Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

03 Apr 2010, 11:31

11

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

75% (01:26) correct
25% (01:54) wrong based on 346 sessions

HideShow timer Statistics

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?

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): \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\)

Re: Help with a probability question. [#permalink]

Show Tags

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).

Re: Help with a probability question. [#permalink]

Show Tags

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.
_________________

Re: Help with a probability question. [#permalink]

Show Tags

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: \(\frac{non-nickel}{all}*\frac{non-nickel -1}{all-1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}\).

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: \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\)

\(P=1-\frac{23}{35}=\frac{12}{35}\)

Answer: B.

so simple and elegant.. +1 for you
_________________

If you like my post, consider giving me a kudos. THANKS!

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

Show Tags

14 May 2014, 13:37

1

This post received KUDOS

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 ... .. .
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Re: Help with a probability question. [#permalink]

Show Tags

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): \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\)

\(P=1-\frac{23}{35}=\frac{12}{35}\)

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?

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): \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\)

\(P=1-\frac{23}{35}=\frac{12}{35}\)

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 non-nickel) = P(two pennies) + P(two dimes) + P(one nickel and one dime).

As for your second question: (not nickel, not nickel) is one case.
_________________

Re: Help with a probability question. [#permalink]

Show Tags

27 May 2014, 16:36

Bunuel wrote:

russ9 wrote:

Bunuel wrote:

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): \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\)

\(P=1-\frac{23}{35}=\frac{12}{35}\)

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 non-nickel) = P(two pennies) + P(two dimes) + P(one nickel and one dime).

As for your second question: (not nickel, not nickel) is one case.

Hi Bunuel,

Second statement is clear but i'd like to challenge the first:

P Non Nickel = 1 - P Nickel = 1 - (P two pennies + P two Dimes + P One Nickel and One Dime). This statement is clear to me.

The solution I was suggestion above was not use the "1-p" methodology, rather a straight P of Non Nickel. Isn't P Non Nickel = (P of Pennies on the first try and P of Dimes on the second try)*2 since order doesn't matter. This seems completely completely different than above since i'm not even accounting for One Nickel in there and therefore following the same logic as (9/15)(8/14). What am I missing here?

Re: Help with a probability question. [#permalink]

Show Tags

27 May 2014, 23:18

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): \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\)

\(P=1-\frac{23}{35}=\frac{12}{35}\)

Answer: B.

Hello Bunuel

I am confused with the wording of the Question It Says: 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?

Is it asking us to find P ( two Pennies back to back) + P ( two Dimes back to back ) OR P(two pennies) + P(two dimes) + P(one nickel and one dime).

To me it seems to be asking P of picking a type of coin (either dime or Penny ) 2 times in a row Only & not Either Penny or Dime also.

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

Show Tags

11 Jun 2015, 13:23

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

14 Aug 2016, 14:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

21 Jan 2017, 02:12

niyantg 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): \(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\)

\(P=1-\frac{23}{35}=\frac{12}{35}\)

Answer: B.

Hello Bunuel

I am confused with the wording of the Question It Says: 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?

Is it asking us to find P ( two Pennies back to back) + P ( two Dimes back to back ) OR P(two pennies) + P(two dimes) + P(one nickel and one dime).

To me it seems to be asking P of picking a type of coin (either dime or Penny ) 2 times in a row Only & not Either Penny or Dime also.

Please Help

Thank you

Bunuel... For me, I took the Q as asking to find the prob of picking a coin in two picks like, penny+penny OR dime+dime OR penny+dime OR dime+penny.... please correct me as I dont take it as asking to pick a nickel+penny OR nicke+dime??? Is my reasoning correct? coz for according to ur solution, Prob(Non Nickels) = Prob(two Pennys) +Prob(two dimes) + Prob(penny and dime) +Prob(dime and penny), since underline part will be same so multiply Prob(dime+penny) with 2....????
_________________

If someone can do it..... Why not can I???

GMAT Clubbers,,,this can be a much better place if we opt for a usual Thanksgiving and Goodwill gesture by endorsing Kudos!!! So, +1 here if you like my posts....

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

Show Tags

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
_________________

What was previously thought to be impossible is now obvious reality. In the past, people used to open doors with their hands. Today, doors open "by magic" when people approach them

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.
_________________

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

Show Tags

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 \(\frac{8}{14}\)

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...