Fifteen tickets are numbered $\displaystyle 1,2,3....,15$ respectively.Seven tickets are selected at random one at a time with replacement.Find the probability that the largest number appearing on a selected ticket is $\displaystyle 9$.
question is with or with-out replacement..
If with replacement...
= (1/15)(9/15)^6
at least once 9 should come = 1/15
after that any thing form 1-9, 6 times = (9/15)^6
!------------------------------------
If with out replacement
= 9C7 / 15C7
selecting combinations = 15C7
selecting from 1-9 is 9C7
ok if it is with replacements... then how come we can "Seven tickets are selected at random one at a time" at a time ?? .. thats why i got confused.....
bye the way can you please give the explainaiton to the answer that you gave ??
It probably means that once a ticket is drawn the number on it is noted and it is put back into the bag.Then another ticket is drawn and so on the process is repeated 7 times.
I only know the answer.It's explanation is still out of reach to me.
I hope somebody will point out the flaw in your answer.I had done it the same way as you did
The following is the answer to the original question is:
$\displaystyle \sum\limits_{k = 1}^7 {\binom{7}{k}}{\left( {\frac{8}{{15}}} \right)^{7 - k} \left( {\frac{1}{{15}}} \right)^k} = 0.057194 = \frac{{9^7 - 8^7 }}{{15^7 }}$
Now why? The summation calculates the probabilities of having from 1 to 7 nines and all the other numbers less that nine.
But note that is the same as the other proposed answer: $\displaystyle \frac{{9^7 - 8^7 }}{{15^7 }}$.
There are $\displaystyle 9^7$ ways to select numbers that do not exceed 9.
There are $\displaystyle 8^7$ ways to select numbers that are less than 9.
So there are $\displaystyle 9^7-8^7$ ways to select numbers that do not exceed 9 having at least one nine.
There are at total of $\displaystyle 15^7$ ways to select the numbers .