In a box there are some numbered balls. The numbers on the ball are

Question

In a box there are some numbered balls. The numbers on the ball are either the prime factor of 102021 or the roots of polynomial equation of \((x^4) - (31x^3) + (321x^2) - 1241x + 1430 = 0.\) If 3 balls are randomly taken out of the box, find the probability that the sum of three numbers is an even number.

(A) \(\frac{6}{7}\)

(B) \(\frac{4}{7}\)

(C) \(\frac{3}{7}\)

(D) \(\frac{1}{7}\)

(E) \(0\)

(A)  \frac{6}{7}

(B)  \frac{4}{7}

(C)  \frac{3}{7}

(D)  \frac{1}{7}

(E)  0

chetan2u · Accepted Answer

PLease follow the rules and mention the source. This in no way is a GMAT type question.

I'll still give a solution as some part of it might help in other questions.
 Although the question is not worded properly, I believe it means that the numbers on the balls are prime factors of 102021 and are roots of given polynomial.

Prime factor of 102021  - 3*31*1097
 Solution of the polynomial : The degree 4 in  x^4  tells us that there are 4 roots, and 1430 is product of these 4 roots. 1430=2*5*11*13
We can also solve the polynomial but it will entail some complex steps.

So we have numbers: 2, 3, 5, 11, 13, 31 and 1097
Sum of three numbers will be EVEN when one is 2 and other two could be anything. Thus 1*6C2=  \frac{6*5}{2}=15 
Total ways = selecting 3 out of given 7 numbers = 7C3 =  \frac{7*6*5}{3*2}=35

Probability(even) =  \frac{15}{35}=\frac{3}{7}

Kinshook · Answer

Given: In a box there are some numbered balls. The numbers on the ball are either the prime factor of 102021 or the roots of polynomial equation of  (x^4) - (31x^3) + (321x^2) - 1241x + 1430 = 0.

Asked: If 3 balls are randomly taken out of the box, find the probability that the sum of three numbers is an even number.

102021 = 3*31*1097
{3,31,1097} are prime factors of 102021.

x^4 - 31x^3 + 321x^2 - 1241x + 1430 = 0 
 x^4 - 31x^3 + 321x^2 - 1241x + 1430 = 0 
x = {2,5,11,13} are the roots of the above polynomial.

We have {2,3,5,11,13,31,1097} as the set of numbers on the balls.

Since only 2 is an even number, for the sum of 3 numbers to be an even number, 1 number must be 2 and other 2 odd numbers.

Total ways of selecting 3 numbers out of {2,3,5,11,13,31,1097} = 7C3 = 35
Ways of selecting 3 numbers such that their sum is an even number = 6C2 = 15

The probability that the sum of three numbers is an even number =  \frac{15}{35} = \frac{3}{7}

IMO C

samarpan.g28 · Answer

Prime factors of 102021=3,31,1097.
The given equation will have 4 roots since the highest power of x is 4. Now, 1430=2*5*11*13.
Therefore, balls will have the numbers 2,3,5,11,13,31,1097 i.e. total 7 numbers.
To get the sum of any three numbers even, we have to take one number as 2, then choose any 2 numbers from the remaining 6 numbers in 6C2=15 ways.
Total ways of selecting 3 numbers 7C3=35 ways.
15/35=3/7(C)

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In a box there are some numbered balls. The numbers on the ball are

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