document.write( "Question 478952: Nine tickets, numbered from 1 to 9, are in a box. If 2 tickets are drawn at random, determine the probability p that both are odd. \r
\n" ); document.write( "\n" ); document.write( "Here is how I did it:
\n" ); document.write( "P (2 tickets = odd, 9 tickets)\r
\n" ); document.write( "\n" ); document.write( "5 odd
\n" ); document.write( "9 tickets\r
\n" ); document.write( "\n" ); document.write( "P (2 tickets = odd, 9 tickets) = desired outcome/total outcome = (5/9) times (5/9) = (5/9)^2 = 25/81 = 0.3086 = 0.31\r
\n" ); document.write( "\n" ); document.write( "However, this is NOT the correct answer. My textbook says this:
\n" ); document.write( "5C2/9C2 = 5/18 = 0.28\r
\n" ); document.write( "\n" ); document.write( "Please note I would have solved this problem the textbook's way; yet, in a similar problem, I solved it the textbook's way and the book solved it the way I did in this problem. So, I corrected myself to proceed through the calculations as stated in my book. Now, the book switched on me. I don't know what is A DIFFERENCE between the two versions of the calculation. Percentages are slightly off, and I assume that my answer must be wrong since a slight difference in math means a huge gap. \r
\n" ); document.write( "\n" ); document.write( "If you do not mind, here is the similar problem I am referring to above: Three cards are drawn from a pack of 52 cards, each card being replaced before the next one is drawn. Compute the probability p that all are spades. \r
\n" ); document.write( "\n" ); document.write( "Here is how I did it: 13C3/52C3 = 0.0129 = 1.29 %
\n" ); document.write( "Here is how the textbook did it: (13/52)times(13/52)times(13/52) = (13/52)^3 = 0.0156 = 1.56%. \r
\n" ); document.write( "\n" ); document.write( "As you notice, the percentages are again just slightly off but still two different values. Please let me what am I missing? What is it that I solve the problem through combinations vs. (desired outcome/total outcome)^(number of drawings)?\r
\n" ); document.write( "\n" ); document.write( "Thank you very much for your time and advice. I really appreciate it. Have a beautiful and safe evening. \r
\n" ); document.write( "\n" ); document.write( "Respectfully, \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Ivanka \n" ); document.write( "
Algebra.Com's Answer #328189 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
the ticket problem assumes no replacement.
\n" ); document.write( "the probability of drawing the first odd ticket is 5/9.
\n" ); document.write( "once you draw that ticket, without replacement, then then there are 4 odd tickets left out of a total of 8, so the probability of getting an odd ticket on the second draw is 4/8.
\n" ); document.write( "5/9 * 4/8 = 20/72 = 5/18
\n" ); document.write( "5C2/9C2 = 10/36 = 5/18
\n" ); document.write( "they're the same.
\n" ); document.write( "with your card problem, there is no replacement.
\n" ); document.write( "using combinations, the answer you show is:
\n" ); document.write( "13C3/52C3 = 286/22100 = .012941176
\n" ); document.write( "using probabilities, the answer would be 13/52 * 13/52 * 13/52 = .015625
\n" ); document.write( "the answer are not the same because replacement was assumed using the probabilties while no replacement was assumed when using the combination formulas.
\n" ); document.write( "that's the difference between each way.
\n" ); document.write( "the two different methods are comparable if you do not replace after every draw.
\n" ); document.write( "they are not comparable if you replace after every draw.
\n" ); document.write( "if you assumed no replacement in the card problem, then the probability would have been calculated as 13/52 * 12/51 * 11/50 = .012941176
\n" ); document.write( "the two methods are comparable in the card problem when you assume no replacement.
\n" ); document.write( "-----
\n" ); document.write( "if you look at 13C3 / 52C3 and work out the math, this is what you get:
\n" ); document.write( "\"13C3+=+13%21+%2F+%283%21+%2A+10%21%29\"
\n" ); document.write( "\"52C3+=+52%21+%2F+%283%21+%2A+49%21%29\"
\n" ); document.write( "13C3 / 52C3 becomes equal to:
\n" ); document.write( "\"%2813%21+%2F+%283%21+%2A+10%21%29%29+%2F+%2852%21+%2F+%283%21+%2A+49%21%29%29\"
\n" ); document.write( "this is equivalent to:
\n" ); document.write( "\"%2813%21+%2F+%283%21+%2A+10%21%29%29+%2A+%28%283%21+%2A+49%21%29%2F52%21%29\"
\n" ); document.write( "this can be rewritten as:
\n" ); document.write( "\"%2813%21+%2A+3%21+%2A+49%21%29+%2F+%283%21+%2A+10%21+%2A+52%21%29\"
\n" ); document.write( "going a little further, this can be rewritten as:
\n" ); document.write( "
\n" ); document.write( "there's a 3! in the numerator and the denominator that cancels out.
\n" ); document.write( "there's a 10! in the numerator and the denominator that cancels out.
\n" ); document.write( "there's a 49! in the numerator and the denominator that cancels out.
\n" ); document.write( "your are left with:
\n" ); document.write( "\"%2813+%2A+12+%2A+11%29+%2F+%2852+%2A+51+%2A+50%29\"
\n" ); document.write( "if you do the problem the alternate way using the probabilities for each event, then you get the same equation without replacement.
\n" ); document.write( "it is:
\n" ); document.write( "\"%2813%2F52%29+%2A+%2812%2F51%29+%2A+%28+12%2F50%29\" which can be rewritten as:
\n" ); document.write( "\"%2813+%2A+12+%2A+11%29+%2F+%2852+%2A+51+%2A+50%29\"
\n" ); document.write( "bottom line:
\n" ); document.write( "the two methods are comparable if you assume no replacement.
\n" ); document.write( "they are not comparable if you assume replacement.\r
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