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

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mathbeginner
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In a box there are some numbered balls. The numbers on the ball are [#permalink] New post   Updated on: 24 Oct 2021, 02:09
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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) \(0\)
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C

Originally posted by mathbeginner on 23 Oct 2021, 22:46.
Last edited by Bunuel on 24 Oct 2021, 02:09, edited 2 times in total.
Renamed the topic and edited the question.
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Re: In a box there are some numbered balls. The numbers on the ball are [#permalink] New post   24 Oct 2021, 00:54
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mathbeginner wrote:
In a box there are some numbered balls. The number at the balls is the prime factor of 102021, and also the roots of polynomial equation of (x^4) - (31x^3) + (321x^2) - 1241x + 1430 = 0.

If 3 balls are taken out of the box, find the probability of the sum of three numbers is an even number!

Posted from my mobile device



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) = \(\frac{15}{35}=\frac{3}{7}\)
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Re: In a box there are some numbered balls. The numbers on the ball are [#permalink] New post   24 Oct 2021, 12:03
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
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Re: In a box there are some numbered balls. The numbers on the ball are [#permalink] New post   24 Jan 2024, 12:46
mathbeginner wrote:
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) \(0\)


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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Re: In a box there are some numbered balls. The numbers on the ball are [#permalink]
24 Jan 2024, 12:46
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