document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1171071>Question 1171071</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The Web site for M&M™ candies claimed that 24% of plain M&M candies are blue, 20% are orange, 16% green, 14% yellow, and 13% each red and brown.\r<BR>\n" );
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document.write( "(a) Pick one M&M at random from a package.\r<BR>\n" );
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document.write( "1.Describe the sample space.\r<BR>\n" );
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document.write( "2.What is the probability that the one you pick is blue or red?\r<BR>\n" );
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document.write( "3.What is the probability that the one you pick is not green?\r<BR>\n" );
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document.write( "(b) You pick three M&M’s in a row randomly from three separate packages.\r<BR>\n" );
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document.write( "1.Describe the sample space for the outcomes of your three choices.\r<BR>\n" );
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document.write( "2.What is the probability that every M&M is blue?\r<BR>\n" );
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document.write( "3.What is the probability that the third M&M is red?4.What is the probability that at least one is blue? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850992 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</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=850992\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's break down this M&M problem.\r<BR>\n" );
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document.write( "**(a) Picking One M&M**\r<BR>\n" );
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document.write( "1.  **Describe the Sample Space:**<BR>\n" );
document.write( "    * The sample space consists of all possible colors of an M&M.<BR>\n" );
document.write( "    * Sample Space = {Blue, Orange, Green, Yellow, Red, Brown}\r<BR>\n" );
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document.write( "2.  **Probability of Blue or Red:**<BR>\n" );
document.write( "    * P(Blue) = 24% = 0.24<BR>\n" );
document.write( "    * P(Red) = 13% = 0.13<BR>\n" );
document.write( "    * Since these are mutually exclusive events (an M&M can't be both blue and red), we add the probabilities:<BR>\n" );
document.write( "    * P(Blue or Red) = P(Blue) + P(Red) = 0.24 + 0.13 = 0.37\r<BR>\n" );
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document.write( "3.  **Probability of Not Green:**<BR>\n" );
document.write( "    * P(Green) = 16% = 0.16<BR>\n" );
document.write( "    * P(Not Green) = 1 - P(Green) = 1 - 0.16 = 0.84\r<BR>\n" );
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document.write( "**(b) Picking Three M&M's**\r<BR>\n" );
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document.write( "1.  **Describe the Sample Space:**<BR>\n" );
document.write( "    * Each M&M can be one of six colors.<BR>\n" );
document.write( "    * The sample space consists of all possible combinations of three colors.<BR>\n" );
document.write( "    * We can represent each outcome as a sequence of three colors (e.g., Blue, Orange, Red).<BR>\n" );
document.write( "    * The sample space would be a list of all possible combinations of three colors, such as (Blue, Blue, Blue), (Blue, Blue, Orange), (Blue, Orange, Blue), etc.<BR>\n" );
document.write( "    * The number of possible outcomes is 6 * 6 * 6 = 216.\r<BR>\n" );
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document.write( "2.  **Probability of Every M&M Being Blue:**<BR>\n" );
document.write( "    * P(Blue) = 0.24<BR>\n" );
document.write( "    * Since the picks are independent, we multiply the probabilities:<BR>\n" );
document.write( "    * P(Blue, Blue, Blue) = P(Blue) * P(Blue) * P(Blue) = (0.24)^3 = 0.013824\r<BR>\n" );
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document.write( "3.  **Probability of the Third M&M Being Red:**<BR>\n" );
document.write( "    * The probability of the third M&M being red is independent of the first two picks.<BR>\n" );
document.write( "    * P(Third M&M is Red) = P(Red) = 0.13\r<BR>\n" );
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document.write( "4.  **Probability of At Least One Blue:**<BR>\n" );
document.write( "    * It's easier to find the probability of no blue M&M's and subtract it from 1.<BR>\n" );
document.write( "    * P(Not Blue) = 1 - P(Blue) = 1 - 0.24 = 0.76<BR>\n" );
document.write( "    * P(No Blue in 3 picks) = (0.76)^3 = 0.438976<BR>\n" );
document.write( "    * P(At Least One Blue) = 1 - P(No Blue) = 1 - 0.438976 = 0.561024<BR>\n" );
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