# Math Help - Probabilty

1. ## Probabilty

Fifteen tickets are numbered $1,2,3....,15$ 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 $9$.

2. 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

3. The question clearly states with replacement.

Answer is $\frac{9^7-8^7}{15^7}$

4. 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 ??

5. Originally Posted by vinay
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

6. The following is the answer to the original question is:
$\sum\limits_{k = 1}^7 {\binom{7}{k}}{\left( {\frac{8}{{15}}} \right)^{7 - k} \left( {\frac{1}{{15}}} \right)^k} = 0.057194 = \frac{{9^7 - 8^7 }}{{15^7 }}$

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: $\frac{{9^7 - 8^7 }}{{15^7 }}$.
There are $9^7$ ways to select numbers that do not exceed 9.
There are $8^7$ ways to select numbers that are less than 9.
So there are $9^7-8^7$ ways to select numbers that do not exceed 9 having at least one nine.
There are at total of $15^7$ ways to select the numbers .

7. Originally Posted by vinay
question is with or with-out replacement..

If with replacement...

= (1/15)(8/15)^6

at least once 9 should come = 1/15
after that any thing form 1-7, 6 times = (8/15)^6
What is wrong with this argument?

8. Originally Posted by pankaj
If with replacement...= (1/15)(8/15)^6
at least once 9 should come = 1/15
after that any thing form 1-7, 6 times = (8/15)^6
What is wrong with this argument?
That is the probability of getting exactly one nine in a particular position.

9. Originally Posted by Plato
That is the probability of getting exactly one nine in a particular position.
So can $\frac{1}{15}\left(\frac{9}{15}\right)^6$ be said to be correct

10. Originally Posted by pankaj
So can $\frac{1}{15}\left(\frac{9}{15}\right)^6$ be said to be correct
Absolutely not. $\frac{9^7 -8^7}{15^7}$ is the correct answer to the OP.

11. does not give the probability of giving 9 alteast once... ????

12. Originally Posted by vinay
does not give the probability of giving 9 alteast once... ???
The answer is no for all the reasons listed above.