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Four integers are randomly selected from the set {1,0,1}, with
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Updated on: 23 Apr 2017, 02:57
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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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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.



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Re: Four integers are randomly selected from the set {1,0,1}, with
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23 Apr 2017, 03:36
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



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Re: Four integers are randomly selected from the set {1,0,1}, with
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30 Apr 2017, 07:39
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



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Re: Four integers are randomly selected from the set {1,0,1}, with
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30 Apr 2017, 11:15
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 eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: Four integers are randomly selected from the set {1,0,1}, with
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30 Jun 2017, 04:28
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 eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts Thank you for your explanation, eGMAT 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. Thank you in advance.




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