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Consider a discrete-time counting process $X_n$ with $n = 0,1,...$ so that $X_0 = 0$ and $X_n = X_{n-1} + \xi_n$ for each $n \geq 1$, where $\xi_1,\xi_2,\ldots$ are iid Bernoulli random variables with the common probability of success $p = 0.4$. Given that $X_8 = 2$ compute: $P(X_{14} = 4 | X_8 = 2)$.

So far I have tried that since we already have 2 successes at 8 trials, this means that we need 2 more in the trials $n = 9$ to $n = 14$ and so this would be with probability $0.4*0.4*0.6*0.6*0.6*0.6$ where there are 2 successes and 4 fails which would be enough since we are trying to get 4 successes in total at $n=14$. However, this was marked as incorrect. I would appreciate if someone could lead me in the right direction.

waterr
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  • What about $(.4)(.6)(.4)(.6)(.6)(.6)$ and $(.6)(.6)(.4)(.6)(.4)(.6)$ and so on? Do you know what a binomial random variable is? – Michael Jan 30 '24 at 01:10
  • @Michael Wouldn't that result in the same decimal number for each permutation since multiplication is associative? Or should I add up the probabilities since there are multiply ways of achieving 2 successes in 6 trials – waterr Jan 30 '24 at 01:17
  • Yes it would. Now count how many such expressions there are (currently you are only including one of the expressions). – Michael Jan 30 '24 at 01:18
  • Since there are 6 trials and we need 2 successes, would it be 6C2 = 15 and then the answer would be the decimal value raised to the 15th power? – waterr Jan 30 '24 at 01:25
  • I follow ${6 \choose 2}$. I do not follow raising something to the 15th power...? – Michael Jan 30 '24 at 01:29
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    Take 3 trials, each independently a "success" with probability $p$. What is the probability there is exactly 1 success? Note the probability the first is successful, the second is not, the third is not, is $p(1-p)(1-p)$. And of course this is not the only way we can have exactly 1 success. – Michael Jan 30 '24 at 01:31
  • So, if there would be 3 ways to have 1 success, p(1-p)(1-p), (1-p)p(1-p), (1-p)(1-p)p and so since we have 3 different permutations, it would be 3 times the value of p(1-p)(1-p) (since they all equal the same decimal value)? – waterr Jan 30 '24 at 01:56
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    Yes, this is how the Binomial random variable works. – Michael Jan 30 '24 at 04:36

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