A die is thrown continuously. Find
- the probability that number 6 will appear before the 10th throw
- the probability that number 6 will appear for a third time on the 10th throw
- expected value on the number of throws needed to observe number 6 for the first time
Here is my approach:
1.1) Consider as we have to fill $n$ places (cause we don't know how many times the die is thrown). Every places can be filled in $6$ ways.
We now have to count the no. of favourable cases, we have to place a six at any of the first $9$ places => $\binom91$ ways
Hence, Total no. of favourable cases is $\binom91 \times 5^{n-1}$.
So $$P = \binom91 \times \frac{5^{n-1}}{6^n}$$
1.2) This time, we fix a $6$ at the $10$th throw, and we have to place $2$ six at any of the first $9$ places => $\binom92$ ways
Hence, total no. favourable cases is: $\binom92 \times 1 \times 5^{n-3}$
1.3) I have no idea about this. Please help me :)
Your answer to number 1 is incorrect. The probability that $6$ appears before the tenth rolls is the complement of the probability that none of the first $9$ rolls is a $6$. You seem to be thinking that the $6$ can appear exactly once. The probability is $$1-\left(\frac56\right)^9$$
Your approach in the second case is correct. You have to divide by $6^{10}$ to get the probability, though.
For the third one, this is the expectation of a geometrically-distributed random variable. Let E be the expected number of throws until a $6$ appears. Then $$E=1+\frac56E$$ because we always have to throw once and $\frac56$ of the time, a $6$ does not appear, and we are back where we started. Thus, $E=6$.