document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=177723>Question 177723</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A local church is studying the amount of offerings in an envelope from their early Sunday mornings services. The church studied 500 envelopes and found the following:<BR>\n" );
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document.write( "Less than $5       200                <BR>\n" );
document.write( "$5 to $9.99         100<BR>\n" );
document.write( "$10  to $19.99      75<BR>\n" );
document.write( "$20  to  $49.99     75<BR>\n" );
document.write( "$50 or more          50<BR>\n" );
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document.write( "a.      what is the probability of opening one envelop and finding $20 or more in it? ______<BR>\n" );
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document.write( "b.     What is the probability of opening one envelop and finding less than $10 in it? ______<BR>\n" );
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document.write( "c.      What is the probability of opening two envelopes and finding less than $5 in<BR>\n" );
document.write( "       each one of them?  ________<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 #132724 by </B><A HREF=/tutors/aboutme.mpl?userid=gonzo><B>gonzo(654)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=gonzo><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=132724\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a = 200 envelopes out of 500 contained less than $5.00.<BR>\n" );
document.write( "b = 100 envelopes out of 500 contained $5.00 to $9.99.<BR>\n" );
document.write( "c = 75 envelopes out of 500 contained $10.00 to $19.99.<BR>\n" );
document.write( "d = 75 envelopes out of 500 contained $20.00 to $49.99.<BR>\n" );
document.write( "e = 50 envelopes out of 500 contained $50.00 or more.<BR>\n" );
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document.write( "the probability of any event occurring is the number of times that event can occur divided by the total number of events possible.<BR>\n" );
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document.write( "given that definition, then:<BR>\n" );
document.write( "p(a) = 200/500<BR>\n" );
document.write( "p(b) = 100/500<BR>\n" );
document.write( "p(c) = 75/500<BR>\n" );
document.write( "p(d) = 75/500<BR>\n" );
document.write( "p(e) = 50/500<BR>\n" );
document.write( "---<BR>\n" );
document.write( "what is the probability of opening one envelope and finding $20 or more in it?<BR>\n" );
document.write( "this would be p(d or e) = p(d) + p(e) = 125/500.<BR>\n" );
document.write( "---<BR>\n" );
document.write( "What is the probability of opening one envelope and finding less than $10 in it?<BR>\n" );
document.write( "this would be p (a or b) = p(a) + p(b) = 300/500.<BR>\n" );
document.write( "---<BR>\n" );
document.write( "What is the probability of opening two envelopes and finding less than $5 in<BR>\n" );
document.write( "each one of them? <BR>\n" );
document.write( "this would be p(a and x) = p(a) * p(x) = (200/500) * (199/499) = (200*199)/(500*499)<BR>\n" );
document.write( "---<BR>\n" );
document.write( "what is p(x)?<BR>\n" );
document.write( "here's what happened.<BR>\n" );
document.write( "you had 500 envelopes.<BR>\n" );
document.write( "you withdrew one envelope and that means you had 499 left to pick from.<BR>\n" );
document.write( "the one you picked had less than $5.00 in it.<BR>\n" );
document.write( "since you originally had 200 envelopes out of 500 with less than $5.00 in it, that means that you now have 199 envelopes out of 499 with less then $5.00 in it.<BR>\n" );
document.write( "---<BR>\n" );
document.write( "x = 199 envelopes out of 499 contained less than $5.00 in it.<BR>\n" );
document.write( "p(x) = 199/499<BR>\n" );
document.write( "---<BR>\n" );
document.write( "the probability of 2 events occurring in succession is the probability of the first event occurring times the probability of the second event occurring.<BR>\n" );
document.write( "this is why probability of getting 2 envelopes with less than $5.00 in it is p(a) * p(x).<BR>\n" );
document.write( "---<BR>\n" );
document.write( "your answers are:<BR>\n" );
document.write( "the probability of opening one envelope and finding $20.00 or more is 125/500<BR>\n" );
document.write( "the probability of opening one envelope and finding less than $10.00 is 300/500.<BR>\n" );
document.write( "the probability of opening two envelopes and finding less than $5.00 in each is (200/500)*(199/499).<BR>\n" );
document.write( "---<BR>\n" );
document.write( "you can stop here if you are satisfied.  otherwise continue for some more explanation.<BR>\n" );
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document.write( "to help you see this, take a smaller example:<BR>\n" );
document.write( "suppose you have 4 envelopes:<BR>\n" );
document.write( "1 envelope contains $1.00<BR>\n" );
document.write( "1 envelope contains $2.00<BR>\n" );
document.write( "1 envelope contains $3.00<BR>\n" );
document.write( "1 envelope contains $4.00<BR>\n" );
document.write( "---<BR>\n" );
document.write( "since there are 4 total envelopes,<BR>\n" );
document.write( "probability of opening 1 envelope and getting $1.00 is 1/4<BR>\n" );
document.write( "probability of opening 1 envelope and getting $2.00 is 1/4<BR>\n" );
document.write( "probability of opening 1 envelope and getting $3.00 is 1/4<BR>\n" );
document.write( "probability of opening 1 envelope and getting $4.00 is 1/4<BR>\n" );
document.write( "---<BR>\n" );
document.write( "what is the probability of getting $1.00 or $2.00 on the first draw.<BR>\n" );
document.write( "this would be 2/4 = 1/2 because your chances of success are 2 out of 4.<BR>\n" );
document.write( "this is the same as p($1.00) + p($2.00) = 1/4 + 1/4 = 2/4 = 1/2.<BR>\n" );
document.write( "---<BR>\n" );
document.write( "what is the probability of opening 1 envelope and getting $1.00 and then opening a second envelope and getting $2.00?<BR>\n" );
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document.write( "once you opened the first envelope and got $1.00, you are left with 3 envelopes, 1 of which contains $2.00.<BR>\n" );
document.write( "---<BR>\n" );
document.write( "p($2.00) once you withdrew $1.00 is 1/3.<BR>\n" );
document.write( "P($1.00 and then $2.00) = p($1.00) * p($2.00) = 1/4 * 1/3 = 1/12<BR>\n" );
document.write( "---<BR>\n" );
document.write( "how does this work?<BR>\n" );
document.write( "well, 1/4 of the time you will get the first envelope containing $1.00.<BR>\n" );
document.write( "1/3 of the time you will get the second envelope containing $2.00 assuming you got $1.00 the first time.<BR>\n" );
document.write( "1/4 of 1/3 of the time you will get $1.00 opening the first envelope and then $2.00 opening the second envelope.<BR>\n" );
document.write( "this is the same as (1/4)*(1/3) = (1/12)<BR>\n" );
document.write( "---<BR>\n" );
document.write( "since i used smaller numbers of envelopes, i can show you how this happens.<BR>\n" );
document.write( "let A = envelope with $1.00 in it.<BR>\n" );
document.write( "let B = envelope with $2.00 in it.<BR>\n" );
document.write( "let C = envelope with $3.00 in it.<BR>\n" );
document.write( "let D = envelope with $4.00 in it.<BR>\n" );
document.write( "let the first time a letter shows up be the opening of the first envelope with the amount of money in it indicated by the letter.<BR>\n" );
document.write( "let the second time a letter shows up be the opening of the second envelope with the amount of money in it indicated by the letter.<BR>\n" );
document.write( "here's all the possibilities with first and second draw.<BR>\n" );
document.write( "AB<BR>\n" );
document.write( "AC<BR>\n" );
document.write( "AD<BR>\n" );
document.write( "BA<BR>\n" );
document.write( "BC<BR>\n" );
document.write( "BD<BR>\n" );
document.write( "CA<BR>\n" );
document.write( "CB<BR>\n" );
document.write( "CD<BR>\n" );
document.write( "DA<BR>\n" );
document.write( "DB<BR>\n" );
document.write( "DC<BR>\n" );
document.write( "---<BR>\n" );
document.write( "total number of possible ways to draw the first and second envelope is 12<BR>\n" );
document.write( "only 1 of this draws A first and then B.<BR>\n" );
document.write( "probability is therefore 1/12 of drawing A first and then B.<BR>\n" );
document.write( "this is the same as p(A) * p(B) = 1/4 * 1/3 = 1/12.<BR>\n" );
document.write( "---<BR>\n" );
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