A miniature roulette wheel is divided into 10 equal sectors,

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by AAPL » Thu Jan 24, 2019 5:17 am

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Veritas Prep

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%

OA A

Source: Beat The GMAT — Problem Solving | Thu Jan 24, 2019 5:17 am

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

A miniature roulette wheel is divided into 10 equal sectors,

by fskilnik@GMATH » Thu Jan 24, 2019 5:26 am

AAPL wrote:Veritas Prep

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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Thu Jan 24, 2019 10:10 am

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%

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8132; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Thu Jan 31, 2019 5:39 pm

AAPL wrote:Veritas Prep

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%

OA A

In order to have an even product, at least one of the numbers must be even.

If the 1st number is even, then it doesn't matter what the 2nd and 3rd numbers are. So P(1st is even) = 1/2 x 1 x 1 = 1/2.

If the 1st number is odd and the second number is even, then it doesn't matter what the 3rd numbers is. So P(1st is even and 2nd iis odd) = 1/2 x 1/2 x 1 = 1/4.

If the 1st number is odd and the second number is odd, then the 3rd numbers must be even so that the product is even. So P(1st is even and 2nd is odd and 3rd is odd) = 1/2 x 1/2 x 1/2 = 1/8.

Since all three cases above are mutually exclusive, the probability that the product of the three winning sectors' integers will be even is 1/2 + 1/4 + 1/8 = 7/8 = 87.5% or approximately 88%.

Alternate solution:

In order to have an even product, at least one of the numbers must be even. The only time we don't have any even number is if the ball lands on an odd number for all three spins. In other words, the ball lands on 3 odd numbers in 3 spins. The probability that this will occur is 1/2 x 1/2 x 1/2 = 1/8. So the probability that this will not occur, i.e., on at least one spin the ball will land on an even number, is 1 - 1/8 = 7/8 = 87.5%, or approximately 88%.

Answer: A

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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