document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1194301>Question 1194301</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A biologist studying a hybrid tomato found that there is a probability of 0.70 that the seeds will germinate. If the biologist plants 10 seeds compute the probability that:\r<BR>\n" );
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document.write( "a) at most 7 seeds will germinate<BR>\n" );
document.write( "b) at least 4 seeds will germinate<BR>\n" );
document.write( "c) between 3 and 7 (inclusive, including 3 and 7) will germinate<BR>\n" );
document.write( "d) between 4 and 9 (exclusive) will germinate<BR>\n" );
document.write( "e) less than 5 will germinate<BR>\n" );
document.write( "f) more than 3 will germinate </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #826490 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=826490\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "n = 10 = sample size<BR>\n" );
document.write( "p = 0.70 = probability of germination\r<BR>\n" );
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document.write( "We have a binomial distribution problem for the following reasoning<ul><li>There are two outcomes: Either a seed germinates, or it does not.</li><li>Each trial (seed) is independent of one another. No one seed affects the germination of another.</li><li>The probability of success (ie germination) is the same for each trial, which is p = 0.70</li><li>There are a fixed number of trials (n = 10).</li></ul>We'll use the aptly named binomial probability formula<BR>\n" );
document.write( "P(x) = (n C x)*(p)^x*(1-p)^(n-x)<BR>\n" );
document.write( "the values of n and p were mentioned earlier<BR>\n" );
document.write( "The n C x refers to the nCr formula\r<BR>\n" );
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document.write( "The x values take on items from the set {0,1,2,3,4,5,6,7,8,9,10}<BR>\n" );
document.write( "because we'll define x like so<BR>\n" );
document.write( "x = number of seeds that germinate\r<BR>\n" );
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document.write( "Let's compute P(x) when x = 0<BR>\n" );
document.write( "P(x) = (n C x)*(p)^x*(1-p)^(n-x)<BR>\n" );
document.write( "P(0) = (10 C 0)*(0.70)^0*(1-0.70)^(10-0)<BR>\n" );
document.write( "P(0) = (1)*(0.70)^0*(1-0.70)^(10-0)<BR>\n" );
document.write( "P(0) = 0.0000059049\r<BR>\n" );
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document.write( "Do the same for x = 1<BR>\n" );
document.write( "Keep n = 10 and p = 0.7 the same<BR>\n" );
document.write( "P(x) = (n C x)*(p)^x*(1-p)^(n-x)<BR>\n" );
document.write( "P(1) = (10 C 1)*(0.70)^1*(1-0.70)^(10-1)<BR>\n" );
document.write( "P(1) = (10)*(0.70)^1*(1-0.70)^(10-1)<BR>\n" );
document.write( "P(1) = 0.000137781\r<BR>\n" );
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document.write( "This process is repeated for the other x values from x = 2 all the way up to x = 10.<BR>\n" );
document.write( "Spreadsheet software makes quick work of the computations, and it gives a natural easy way to display it as a table.<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >x</TD><TD >P(x)</TD></tr><tr><TD >0</TD><TD >0.0000059049</TD></tr><tr><TD >1</TD><TD >0.000137781</TD></tr><tr><TD >2</TD><TD >0.0014467005</TD></tr><tr><TD >3</TD><TD >0.009001692</TD></tr><tr><TD >4</TD><TD >0.036756909</TD></tr><tr><TD >5</TD><TD >0.1029193452</TD></tr><tr><TD >6</TD><TD >0.200120949</TD></tr><tr><TD >7</TD><TD >0.266827932</TD></tr><tr><TD >8</TD><TD >0.2334744405</TD></tr><tr><TD >9</TD><TD >0.121060821</TD></tr><tr><TD >10</TD><TD >0.0282475249</TD></tr>\n" );
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document.write( "The previous section is a lot of set up if you aren't quite familiar with the binomial distribution. <BR>\n" );
document.write( "Fortunately, software can be used to quickly generate such a table.\r<BR>\n" );
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document.write( "We'll use that table to answer parts (a) through (f)\r<BR>\n" );
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document.write( "For part (a), the phrasing \"at most\" means \"that is the highest we can go\". It is the ceiling value.<BR>\n" );
document.write( "\"At most 7\" means \"7 is the highest we can go\".\r<BR>\n" );
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document.write( "We're tasked to find P(x ≤ 7)<BR>\n" );
document.write( "This is the same as adding P(0) all the way through to P(7)<BR>\n" );
document.write( "That's 8 numbers we have to add up.\r<BR>\n" );
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document.write( "But we can take a shortcut. Notice that<BR>\n" );
document.write( "P(x ≥ 8) = P(8) + P(9) + P(10)<BR>\n" );
document.write( "P(x ≥ 8) = 0.2334744405 + 0.121060821 + 0.0282475249<BR>\n" );
document.write( "P(x ≥ 8) = 0.3827827864\r<BR>\n" );
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document.write( "Then we can say,<BR>\n" );
document.write( "P(x ≤ 7) + P(x ≥ 8) = 1<BR>\n" );
document.write( "P(x ≤ 7) = 1 - P(x ≥ 8)<BR>\n" );
document.write( "P(x ≤ 7) = 1 - 0.3827827864<BR>\n" );
document.write( "P(x ≤ 7) = 0.6172172136\r<BR>\n" );
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document.write( "Answer: Approximately 0.6172172136<BR>\n" );
document.write( "Round this however you need to\r<BR>\n" );
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document.write( "Part (b)\r<BR>\n" );
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document.write( "\"At least 4\" means \"4 or more\".<BR>\n" );
document.write( "We want to compute P(x ≥ 4)\r<BR>\n" );
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document.write( "Like before we can use a shortcut to find P(x ≤ 3)<BR>\n" );
document.write( "P(x ≤ 3) = P(0) + P(1) + P(2) + P(3)<BR>\n" );
document.write( "P(x ≤ 3) = 0.0000059049 + 0.000137781 + 0.0014467005 + 0.009001692<BR>\n" );
document.write( "P(x ≤ 3) = 0.0105920784\r<BR>\n" );
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document.write( "Then,<BR>\n" );
document.write( "P(x ≤ 3) + P(x ≥ 4) = 1<BR>\n" );
document.write( "P(x ≥ 4) = 1 - P(x ≤ 3)<BR>\n" );
document.write( "P(x ≥ 4) = 1 - 0.0105920784<BR>\n" );
document.write( "P(x ≥ 4) = 0.9894079216<BR>\n" );
document.write( "The longer alternative method would be to add up P(4) all the way up to P(10) which means you have to add up 7 values. \r<BR>\n" );
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document.write( "Answer: Approximately 0.9894079216\r<BR>\n" );
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document.write( "Part (c)\r<BR>\n" );
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document.write( "P(3 ≤ x ≤ 7) = P(3) + P(4) + P(5) + P(6) + P(7)<BR>\n" );
document.write( "P(3 ≤ x ≤ 7) = 0.009001692 + 0.036756909 + 0.1029193452 + 0.200120949 + 0.266827932<BR>\n" );
document.write( "P(3 ≤ x ≤ 7) = 0.6156268272\r<BR>\n" );
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document.write( "Answer: Approximately 0.6156268272\r<BR>\n" );
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document.write( "Part (d)\r<BR>\n" );
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document.write( "P(4 < x < 9) = P(5 ≤ x ≤ 8)<BR>\n" );
document.write( "P(4 < x < 9) = P(5) + P(6) + P(7) + P(8)<BR>\n" );
document.write( "P(4 < x < 9) = 0.1029193452 + 0.200120949 + 0.266827932 + 0.2334744405<BR>\n" );
document.write( "P(4 < x < 9) = 0.8033426667\r<BR>\n" );
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document.write( "Answer: Approximately 0.8033426667\r<BR>\n" );
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document.write( "I'll let you tackle parts (e) and (f)\r<BR>\n" );
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document.write( "Hints:<BR>\n" );
document.write( "P(x < 5) = P(0) + P(1) + P(2) + P(3) + P(4)<BR>\n" );
document.write( "P(x > 3) = P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10)<BR>\n" );
document.write( "Or you can take the shortcut that<BR>\n" );
document.write( "P(x ≤ 3) + P(x > 3) = 1<BR>\n" );
document.write( "where<BR>\n" );
document.write( "P(x ≤ 3) = P(0) + P(1) + P(2) + P(3)<BR>\n" );
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document.write( " </SPAN>\n" );
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