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Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato ch [#permalink]
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09 Apr 2018, 23:56
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85% (00:46) correct 15% (01:36) wrong based on 33 sessions
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Re: Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato ch [#permalink]
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10 Apr 2018, 00:08
Bunuel wrote: Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato chips, and 5 bags of popcorn. Jenny reaches into the cabinet and draws a bag at random, places it on the counter, and returns to draw one more bag at random. What is the probability that she pulls out two bags of popcorn?
A. 1/13 B. 5/39 C. 2/13
D. 3/13 E. 10/39 As there is an explicit rule for calculating the probability of multiple independent experiments (in this case, taking bags from the cabinet), we'll use it. This is a Precise approach. The probability that the first bag is a popcorn bag is 5/(4+4+5) = 5/13. As there are now 12 bags left, 4 of which contain popcorn, the probability of drawing another bag of popcorn is 4/12 = 1/3 So, the probability of drawing 2 bags of popcorn is (5/13)*(1/3) = 5/39. (B) is our answer.
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Re: Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato ch [#permalink]
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10 Apr 2018, 04:52
Total number of ways of selecting 2 packets out of 13 packets is 13C2. Selection of first packet which is popcorn 5C1 Selection of second subsequent packet which is popcorn is 4C1 So probablity should be (5C1 * 4C1)/ 13C2 Solving which we get 10/39, Ans E Sent from my SMG935F using GMAT Club Forum mobile app



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Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato ch [#permalink]
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Updated on: 11 Apr 2018, 19:33
1. At the beginning you have 5 popcorn and 13 different items so 5/13. Assuming you've already selected a popcorn, you will only have 4/12 chance of selecting a second popcorn. In total, this would be (5/13)*(4/12) = (5/39). 2. Since there are 4 popcorn and we want to choose 2 popcorn, there are 5C2 different ways to do this (numerator). Since there are 13 total items and we want to choose to the denominator is 13C2. So 5C2/13C2 = (5*4)/(13*12) = 5/39. Edit since I previously thought problem said 4 popcorn instead of 5
Originally posted by dracobook on 10 Apr 2018, 05:05.
Last edited by dracobook on 11 Apr 2018, 19:33, edited 1 time in total.



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Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato ch [#permalink]
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10 Apr 2018, 05:19
Terrifficm wrote: Total number of ways of selecting 2 packets out of 13 packets is 13C2. Selection of first packet which is popcorn 5C1 Selection of second subsequent packet which is popcorn is 4C1 So probablity should be (5C1 * 4C1)/ 13C2 Solving which we get 10/39, Ans E Sent from my SMG935F using GMAT Club Forum mobile appHi, In your calculation, the numerator is 5*4 and the denominator is (13*12/2). Notice that this is the same as I did except for the denominator. Also, notice that your denominator is 'all the ways to choose two packets without respect to order' and your numerator is 'all the ways to choose the first packet and then choose the second packet'. This is what causes the error; if you were to multiply your denominator by 2 (because you can choose each pair two ways) you would get to answer (B).
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Re: Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato ch [#permalink]
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10 Apr 2018, 05:28
DavidTutorexamPAL wrote: Terrifficm wrote: Total number of ways of selecting 2 packets out of 13 packets is 13C2. Selection of first packet which is popcorn 5C1 Selection of second subsequent packet which is popcorn is 4C1 So probablity should be (5C1 * 4C1)/ 13C2 Solving which we get 10/39, Ans E Sent from my SMG935F using GMAT Club Forum mobile appHi, In your calculation, the numerator is 5*4 and the denominator is (13*12/2). Notice that this is the same as I did except for the denominator. Also, notice that your denominator is 'all the ways to choose two packets without respect to order' and your numerator is 'all the ways to choose the first packet and then choose the second packet'. This is what causes the error; if you were to multiply your denominator by 2 (because you can choose each pair two ways) you would get to answer (B). Thanks i understand the issue in my answer Sent from my SMG935F using GMAT Club Forum mobile app



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Re: Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato ch [#permalink]
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10 Apr 2018, 16:40
total bags = 13 popcorm = 5 tortilla = 4 potato = 4
First popcorn probability = 5/13 Second popcorn probability = 4/12
(5/13)* (4/12) = 5/39
Answer b



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Re: Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato ch [#permalink]
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11 Apr 2018, 16:16
Bunuel wrote: Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato chips, and 5 bags of popcorn. Jenny reaches into the cabinet and draws a bag at random, places it on the counter, and returns to draw one more bag at random. What is the probability that she pulls out two bags of popcorn?
A. 1/13 B. 5/39 C. 2/13
D. 3/13 E. 10/39 On the first draw, the probability of pulling a bag of popcorn is 5/13. Because Jenny doesn‘t put that bag back, there are now 12 bags in the drawer, 4 of which are popcorn, so the probability of popcorn on the second draw is 4/12. Because we want to calculate the probability of both outcomes happening, we multiply their respective probabilities. Thus, the probability of pulling two bags of popcorn is: 5/13 x 4/12 = 5/13 x 1/3 = 5/39 Answer: B
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Re: Jenny's cabinet contains 4 bags of tortilla chips, 4 bags of potato ch
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