1 - e^(-λ2) <BR>\n" );
document.write( "   Given P(Y > 0) = 0.43:<BR>\n" );
document.write( "   0.43 = 1 - e^(-λ2) <BR>\n" );
document.write( "   e^(-λ2) = 0.57 <BR>\n" );
document.write( "   λ2 = -ln(0.57) ≈ 0.5621\r<BR>\n" );
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document.write( "**2. Calculate P(X = 0) and P(Y = 0):**\r<BR>\n" );
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document.write( "* P(X = 0) = e^(-λ1) = e^(-0.8675) ≈ 0.42<BR>\n" );
document.write( "* P(Y = 0) = e^(-λ2) = e^(-0.5621) ≈ 0.57\r<BR>\n" );
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document.write( "**3. Calculate P(X = 1) and P(Y = 1):**\r<BR>\n" );
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document.write( "* For a Poisson distribution, P(X = k) = (e^(-λ) * λ^k) / k!<BR>\n" );
document.write( "* P(X = 1) = (e^(-λ1) * λ1^1) / 1! = e^(-0.8675) * 0.8675 ≈ 0.3698<BR>\n" );
document.write( "* P(Y = 1) = (e^(-λ2) * λ2^1) / 1! = e^(-0.5621) * 0.5621 ≈ 0.3155\r<BR>\n" );
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document.write( "**4. Calculate P(X + Y = 1):**\r<BR>\n" );
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document.write( "* P(X + Y = 1) can occur in two ways:<BR>\n" );
document.write( "    * X = 1 and Y = 0<BR>\n" );
document.write( "    * X = 0 and Y = 1<BR>\n" );
document.write( "* Since X and Y are independent:<BR>\n" );
document.write( "   P(X + Y = 1) = P(X = 1) * P(Y = 0) + P(X = 0) * P(Y = 1)<BR>\n" );
document.write( "   P(X + Y = 1) = 0.3698 * 0.57 + 0.42 * 0.3155 ≈ 0.3423\r<BR>\n" );
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document.write( "**Therefore, the probability of P(X + Y = 1) is approximately 0.3423.**<BR>\n" );
document.write( " </SPAN>\n" );
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