MHT CET · Maths · Probability

Let \(\omega\) be a complex cube root of unity with \(\omega \neq 1\). A fair die is thrown three times. If \(r_1, r_2\) and \(r_3\) are the numbers obtained on the die, then the probability that \(\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0\), is

A \(\frac{1}{36}\)
B \(\frac{1}{8}\)
C \(\frac{1}{9}\)
D \(\frac{2}{9}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{2}{9}\)

Step-by-step Solution

Detailed explanation

Solution:
\(\because \omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0\)
\(\Rightarrow \text { each of } r_1, r_2, r_3 \text { belongs to each of the categories}\) \(3 k, 3 k+1, 3 k+2\)
So the required probability
\(3 \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}=\frac{2}{9}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app