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Senior Manager  G
Joined: 02 Mar 2017
Posts: 255
Location: India
Concentration: Finance, Marketing
Four integers are randomly selected from the set {-1,0,1}, with  [#permalink]

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2
8 00:00

Difficulty:   85% (hard)

Question Stats: 56% (02:22) correct 44% (02:14) wrong based on 112 sessions

HideShow timer Statistics Four integers are randomly selected from the set {-1,0,1},with repetitions allowed.What is the probability that the product of the four integers chosen will be its least value possible

A. 1/81
B. 4/81
C. 8/81
D. 10/81
E. 16/81

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Kudos-----> If my post was Helpful

Originally posted by VyshakhR1995 on 23 Apr 2017, 00:20.
Last edited by Bunuel on 23 Apr 2017, 02:57, edited 1 time in total.
Renamed the topic and edited the question.
Current Student G
Joined: 18 Jan 2017
Posts: 81
Location: India
Concentration: Finance, Economics
GMAT 1: 700 Q50 V34 Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink]

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1
1
The total probability is selecting any of the 3 integers in the 4 turns= 3*3*3*3= 81

Minimum Product (Least Value) = -1

Case 1: Selecting three "-1" and one "1"
=4!/3!=4

Case 2:Selecting three "1" and one "-1"
=4!/3!=4

Required Probability= (4+4)/81= 8/81
Intern  S
Joined: 05 Dec 2016
Posts: 7
Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink]

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Minimum Product of four items on the list = -1
Number of ways to achieve the result= 2 {(3 Nos -1 + 1 Nos 1) OR (3 Nos +1 + 1 Nos -1)}

Probability for case 1 (3 Nos -1 + 1 Nos 1)
Number of ways= 4 (+1,+1,+1,-1/+1,+1,-1,+1/.........)
Probability = 4x (1/3)^4 = 4/81

Probability for case 2 (3 Nos +1 + 1 Nos -1)
Number of ways= 4 Similar explanations as earlier
Probability = 4x (1/3)^4= 4/81

Hence total probability= 4/81 +4/81= 8/81
e-GMAT Representative D
Joined: 04 Jan 2015
Posts: 2893
Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink]

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VyshakhR1995 wrote:
Four integers are randomly selected from the set {-1,0,1},with repetitions allowed.What is the probability that the product of the four integers chosen will be its least value possible

A. 1/81
B. 4/81
C. 8/81
D. 10/81
E. 16/81

Solution

• We have a negative number in the set, so the least value will be possible, only when the product is negative.

• If the product is least and negative, we cannot select 0, so we need to select a combination of -1 and 1 to get a product of -1.

• Since 4 numbers are selected –

o We get a product of -1 if we select one -1 and three 1s.

 We can do that in 4C3 ways = 4 ways

o OR we can select three -1s and one 1.

 We can do that same in 4C3 = 4 ways.

• Thus, the total ways to get a least value $$= 4 + 4 = 8$$ ways

• Now total ways of selecting a number $$= 3*3*3*3 = 81$$

o Thus, the probability of choosing the 4 integers whose product has least value $$= \frac{8}{81}$$

• Hence, the correct answer is Option C.

Thanks,
Saquib
Quant Expert
e-GMAT

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts  _________________
Manager  S
Joined: 24 Jan 2017
Posts: 142
GMAT 1: 640 Q50 V25 GMAT 2: 710 Q50 V35 GPA: 3.48
Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink]

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EgmatQuantExpert wrote:
VyshakhR1995 wrote:
Four integers are randomly selected from the set {-1,0,1},with repetitions allowed.What is the probability that the product of the four integers chosen will be its least value possible

A. 1/81
B. 4/81
C. 8/81
D. 10/81
E. 16/81

Solution

• We have a negative number in the set, so the least value will be possible, only when the product is negative.

• If the product is least and negative, we cannot select 0, so we need to select a combination of -1 and 1 to get a product of -1.

• Since 4 numbers are selected –

o We get a product of -1 if we select one -1 and three 1s.

 We can do that in 4C3 ways = 4 ways

o OR we can select three -1s and one 1.

 We can do that same in 4C3 = 4 ways.

• Thus, the total ways to get a least value $$= 4 + 4 = 8$$ ways

• Now total ways of selecting a number $$= 3*3*3*3 = 81$$

o Thus, the probability of choosing the 4 integers whose product has least value $$= \frac{8}{81}$$

• Hence, the correct answer is Option C.

Thanks,
Saquib
Quant Expert
e-GMAT

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts  Thank you for your explanation, e-GMAT expert.

Btw I have a general question, hope that you will help me clear this concern:
How can we know whether order of selection matters in a probability question? In other words, what are the indicators of order of selection?

In the above question, initially I didn't consider the order of 4 numbers chosen, so I worked out an answer different from all 5 choices. But then I realized the denominator is very big, I thought I might overlook many cases... the selecting order could be the key... so I calculated again and figured out the correct answer.

Since the above question is not an official one, I cannot draw any conclusion as to whether there is any OG/GMATPrep question in which no indicator is provided, and we have to test each case. I believe that such an experience expert as you could give me a reliable answer.

Non-Human User Joined: 09 Sep 2013
Posts: 11430
Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink]

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_________________ Re: Four integers are randomly selected from the set {-1,0,1}, with   [#permalink] 15 Oct 2018, 15:38
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