document.write( "Question 1193829: 12 balls — 4 red, 4 white, and 4 blue.\r
\n" ); document.write( "\n" ); document.write( "If three balls are drawn at random out of this box one at a time without replacement, what is the probability of obtaining\r
\n" ); document.write( "\n" ); document.write( "3 white balls?
\n" ); document.write( "Answer: 1/55
\n" ); document.write( "4/12 * 3/12 * 2/10\r
\n" ); document.write( "\n" ); document.write( "3 balls of the same color?
\n" ); document.write( "Answer: 3/55
\n" ); document.write( "p{(w,w,w),(R,R,R),(B,B,B)}
\n" ); document.write( " 1/55 + 1/55 + 1/55\r
\n" ); document.write( "\n" ); document.write( "Are my answers correct? Thanks
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Algebra.Com's Answer #825860 by math_tutor2020(3817)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Question 1: What is the probability of getting 3 white balls?\r
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\n" ); document.write( "\n" ); document.write( "If we select without replacement, then,
\n" ); document.write( "4/12 is the probability of getting a white ball
\n" ); document.write( "3/11 is the probability of getting a second white ball
\n" ); document.write( "2/10 is the probability of getting a third white ball
\n" ); document.write( "Each time we count down by 1 for the numerator and denominator separately.\r
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\n" ); document.write( "\n" ); document.write( "Multiply out those fractions:
\n" ); document.write( "(4/12)*(3/11)*(2/10)
\n" ); document.write( "(4*3*2)/(12*11*10)
\n" ); document.write( "(4*3*2)/(4*3*11*2*5)
\n" ); document.write( "1/(11*5)
\n" ); document.write( "1/55
\n" ); document.write( "You have the correct answer. Nice job.
\r
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\n" ); document.write( "\n" ); document.write( "Here's another way to get the answer.
\n" ); document.write( "There are n = 4 white balls and we select r = 3 of them.
\n" ); document.write( "Using the nCr combination formula leads to 4C3 = 4 ways to do this. In other words, there are 4 ways to not select a particularly unlucky white ball.\r
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\n" ); document.write( "\n" ); document.write( "Now let's count the number of ways to select any 3 balls regardless of color.
\n" ); document.write( "We have n = 12 of them and we select r = 3
\n" ); document.write( "12 C 3 = 220 ways to pick any 3 balls
\n" ); document.write( "Then notice how 4/220 = (1*4)/(55*4) = 1/55\r
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\n" ); document.write( "\n" ); document.write( "Here's some scratch work to show how to compute 12 C 3
\n" ); document.write( "n C r = (n!)/(r!(n-r)!)
\n" ); document.write( "12 C 3 = (12!)/(3!*(12-3)!)
\n" ); document.write( "12 C 3 = (12!)/(3!*9!)
\n" ); document.write( "12 C 3 = (12*11*10*9!)/(3!*9!)
\n" ); document.write( "12 C 3 = (12*11*10)/(3!)
\n" ); document.write( "12 C 3 = (12*11*10)/(3*2*1)
\n" ); document.write( "12 C 3 = (1320)/(6)
\n" ); document.write( "12 C 3 = 220
\n" ); document.write( "Use this template to compute 4C3 as well if necessary.\r
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\n" ); document.write( "\n" ); document.write( "As you can tell or already guess, order does not matter when selecting the balls which is why the nCr formula is used (rather than the nPr one).\r
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\n" ); document.write( "\n" ); document.write( "Question 2: What is the probability of getting 3 balls of the same color?\r
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\n" ); document.write( "\n" ); document.write( "You are correct in thinking that 1/55 is the probability of getting all red, and also all blue. This works because we have 4 of each color.
\n" ); document.write( "This leads to the correct answer of 3/55. Once again, nice work.\r
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\n" ); document.write( "\n" ); document.write( "Side note: I appreciate you showing your work and thought process.
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