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Let ωω be a cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ωr1 + ωr2 + ωr3 = 0 is
  • a)
    1/18
  • b)
    1/9
  • c)
    2/9
  • d)
    1/36
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Let ωω be a cube root of unity with ω ≠ 1. A fair die is thrown three...
Given:
Let ω be a cube root of unity with ω ≠ 1.
A fair die is thrown three times.
r1, r2, and r3 are the numbers obtained on the die.

To find:
The probability that ωr1 ωr2 ωr3 = 0

Solution:
The cube roots of unity are 1, ω, and ω^2, where ω^3 = 1.

Let's consider the possible values of ωr1 ωr2 ωr3. Since ω^3 = 1, we have:

ωr1 ωr2 ωr3 = (ω^3)^k * r1r2r3

This can be simplified as:

ωr1 ωr2 ωr3 = ω^3k * r1r2r3

Now, let's analyze the possible values of r1r2r3:

- If r1r2r3 = 0, then ωr1 ωr2 ωr3 = ω^3k * 0 = 0.
- If r1r2r3 ≠ 0, then ωr1 ωr2 ωr3 = ω^3k * (r1r2r3) ≠ 0.

Therefore, in order for ωr1 ωr2 ωr3 to equal 0, r1r2r3 must be equal to 0.

Calculating the probability:
Since a fair die is thrown three times, there are a total of 6^3 = 216 possible outcomes.

Out of these 216 outcomes, there are 6 outcomes where r1, r2, or r3 equals 0 (i.e., (0, a, b), (a, 0, b), (a, b, 0), where a and b can take values from 1 to 6).

Therefore, the probability of obtaining r1r2r3 = 0 is 6/216 = 1/36.

However, we need to consider the condition that ωr1 ωr2 ωr3 = 0. Since ω ≠ 1, we can only have ωr1 ωr2 ωr3 = 0 if r1r2r3 = 0.

Out of the 6 possible outcomes where r1r2r3 = 0, only 2 of them satisfy the condition ωr1 ωr2 ωr3 = 0 (i.e., (0, ω, ω^2) and (0, ω^2, ω)).

Therefore, the probability that ωr1 ωr2 ωr3 = 0 is 2/216 = 1/108.

Since the given options do not match this probability, it seems there might be an error in the given answer options.
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Community Answer
Let ωω be a cube root of unity with ω ≠ 1. A fair die is thrown three...
Total number of events = 6 × 6 × 6 = 216.
Given, ωr1 + ωr2 + ωr3 = 0
If one of r1​,r2​,r3​ ​ takes value from the set {3, 6} other takes values from the set {1, 4} and the third takes value from the set {2, 5}
The total number of these ways
=(​2C1 × ​2C1 × ​2C1​) × 3!.
So, favourable number of elementary events
=(​2C1 × ​2C1 × ​2C1​) × 3! = 48
Hence, required probability will be = 48 / 216 ​ = 2/9
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Let ωω be a cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ωr1 + ωr2 + ωr3 = 0 isa)1/18b)1/9c)2/9d)1/36Correct answer is option 'C'. Can you explain this answer?
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Let ωω be a cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ωr1 + ωr2 + ωr3 = 0 isa)1/18b)1/9c)2/9d)1/36Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let ωω be a cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ωr1 + ωr2 + ωr3 = 0 isa)1/18b)1/9c)2/9d)1/36Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let ωω be a cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ωr1 + ωr2 + ωr3 = 0 isa)1/18b)1/9c)2/9d)1/36Correct answer is option 'C'. Can you explain this answer?.
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