document.write( "Question 214683: if a jar contains 23 red marbles and 4 green marbles, the probability that a marble selectted from the jar at random will be green is?\r
\n" ); document.write( "\n" ); document.write( "If the above green is not replaced, what is the probability that the next pulled marbles is red?\r
\n" ); document.write( "\n" ); document.write( "if you reached into the jar 4 times, what is the probability that you would never choose a green marble (after each pullout you do not replace the chosen marble?
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Algebra.Com's Answer #162121 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
a)\r
\n" ); document.write( "\n" ); document.write( "Total number of marbles = 23 + 4 = 27 marbles\r
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\n" ); document.write( "\n" ); document.write( "P(Green Marble) = (# of green marbles)/(# of marbles total) = 4/27 = 0.148\r
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\n" ); document.write( "\n" ); document.write( "So the probability is 4/27 which is about a 14.8% chance\r
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\n" ); document.write( "\n" ); document.write( "b)\r
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\n" ); document.write( "\n" ); document.write( "Since the selected marble from part a) is not replaced, the total number of marbles is now 27-1=26\r
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\n" ); document.write( "\n" ); document.write( "P(Red Marble) = (# of red marbles)/(# of marbles total) = 23/26 = 0.885\r
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\n" ); document.write( "\n" ); document.write( "So the probability is 23/27 which is about a 88.5% chance\r
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\n" ); document.write( "\n" ); document.write( "c)\r
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\n" ); document.write( "\n" ); document.write( "To find the probability of NOT choosing green after four selections, you simply calculate the probability of choosing red 4 times in a row (without replacement). This works because there are only two choices in marble selection.\r
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\n" ); document.write( "\n" ); document.write( "So the basic outline is\r
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\n" ); document.write( "\n" ); document.write( "P(4 red) = P(red)*P(red with 1 less from total)*P(red with 2 less from total)*P(red with 3 less from total)\r
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\n" ); document.write( "\n" ); document.write( "I'll let you take it from here. Let me know if you still need help.
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