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v_gurucharan

Q1. WHAT IS THE PROBABILITY OF THROWING ATLEAST TWO SEVENS IN THREE THROWS OF A PAIR OF DICE ?

ANS: 2/27

Q2. SHOW THAT THE SUM OFCOMPLEX CUBE ROOT OF 1 , TO THE POWER OF n AND ANOTHER COMPLEX CUBE ROOT OF 1, TO THE POWER OF n , EQUAL -1 , IF n IS NOT A MULTIPLE OF 3.

{ HINT: COMPLEX CUBE ROOT OF 1 IS ----- OMEGA}

KINDLY SHOW ALL THE STEPS.

THANKYOU.

sss

we know that, 1+w+w^2=0

now, w^n is not 1,bcoz n=not the integral multiple of 3

therefore w^n=either w or w^2

now , w^2n=(w^n)^n i.e. w^n or (w^2) ^ n.......................(from above eqn.)

it can not be = w^n bcoz n= not the integral multiple of 3

therefore it is = (w^2)^n =either w^2 0r w

1+w^n +w^2n=1+w+w^2 or 1+w^2+w

and both of them are equal to 0

threfore w^n+w^2n= -1

Moderator

Please refer to the post below:
http://www.goiit.com/posts/list/3729.htm#16510

v_gurucharan

thank u SACHIN!!!!

can anyone solve the first question????

v_gurucharan

Q1. WHAT IS THE PROBABILITY OF THROWING ATLEAST TWO SEVENS IN THREE THROWS OF A PAIR OF DICE ?

v_gurucharan

sorry moderator but pls. get my ques. answered

hhvivek

Sorry yar but how could u get a seven when u throw a dice.Do u mean getting two sevens as sum from the two or the three dice?

amar.gupta

Dear,

see, if you throw a pair of dies total no. of possibilities will be:

(1,6) (2,6)................(6,6) (1,5).......................=6*6=36

no. of favourable cases will be : (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)=6 cases

so probability of getting the sum seven will be 6/36=1/6

now you throw dices three times:

so prob. of getting seven atleast two times = probability of getting seven exactly 2 times+ probability of getting seven all the three times

= ³c₂(1/6)² (5/6) + ³c₃(1/6)³ where 5/6 is the probability of not getting seven =(15/216)+(1/216)

=16/216

=2/27

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