Four integers are randomly selected from the set {-1,0,1}, with

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

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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.

Thank you in advance.

guys, ssory for the stupid question, but i do not get how the total ways of selecting number is 81?

I didn’t follow the combinations approach. I used Probability of independent events and the binomial theorem.


The minimum value we can get as a Product of 4 values picked from the given Set is ———> (-1)

Given that we are pulling 4 numbers with NO REPLACEMENT - we can accomplish this in 2 ways


Scenario 1:

Pull -1
Pull -1
Pull -1
Pull +1


OR

Scenario 2:

Pull -1
Pull +1
Pull +1
Pull +1


Call the prob of pulling -1 ——-> P(A) = 1/3

Call the prob of pulling +1 ———> P(B) = 1/3

Scenario 1:
A - A - A - B

The different ways this event can occur is given by: 4!/3! = 4 ways

AND

P(A) * P(A) * P(A) * P(B) = (1/3) (1/3) (1/3) (1/3) = 1/81

Probability that scenario 1 occurs = (4) (1/81) = 4/81


OR

Scenario 2:

A - B - B - B

The logic will be the same, as well as the calculation

(4!/3!) * (1 / 3^4) = 4/81


Probability of getting a Product of (-1) from 4 picks from the set with No Replacement is equal to:

(4/81) + (4/81) =

8/81

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