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555-605 Level|   Probability|      
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changhiskhan
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In United States currency, a nickel is worth 5 cents 03 Apr 2010, 11:31
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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25% (medium)

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80% (02:09) correct 20% (02:47) wrong based on 691 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

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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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Bunuel
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In United States currency, a nickel is worth 5 cents 03 Apr 2010, 12:01
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changhiskhan
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.
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nickk
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In United States currency, a nickel is worth 5 cents 04 Apr 2010, 05:48
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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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In United States currency, a nickel is worth 5 cents 04 Apr 2010, 08:51
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Bunuel
changhiskhan
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
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In United States currency, a nickel is worth 5 cents 14 May 2014, 13:37
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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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In United States currency, a nickel is worth 5 cents 26 May 2014, 14:17
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Bunuel
changhiskhan
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?
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In United States currency, a nickel is worth 5 cents 27 May 2014, 01:08
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russ9
Bunuel
changhiskhan
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 penny and one dime).

As for your second question: (not nickel, not nickel) is one case.
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In United States currency, a nickel is worth 5 cents 17 Jul 2017, 04:19
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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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In United States currency, a nickel is worth 5 cents 24 Aug 2017, 21:42
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Can anyone please clarify why we don't multiply (9/15) * (8/14) by 2! ?
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In United States currency, a nickel is worth 5 cents 24 Aug 2017, 21:49
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ranaazad
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 26 Sep 2019, 22:06
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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
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In United States currency, a nickel is worth 5 cents 26 Sep 2019, 22:28
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nishthagupta
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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