IIT JEE , AIEEE Forum - Mathematics , Algebra

nikita19

THE BEST QUESTION ON PROBABILITY

the sum of a seven digit number is 59 find the probability that it is divisible by 11

please some one give me the full solution with explantion .

akhil_o

Hey!

I've attemptd the same q here

go to

http://www.goiit.com/posts/list/algebra-probability-39887.htm

mukulss

me too???????

akhil_o

@mukulss whats ur answer?

rv_hbk

For a no. to be divisible by 11,sum(even)-sum(odd)=multiple of 11.For this a minimum of 3 9's are required.

Case 1:

3 9's:hence sum left=59-27=32

4 places left,this can be attained if we take 4 8's.

eg:9898988

the above no. has been found by hit & trial. Now we've to arrange the even place digits and the odd place digits.Sample space=7!/4!3!=S

Probability=P(even)*P(odd)/S=(3!/3!)*(4!/3!)/S=4/35

Case 2:

4 9's: hence,sum left=59-36=23;

Only way to do this:4 9's,2 8's,one 7

eg:9998978

Probability=((4!/3!)*(3!)) / (7! /4!2!)=8/35

Case 3:5 9's,hence sum left=59-45=14

Subcase (a):5 9's,one 8,one 6

eg:9996998

probability=(4!/3!)*(3!/2!) / (7!/5!)=6/21

Subcase (b):5 9's,2 7's

eg:9997997

probability=(4!/3!)*(3!/2!) / (7!/5!2!)=12/21

Sum=6/7

The total probability cannot be found by simply adding these cases as each case can be rearranged among the nos.

Ans=4/21

akhil_o

if u check my link i got 6/65

nikita19

please someone help and explain which is the correct solution

saurabh_reincarnated

to me what rv_hbk said seems correct

he got the right point - to know possible criterias for divisibility.... and that a no. is div. by 11 only if sum of even and sum of odd -even plces must be a multiple of 11...... but i m not sure that he proceeded in right way........

check if i am correct

"abcd" be odd placed no. and "xyz" even placed .....

so no is axbyczd ...

now multiple of 11 means at most it can go upto 33 as if u put a=b=c=d and x=y=z=0 we get 36

now using multinomial theorm values of b,c,d can go from 0 to 9and a from 1 to9

now sum of xyz = 59- (a+b+c+d)

using multinomial theorm..... their diff. is either 11 or 22 or 33....

and for each value or abcd there are corresponding 3! values of xyz as their places can be interchanged

so [{(t^0+t^1 +........t^9)^3} (t^1+ t^2 +..........t^9) ( t^59)] [{(t^0+t^1 +........t^9)^3}{(t^1+ t^2 +..........t^9)}]^-1= 11 or 22 or 33 add all cases...

now we sud get it but dont forget to mutiply 3! to favourble cases... then find prob.

if u like the approach then plz. rate me.....

saurabh_reincarnated

common goiitians check this............

any one do this plz.................................................

rv_hbk

You can refer TMH.The same question is given.ANS=4/21.

I m dead sure about it!!!!!

BEGGING FOR RATES........

Please read my solution

Ask & Discuss Questions with Community & Experts