Question

Find the probaility that 1 , 3 , 3 , 5 , 5 ,5 picked at random; does not end with 33

Answer 1

a 2 digit number picked at random , right??

Answer 2

a 2 digit number picked at random..?

I think that you use all 6 digits

Answer 3

ok,a 6 digit number picked at random does not end with 33....is it?

Answer 4

I guess so. I tried it but i did not get the right answer.

Answer 5

I got \[\frac{4!}{2!} *\frac{3*4}{3!} \]in which failed

Answer 6

what do u get as total numbers formed? and obviously, repetition is allowed here isn't it ?

Answer 7

total numbered form is obv. 60. and repetition is allowed i suppose.

Answer 8

ok,answer my this question , how many 4 digit numbers u can form with 1,5,5,5.

Answer 9

4!/3!

Answer 10

so is the final answer 56 ?

Answer 11

the answer is 56. but how

Answer 12

i mean 56/60 is the probability.
i just subtracted the total 'numbers' not valid (i.e. one with 33 in the end) from total numbers, so 60-4=56

Answer 13

i found total numbers invalid by fixing 33 in the last 2 digits of 6 digit number.
then the possible combinations are there only for 1st 4 places with 1,5,5,5.
did u get it?

Answer 14

tell me if u didn't get @Omniscience

Answer 15

yes i do. \[60 - \frac{4!}{3!} *\frac{2!}{2!} = 56\]
but is there by change another method; wher i dont need to minus the invalid by the total?