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In United States currency, a nickel is worth 5 cents [#permalink]
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03 Apr 2010, 11:31
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In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back? A. 8/25 B. 12/35 C. 13/35 D. 9/25 E. 17/25 My approach was calculating what the probability of nickel getting picked twice in a row is and subtracting that probability by 1. Probability of a nickel getting picked the first time is 2/5 and second time is 5/14. (2/5)*(5/14) = 1/7. So I was expecting 6/7 to show up but that choice is not given. I understand the OA and its methodology but what's wrong with my approach? Thanks!
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Re: Help with a probability question. [#permalink]
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03 Apr 2010, 12:01
changhiskhan wrote: In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?
a) 8/25 b) 12/35 c) 13/35 d) 9/25 e) 17/25
My approach was calculating what the probability of nickel getting picked twice in a row is and subtracting that probability by 1. Probability of a nickel getting picked the first time is 2/5 and second time is 5/14. (2/5)*(5/14) = 1/7. So I was expecting 6/7 to show up but that choice is not given. I understand the OA and its methodology but what's wrong with my approach?
Thanks! In this case it would be easier to calculate the probability in direct way: \(\frac{nonnickel}{all}*\frac{nonnickel 1}{all1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}\). The problem in your solution is that the opposite probability of two nonnickel coins in a row is not two nickel in a row, it's two nickels in a row plus either of two coins from two is nickel, (so basically the probability of at least one nickel from two picks): \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\) \(P=1\frac{23}{35}=\frac{12}{35}\) Answer: B.
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Re: Help with a probability question. [#permalink]
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04 Apr 2010, 05:24
Here's my solution  Can be explained if anyone want 5/15*4/14 + 4/15*3/14 +5/15*4/14 + 4/15*5/14 = 12/35
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Re: Help with a probability question. [#permalink]
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04 Apr 2010, 05:48
Is it just me or is the phrasing of the questions pretty bad?
First of all it tells you the values of the coins and that 1 dollar = 100 cents. Why? Just to throw you off?
Next, it says "probability of picking a coin other than a nickel twice in a row" which I took to mean either 2 pennies in a row, or 2 dimes in a row (not correct). Also it isnt clear that they mean that the 2 coins have to be the same right at the beginning (we dont know how many are taken out in total).



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Re: Help with a probability question. [#permalink]
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04 Apr 2010, 06:03
nickk first i got confused as well but if you take 2 dimes or 2 pennies then the answer is not one of the choice given So i decided to go with any of the coin apart from nickel.
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Re: Help with a probability question. [#permalink]
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04 Apr 2010, 08:51
Bunuel wrote: changhiskhan wrote: In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?
a) 8/25 b) 12/35 c) 13/35 d) 9/25 e) 17/25
My approach was calculating what the probability of nickel getting picked twice in a row is and subtracting that probability by 1. Probability of a nickel getting picked the first time is 2/5 and second time is 5/14. (2/5)*(5/14) = 1/7. So I was expecting 6/7 to show up but that choice is not given. I understand the OA and its methodology but what's wrong with my approach?
Thanks! In this case it would be easier to calculate the probability in direct way: \(\frac{nonnickel}{all}*\frac{nonnickel 1}{all1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}\). The problem in your solution is that the opposite probability of two nonnickel coins in a row is not two nickel in a row, it's two nickels in a row plus either of two coins from two is nickel, which is at least one nickel from two picks: \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\) \(P=1\frac{23}{35}=\frac{12}{35}\) Answer: B. so simple and elegant.. +1 for you
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Re: In United States currency, a nickel is worth 5 cents [#permalink]
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14 May 2014, 13:37
Language of this question is very confusing to me. I taught we can draw continuously all coins one by one in sequence. Possible combinations as: NPNDNPNDNPNDDPP NPNDNPNDNPNPPDD PNDNPNDNPNDNDPP ... .. .
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Re: Help with a probability question. [#permalink]
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26 May 2014, 14:17
Bunuel wrote: changhiskhan wrote: In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?
a) 8/25 b) 12/35 c) 13/35 d) 9/25 e) 17/25
My approach was calculating what the probability of nickel getting picked twice in a row is and subtracting that probability by 1. Probability of a nickel getting picked the first time is 2/5 and second time is 5/14. (2/5)*(5/14) = 1/7. So I was expecting 6/7 to show up but that choice is not given. I understand the OA and its methodology but what's wrong with my approach?
Thanks! In this case it would be easier to calculate the probability in direct way: \(\frac{nonnickel}{all}*\frac{nonnickel 1}{all1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}\). The problem in your solution is that the opposite probability of two nonnickel coins in a row is not two nickel in a row, it's two nickels in a row plus either of two coins from two is nickel, (so basically the probability of at least one nickel from two picks): \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\) \(P=1\frac{23}{35}=\frac{12}{35}\) Answer: B. Hi Bunuel, I tried a different approach but didn't end up with the same answer. Can you please clarify? Prob of pennies * Prob of Dimes * Permutation since order doesn't matter (5/15)(4/14)(2) = 4/21? Additionally, in your method above (9/15)(8/14)  Why don't we factor in permutation?



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Re: Help with a probability question. [#permalink]
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27 May 2014, 01:08
russ9 wrote: Bunuel wrote: changhiskhan wrote: In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?
a) 8/25 b) 12/35 c) 13/35 d) 9/25 e) 17/25
My approach was calculating what the probability of nickel getting picked twice in a row is and subtracting that probability by 1. Probability of a nickel getting picked the first time is 2/5 and second time is 5/14. (2/5)*(5/14) = 1/7. So I was expecting 6/7 to show up but that choice is not given. I understand the OA and its methodology but what's wrong with my approach?
Thanks! In this case it would be easier to calculate the probability in direct way: \(\frac{nonnickel}{all}*\frac{nonnickel 1}{all1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}\). The problem in your solution is that the opposite probability of two nonnickel coins in a row is not two nickel in a row, it's two nickels in a row plus either of two coins from two is nickel, (so basically the probability of at least one nickel from two picks): \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\) \(P=1\frac{23}{35}=\frac{12}{35}\) Answer: B. Hi Bunuel, I tried a different approach but didn't end up with the same answer. Can you please clarify? Prob of pennies * Prob of Dimes * Permutation since order doesn't matter (5/15)(4/14)(2) = 4/21? Additionally, in your method above (9/15)(8/14)  Why don't we factor in permutation? Hint: P(two nonnickel) = P(two pennies) + P(two dimes) + P(one nickel and one dime). As for your second question: (not nickel, not nickel) is one case.
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Re: In United States currency, a nickel is worth 5 cents [#permalink]
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17 Jul 2017, 04:19
My input: This problem has a lot of unnecessary information. Typical example: Fact that 100 cents is equals to 1 dollar is not a necessary information. It can be reformalute as follows: A bag contains 6 Red pens and 9 Blue pens. 2 pens are consecutively drawn from the bag; the first drawn is not put back in the bag. What the probability of drawing 2 Blue pens in a row? Then it becomes easier: (9/15) * (8/14) = 12/35
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Re: In United States currency, a nickel is worth 5 cents [#permalink]
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24 Aug 2017, 21:42
Can anyone please clarify why we don't multiply (9/15) * (8/14) by 2! ?
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Re: In United States currency, a nickel is worth 5 cents [#permalink]
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24 Aug 2017, 21:49



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In United States currency, a nickel is worth 5 cents [#permalink]
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25 Aug 2017, 18:28
changhiskhan wrote: In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?
A. 8/25 B. 12/35 C. 13/35 D. 9/25 E. 17/25
There are 6 nickels = 6N 5 pennies = 5P 4 dimes D = 4D What is the probability of picking [any] a coin other than that is different from a nickel twice in a row (no replacement)? I do not think that this question is poorly written except for the extra bits about what each coin is worth. I do think its language might be colloquial. I've put some different language in to show what "a" and "other than" signify. "Not nickel" = anything other than, or different from, a nickel = pennies and dimes. Change "not N" to the "affirmative" case: yes to pennies and dimes. When defining "favorable outcome," both P and D are included. Both are different from N. I think FCP is easiest. There are 15 coins total. 5P + 4D = 9 coins that give me the desired outcome. For first pick, total outcomes = 15. Desired outcomes = 9. Probability of desired outcome is \(\frac{9}{15}\): I have 9 chances out of 15 total to get a D or a P. The total number of coins is now 14. And there is one fewer coin in the desired group. So for second pick, probability of desired outcome is \(\frac{8}{14}\) \(\frac{9}{15}\) * \(\frac{8}{14}\) = \(\frac{12}{35}\) Answer B
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Re: In United States currency, a nickel is worth 5 cents [#permalink]
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22 Jan 2018, 09:37
sometimes, gmat gives unnecessary information, we have to pay attention to the right detail. "100 cents equals one dollar." is just a good distractor.




Re: In United States currency, a nickel is worth 5 cents
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