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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, 10:31
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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, 11:01
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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: Help with a probability question. [#permalink]

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04 Apr 2010, 04:24
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, 04:48
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, 05:03
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, 07:51
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:
$$\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}$$

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

so simple and elegant.. +1 for you
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Re: Help with a probability question. [#permalink]

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08 Oct 2010, 09:16
B it is. 9/15 x 8/14 = 12/35
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Re: In United States currency, a nickel is worth 5 cents [#permalink]

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14 May 2014, 12:37
1
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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: Help with a probability question. [#permalink]

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26 May 2014, 13:17
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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Re: Help with a probability question. [#permalink]

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27 May 2014, 00:08
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 nickel and one dime).

As for your second question: (not nickel, not nickel) is one case.
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Re: Help with a probability question. [#permalink]

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27 May 2014, 15:36
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):
$$\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}$$

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

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?

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

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27 May 2014, 22:18
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):
$$\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}$$

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

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.

Thank you

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

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

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

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21 Jan 2017, 01:12
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):
$$\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}$$

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

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.

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

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17 Jul 2017, 03:19
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, 20:42
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, 20:49
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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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In United States currency, a nickel is worth 5 cents [#permalink]

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25 Aug 2017, 17:28
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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29 Aug 2017, 05:17
hardnstrong wrote:
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

i tried by similar approach but could you please explain why 5/15*4/14 + 4/15*5/14 this was repeated twice?

Thanks,
Arpit

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Re: In United States currency, a nickel is worth 5 cents   [#permalink] 29 Aug 2017, 05:17
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