document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1192723>Question 1192723</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Records show that the probability of seeing a hawk migrating on a day in <BR>\n" );
document.write( "September is about 35%. What is the minimum number of days a person must<BR>\n" );
document.write( "watch to be at least 96.8% sure of seeing one or more hawks migrating?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848953 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=848953\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To solve this, we'll use the concept of the complement of the binomial distribution.\r<BR>\n" );
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document.write( "* **Probability of not seeing a hawk on a single day:** 1 - 0.35 = 0.65\r<BR>\n" );
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document.write( "* **Probability of not seeing a hawk on 'n' consecutive days:** 0.65^n\r<BR>\n" );
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document.write( "* **Probability of seeing at least one hawk in 'n' days:** 1 - 0.65^n\r<BR>\n" );
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document.write( "We want to find the minimum 'n' such that:\r<BR>\n" );
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document.write( "1 - 0.65^n >= 0.968\r<BR>\n" );
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document.write( "0.65^n <= 0.032\r<BR>\n" );
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document.write( "Taking the natural logarithm of both sides:\r<BR>\n" );
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document.write( "n * ln(0.65) <= ln(0.032)\r<BR>\n" );
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document.write( "n >= ln(0.032) / ln(0.65) \r<BR>\n" );
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document.write( "n >= 7.85\r<BR>\n" );
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document.write( "Since 'n' must be an integer, the minimum number of days a person must watch is **8 days**.<BR>\n" );
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