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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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  45% (medium)

Question Stats:

67% (02:15) correct 33% (02:14) wrong based on 309 sessions

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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 &nbs [#permalink] 11 Sep 2018, 09:29
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A miniature roulette wheel is divided into 10 equal sectors

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