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A die is thrown continuously. Find

  1. the probability that number 6 will appear before the 10th throw
  2. the probability that number 6 will appear for a third time on the 10th throw
  3. 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 :)

Vasily Mitch
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Dot Point
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  • Welcome to MSE. Here is a tutorial on how to format math on this site. – saulspatz Oct 01 '19 at 14:40
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    For problem 1.1) you seem to have been looking only at the case where a six occurs exactly once within the first nine throws, but wouldn't you consider the cases where a six occurs multiple times within the first nine throws as also satisfying "the number 6 will appear before the tenth throw"? You also use the number $n$ in place of nine or ten throughout your attempts. – JMoravitz Oct 01 '19 at 14:48
  • As for 1.3) this is a geometric distribution and is well known and widely available in literature and resources. I strongly suggest deriving the formula yourself, but if you wish to see someone else's work explaining the solution see here. – JMoravitz Oct 01 '19 at 14:51
  • @saulspatz thank you for your comment, I'll check it out – Dot Point Oct 01 '19 at 15:02

1 Answers1

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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, $E=6$.

saulspatz
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    which means for the first one, I'll have smt like: 1 - 5^9 x 6^(n-9)/ 6^n ? – Dot Point Oct 01 '19 at 15:02
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    @DotPoint I can't understand what that is supposed to mean. You really need to learn MathJax. I updated my answer to show the probability. – saulspatz Oct 01 '19 at 15:12