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# In United States currency, a nickel is worth 5 cents

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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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25% (medium)

Question Stats:

78% (02:08) correct 22% (02:38) wrong based on 348 sessions

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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: In United States currency, a nickel is worth 5 cents  [#permalink]

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03 Apr 2010, 12:01
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5
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:
$$\frac{non-nickel}{all}*\frac{non-nickel -1}{all-1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}$$.

The problem in your solution is that the opposite probability of two non-nickel 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):
$$\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}$$

$$P=1-\frac{23}{35}=\frac{12}{35}$$

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Re: In United States currency, a nickel is worth 5 cents  [#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: In United States currency, a nickel is worth 5 cents  [#permalink]

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04 Apr 2010, 05:48
1
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: In United States currency, a nickel is worth 5 cents  [#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: In United States currency, a nickel is worth 5 cents  [#permalink]

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04 Apr 2010, 08:51
1
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:
$$\frac{non-nickel}{all}*\frac{non-nickel -1}{all-1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}$$.

The problem in your solution is that the opposite probability of two non-nickel 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:
$$\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}$$

$$P=1-\frac{23}{35}=\frac{12}{35}$$

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
1
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:
N-P-N-D-N-P-N-D-N-P-N-D-D-P-P
N-P-N-D-N-P-N-D-N-P-N-P-P-D-D
P-N-D-N-P-N-D-N-P-N-D-N-D-P-P
...
..
.
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Re: In United States currency, a nickel is worth 5 cents  [#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:
$$\frac{non-nickel}{all}*\frac{non-nickel -1}{all-1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}$$.

The problem in your solution is that the opposite probability of two non-nickel 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):
$$\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}$$

$$P=1-\frac{23}{35}=\frac{12}{35}$$

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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In United States currency, a nickel is worth 5 cents  [#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:
$$\frac{non-nickel}{all}*\frac{non-nickel -1}{all-1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}$$.

The problem in your solution is that the opposite probability of two non-nickel 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):
$$\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}$$

$$P=1-\frac{23}{35}=\frac{12}{35}$$

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 non-nickel) = P(two pennies) + P(two dimes) + P(one penny 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
1
Can anyone please clarify why we don't multiply (9/15) * (8/14) by 2! ?

We have 6 nickels and 9 non-nickels. We want the probability P(non-nickel, non-nickel). non-nickel, non-nickel can be arranged only in one way.
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Re: 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 $$\frac{8}{14}$$

$$\frac{9}{15}$$ * $$\frac{8}{14}$$ = $$\frac{12}{35}$$

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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.
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Re: In United States currency, a nickel is worth 5 cents  [#permalink]

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26 Sep 2019, 07:30
Why is it incorrect to do
(5C2 + 4C2)/(15C2)?
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Re: In United States currency, a nickel is worth 5 cents  [#permalink]

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26 Sep 2019, 07:33
1
ghnlrug wrote:
Why is it incorrect to do
(5C2 + 4C2)/(15C2)?

I think this is addressed here: https://gmatclub.com/forum/in-united-st ... l#p1368423
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Re: In United States currency, a nickel is worth 5 cents  [#permalink]

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26 Sep 2019, 07:37
Bunuel wrote:
ghnlrug wrote:
Why is it incorrect to do
(5C2 + 4C2)/(15C2)?

I think this is addressed here: https://gmatclub.com/forum/in-united-st ... l#p1368423

Thanks Bunuel, I appreciate it. So the correct way is (9C2)/(15C2), but could you explain how the question would be so that (5C2 + 4C2)/(15C2) is actually the right answer?
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In United States currency, a nickel is worth 5 cents  [#permalink]

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26 Sep 2019, 08:18
ghnlrug wrote:
Bunuel wrote:
ghnlrug wrote:
Why is it incorrect to do
(5C2 + 4C2)/(15C2)?

I think this is addressed here: https://gmatclub.com/forum/in-united-st ... l#p1368423

Thanks Bunuel, I appreciate it. So the correct way is (9C2)/(15C2), but could you explain how the question would be so that (5C2 + 4C2)/(15C2) is actually the right answer?

P(two non-nickel) = P(two pennies) + P(two dimes) + P(one penny and one dime) = $$\frac{5C2 + 4C2 + 5C1*4C1}{15C2}=\frac{10+6+5*4}{105}=\frac{12}{35}$$
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Re: In United States currency, a nickel is worth 5 cents  [#permalink]

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26 Sep 2019, 22:06
3
either of two coins from two is nickel, (so basically the probability of at least one nickel from two picks)
Could you please explain this part?
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Re: In United States currency, a nickel is worth 5 cents  [#permalink]

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26 Sep 2019, 22:28
nishthagupta wrote:
either of two coins from two is nickel, (so basically the probability of at least one nickel from two picks)
Could you please explain this part?
Bunuel

We want to find the probability of picking a coin other than a nickel twice in a row. When picking two coins we can get:
{penny, penny}
{penny, dime}
{dime, dime}

{penny, nickel}
{dime, nickel}
{nickel, nickel}

So, we want to find the probability of green events, and the red events are the events we don't want. Basically we want any event which does not have a nickel and don't want any event which has at least one nickel.
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Re: In United States currency, a nickel is worth 5 cents   [#permalink] 26 Sep 2019, 22:28
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