document.write( "Question 1196701: #1 spinner circle is half green, 1/4 red and 1/4 blue. #2 spinner circle is half red, the other half has green, red, and yellow in equal sections.
\n" ); document.write( "If you get the same color twice, what is the probability it was red? green?\r
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Algebra.Com's Answer #829648 by math_tutor2020(3817) $\"\"$ $\"About$

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\n" ); document.write( "spinner A: 1/2 green, 1/4 red, 1/4 blue
\n" ); document.write( "spinner B: 2/3 red, 1/6 green, 1/6 yellow\r
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\n" ); document.write( "\n" ); document.write( "Note: if spinner B has half red on one side, and a 1/6 slice of red on the other side, then 1/2+1/6 = 3/6+1/6 = 4/6 = 2/3 is how much of a fraction red takes up in total.\r
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\n" ); document.write( "\n" ); document.write( "P(2 green) = P(A is green)*P(B is green)
\n" ); document.write( "P(2 green) = (1/2)*(1/6)
\n" ); document.write( "P(2 green) = 1/12\r
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\n" ); document.write( "\n" ); document.write( "P(2 red) = P(A is red)*P(B is red)
\n" ); document.write( "P(2 red) = (1/4)*(2/3)
\n" ); document.write( "P(2 red) = 2/12\r
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\n" ); document.write( "\n" ); document.write( "P(2 blue) = P(A is blue)*P(B is blue)
\n" ); document.write( "P(2 blue) = (1/4)*(0)
\n" ); document.write( "P(2 blue) = 0
\n" ); document.write( "It is impossible to get 2 blue since spinner B does not have blue.\r
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\n" ); document.write( "\n" ); document.write( "P(2 yellow) = P(A is yellow)*P(B is yellow)
\n" ); document.write( "P(2 yellow) = (0)*(2/3)
\n" ); document.write( "P(2 yellow) = 0
\n" ); document.write( "It is impossible to get 2 yellow since spinner A does not have yellow.\r
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\n" ); document.write( "\n" ); document.write( "So either we get 2 red or 2 green if we want two of the same color.
\n" ); document.write( "P(2 same color) = P(2 red OR 2 green)
\n" ); document.write( "P(2 same color)= P(2 red) + P(2 green)
\n" ); document.write( "P(2 same color) = 2/12 + 1/12
\n" ); document.write( "P(2 same color) = (2 + 1)/12
\n" ); document.write( "P(2 same color) = 3/12
\n" ); document.write( "P(2 same color) = 1/4
\n" ); document.write( "Note: The events \"2 red\" and \"2 green\" are mutually exclusive. \r
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\n" ); document.write( "\n" ); document.write( "Then we can say
\n" ); document.write( "P(2 red, given 2 of the same color) = P(2 red)/P(2 same color)
\n" ); document.write( "P(2 red, given 2 of the same color) = (2/12)/(1/4)
\n" ); document.write( "P(2 red, given 2 of the same color) = (2/12)*(4/1)
\n" ); document.write( "P(2 red, given 2 of the same color) = 8/12
\n" ); document.write( "P(2 red, given 2 of the same color) = 2/3\r
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\n" ); document.write( "\n" ); document.write( "And,
\n" ); document.write( "P(2 green, given 2 of the same color) = P(2 green)/P(2 same color)
\n" ); document.write( "P(2 green, given 2 of the same color) = (1/12)/(1/4)
\n" ); document.write( "P(2 green, given 2 of the same color) = (1/12)*(4/1)
\n" ); document.write( "P(2 green, given 2 of the same color) = 4/12
\n" ); document.write( "P(2 green, given 2 of the same color) = 1/3\r
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\n" ); document.write( "\n" ); document.write( "The results 2/3 and 1/3 are complementary to represent all the possible ways to get 2 of the same color. For more information, check out the concept of conditional probability.\r
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\n" ); document.write( "\n" ); document.write( "Another approach:\r
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\n" ); document.write( "\n" ); document.write( "We can use table to lay out the possible outcomes.
\n" ); document.write( "Along the top row are the four slices of green, green, red and blue for spinner A.
\n" ); document.write( "The outcomes for spinner B are along the left hand side.\n" ); document.write( "\n" ); document.write( "

		1	2	3	4
		green	green	red	blue
1	red			X
2	red			X
3	red			X
4	red			X
5	red			X
6	red			X
7	green	X	X
8	green	X	X
9	red			X
10	red			X
11	yellow
12	yellow

\n" ); document.write( "Each X represents a situation where we get 2 of the same color. Either 2 red or 2 green. The cells left blank are when we get a mix of two different colors, so we'll ignore those cases. The numbers along the left and top are there to help keep track of the slices.\r
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\n" ); document.write( "\n" ); document.write( "There are 12 X's total.\r
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\n" ); document.write( "\n" ); document.write( "Four of the Xs are when we get 2 greens
\n" ); document.write( "4/12 = 1/3
\n" ); document.write( "while the remaining eight are 2 reds
\n" ); document.write( "8/12 = 2/3\r
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\n" ); document.write( "\n" ); document.write( "Answers: \r
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\n" ); document.write( "\n" ); document.write( "If we get 2 of the same color, then,
\n" ); document.write( "probability of 2 reds = 2/3
\n" ); document.write( "probability of 2 greens = 1/3
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