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A miniature roulette wheel is divided into 10 equal sectors

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A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 03 Mar 2015, 14:05
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A miniature roulette wheel is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly determines the winning sector by settling in that sector. If the wheel is spun three times, approximately what is the probability that the product of the three winning sectors’ integers will be even?

A 88%

B 75%

C 67%

D 63%

E 50%
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 03 Mar 2015, 14:11
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A very nice problem from veritas prep that is much easier than it seems.

So, I actually drew the roulette, just to visualize it. I divided the circle in 10 sectors and wrote one number from 1 to 10 in each sector.

The question asks about the probability that in 3 spins the product of the 3 spins will be even.
What we know is that it only takes 1 even number in any multiplication for the product to result in an even number.

So, if we want at least 1 even, we can calculate the probability of not one even, or put differently all 3 spins end up in an odd number, and subtract this from 1.

5 out of 10 numbers are odd. So, the probability of odd is 5/10, which is 1/2.

We have 3 spins, so we calculate (1/2)^3 = 1/8 = 0.125 and 0.125*100 = 12.5 %.

100 - 12.5 = 87.5, so I guess we go with A, which is the closest.
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 03 Mar 2015, 14:19
1- Pr(three odds) = 1- (.5*.5*.5)=1-0.125=0.875
88%
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 16 Aug 2015, 11:50
My take, 50% chances in first, half of 50 or 25% chances in 2nd, half of that and its a 12.5% chances in third: add it up to get 88%, in the test I stopped at 75% since the number was clearly bigger than that and I had another half to still consider.
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 21 Apr 2017, 15:46
P(even) = 1/2 each time.
so let's find the P of no even numbers...i thought it's easier.
1 spin P(not even) = 1/2
2 spin P(not even) = 1/2
3 spin P(not even) = 1/2
P not even all 3 spins = 1/8
p to be even = 7/8 or 87.5 ~88%

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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 26 Jan 2018, 10:48
pacifist85 wrote:
A miniature roulette wheel is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly determines the winning sector by settling in that sector. If the wheel is spun three times, approximately what is the probability that the product of the three winning sectors’ integers will be even?

A 88%

B 75%

C 67%

D 63%

E 50%


We can use the formula:

P(even product) = 1 - P(odd product)

To get an odd product we must have all odd numbers, to the probability of 3 odds is:

1/2 x 1/2 x 1/2 = 1/8.

So P(even product) = 1 - 1/8 = 7/8 ≈ 88%

Answer: A
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 05 Feb 2018, 23:29
JeffTargetTestPrep wrote:
pacifist85 wrote:
A miniature roulette wheel is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly determines the winning sector by settling in that sector. If the wheel is spun three times, approximately what is the probability that the product of the three winning sectors’ integers will be even?

A 88%

B 75%

C 67%

D 63%

E 50%


We can use the formula:

P(even product) = 1 - P(odd product)

To get an odd product we must have all odd numbers, to the probability of 3 odds is:

1/2 x 1/2 x 1/2 = 1/8.

So P(even product) = 1 - 1/8 = 7/8 ≈ 88%

Answer: A



Hi, just a query,

Why are we not considering the probablitity of a winning sector , i.e 1/10?

why only probability of getting even or odd on winner?
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 09 Sep 2018, 14:11
pacifist85 wrote:
A very nice problem from veritas prep that is much easier than it seems.

So, I actually drew the roulette, just to visualize it. I divided the circle in 10 sectors and wrote one number from 1 to 10 in each sector.

The question asks about the probability that in 3 spins the product of the 3 spins will be even.
What we know is that it only takes 1 even number in any multiplication for the product to result in an even number.

So, if we want at least 1 even, we can calculate the probability of not one even, or put differently all 3 spins end up in an odd number, and subtract this from 1.

5 out of 10 numbers are odd. So, the probability of odd is 5/10, which is 1/2.

We have 3 spins, so we calculate (1/2)^3 = 1/8 = 0.125 and 0.125*100 = 12.5 %.

100 - 12.5 = 87.5, so I guess we go with A, which is the closest.




Vertias has this as a 550-600 level problem
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 11 Sep 2018, 09:29
ok , i know the previous answers are straight to the point ;
my take :
1-P(having an odd product)=1-P(the product of 3 odd integers)
P(the product of 3 odd integers) = number of choices of having 3 odd integers / number of all choices
number of choices of having 3 odd integers = OOO = 5^3 (Repetition and order matter)
number of all choices = 10^3
1-P (odd product) =(10^3-5^3)/10^3 =7/8=0.875
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 24 Jan 2019, 07:17
pacifist85 wrote:
A miniature roulette wheel is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly determines the winning sector by settling in that sector. If the wheel is spun three times, approximately what is the probability that the product of the three winning sectors’ integers will be even?

A 88%

B 75%

C 67%

D 63%

E 50%

\(?\,\,\, \cong \,\,\,1 - P\left( {\underbrace {{\text{all}}\,\,{\text{odd}}\,\,{\text{sectors}}}_{{\text{unfavorable}}}} \right)\)


\(P\left( {{\rm{all}}\,\,{\rm{odd}}\,\,{\rm{sectors}}} \right)\,\,\,\mathop = \limits^{{\rm{independency}}} \,\,\,{1 \over 2} \cdot {1 \over 2} \cdot {1 \over 2} = {1 \over 8}\)


\(? = 1 - \frac{1}{8} = \frac{7}{8} = \underleftrightarrow {7 \cdot \frac{{0.25}}{2} > 7 \cdot \frac{{0.24}}{2}} = 84\%\)

(Please note that our approximation is good enough for the alternative choices offered.)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 26 Jan 2019, 13:35
Hi All,

We're told that a miniature roulette wheel is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly determines the winning sector by settling in that sector and the wheel is spun three times. We're asked for the approximate probability that the PRODUCT of the three winning sectors’ integers will be even. This question is based on a few Number Properties and can be solved in a couple of different ways, depending on how you want to organize your work.

To start, the following Number Properties are worth knowing:
(Even)(Even) = Even
(Even)(Odd) = Even
(Odd)(Odd) = Odd

When multiplying integers, to end up with a PRODUCT that is EVEN, we need AT LEAST ONE of the integers to be Even. If ALL of the integers are ODD, then the product will be ODD. On this wheel, the probability of landing on an even number or landing on an odd number is the same: 1/2. We can either calculate the probability of ending up with an even product or the probability of ending with an odd product (and then subtracting that fraction from 1). As it stands, there are only 8 possible arrangements of the 3 winning results:

(Even)(Even)(Even) = Even
(Even)(Even)(Odd) = Even
(Even)(Odd)(Even) = Even
(Odd)(Even)(Even) = Even
(Even)(Odd)(Odd) = Even
(Odd)(Even)(Odd) = Even
(Odd)(Odd)(Even) = Even
(Odd)(Odd)(Odd) = Odd

The probability of ending with an EVEN product is 7/8 = .875 = 87.5% --> approximately 88%

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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 12 Mar 2019, 06:49
Hi Bunuel
I got that this can be easily done if we think of it is as "Atleast" problem, thus substracting it from one to get the answer. However can you please explain, suppose if we want to do it in normal way, I mean calculating the probability of first(EEE), then(OOE) and lastly(OEE), how can we do it?
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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 12 Mar 2019, 11:29
Hi rraman,

If you want to calculate the probability of each individual option that ends in an EVEN product, then you will have to do more than just the 3 calculations that you listed - there are actually SEVEN outcomes here that will lead to an EVEN product. They are:

(Even)(Even)(Even) = Even
(Even)(Even)(Odd) = Even
(Even)(Odd)(Even) = Even
(Odd)(Even)(Even) = Even
(Even)(Odd)(Odd) = Even
(Odd)(Even)(Odd) = Even
(Odd)(Odd)(Even) = Even

We know that HALF of the given sectors are EVEN and HALF are ODD, there is a 1/2 chance of a ball landing on an even or odd sector. Thus, each of the above calculations will be the SAME:

(1/2)(1/2)(1/2) = 1/8

With 7 possible ways to get an EVEN product, he probability of that outcome is (7))(1/8) = 7/8 = .875 = 87.5% --> approximately 88%

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A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 13 Mar 2019, 00:30
Hi Rich,
Thank you for the explanation, however I have one more query

EMPOWERgmatRichC wrote:
Hi rraman,

If you want to calculate the probability of each individual option that ends in an EVEN product, then you will have to do more than just the 3 calculations that you listed - there are actually SEVEN outcomes here that will lead to an EVEN product. They are:

(Even)(Even)(Even) = Even

(Even)(Even)(Odd) = Even
(Even)(Odd)(Even) = Even
(Odd)(Even)(Even) = Even


This contains 2 evens and one odd, only the order and number will change, can it be find out by using combinations ??

(Even)(Odd)(Odd) = Even
(Odd)(Even)(Odd) = Even
(Odd)(Odd)(Even) = Even


same with this one that contains 2 odds and one even


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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 13 Mar 2019, 15:56
Hi rraman,

Yes - if you recognize that each of those options can be 'grouped' into a Combination, then you can certainly approach the question that way. However, that requires a higher 'level' of math that's not really necessary once you recognize that the probability of ANY of those individual outcomes is 1/8.

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Re: A miniature roulette wheel is divided into 10 equal sectors  [#permalink]

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New post 14 Mar 2019, 07:31
pacifist85 wrote:
A miniature roulette wheel is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly determines the winning sector by settling in that sector. If the wheel is spun three times, approximately what is the probability that the product of the three winning sectors’ integers will be even?

A 88%

B 75%

C 67%

D 63%

E 50%


Hi Folks,
In order to find the probability that the product of the three winning sectors’ integers will be even, we need to find what are the chances that at least in one out of three times, the ball would stop at an even number. Why? because Even * Odd = Even.
So,
At least one = 1-None
None means all three are odd.
i.e. P(odd) = #Chances of odd/#total selections = 1/2
for all three times = (1/2)^3 = 1/8

Atleast one = 1-1/8 = 7/8

Since the question statement has asked percentage, Atleast one % = 7/8 * 100 = 87.5% = 88%(Round off)

Option A.

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Re: A miniature roulette wheel is divided into 10 equal sectors   [#permalink] 14 Mar 2019, 07:31
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