Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Let w be a complex cube root of unity,w is not equal to 1.A fair die is thrown three times.If r1,r2 and r3 are the numbers obtained on the die,then the probability that w^r1+w^r2+w^r3=0 is?

Let w be a complex cube root of unity,w is not equal to 1.A fair die is thrown three times.If r1,r2 and r3 are the numbers obtained on the die,then the probability that w^r1+w^r2+w^r3=0 is?

Grade:12th pass

1 Answers

Arun
25763 Points
2 years ago
w^1 + w^2 + w^3 = 0 
w^1 = w^4 
w^2 = w^5 
1 = w^3 = w^6 

Probability that r1 = 1 or 4, r2 = 2 or 5, r3 = 3 or 6 
= 2/6 * 2/6 * 2/6 
= 1/27 
If the values of r1, r2, r3 are reshuffled, then also we will get w^r1 + w^r2 + w^r3 = 0 
The values of r1, r2, r3 can be reshuffled in 3! = 6 wats 
Therefore, the required probability = 6 * 1/27 = 2/9 
Ans: 2/9 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free