I would appreciate if you could check if I obtained the right results:
The problem:
A drunk man is walking in random steps along an axis with the points +-1, +-2,+-3,... each steps he does is in the length of 1 unit with the probability of 0.4 forward and 0.6 backwards(the steps are undependable). X will mark his placement on the axis after 50 steps.
1. What is $p\{x=-10\}$
My answer: The probability that the drunken man will be at position $x=-10$ after 50 steps is $0.053$
2. What is the variance of $x$?
My answer: Using the binomial variance distribution $var\left(x\right)=\binom{n}{k}p^k\left(1-p\right)^{n-k}$ i obtained that the variance is 24
3. What are the odds that his last step ($50$th) will be at $x = -27$?
My answer: The odds of his last step, the $50$th, will be at position $x = -27$ is $0.130$
4. Assuming the the chances of the drunken man of falling in each step is $0.01$: if the drunken man walks for $2000$ steps, what are the odds in estimation that he will fall exactly $23$ times?
My answer: I am quite sure I got this one wrong, but after 2000 steps, the probability he'll fall exactly 23 times is $0.72$
Can anyone check if I got the correct results?
EDIT: for some reason when i try to write the full calculations it goes wrong, so i'll explain what i did and which formula i used.
for 1:$C_50,20$ with the given probabilities times $20$ and $30$ accordingly
for 2: using the variance distribution formula along with the given details: $var\left(x\right)=\binom{n}{k}p^k\left(1-p\right)^{n-k}$
for 3: for some reason now i can't find a way to land on $-27$ on the last step(the 50th)
for 4:using the same formula, $c_2000,23$ with 0.1 times 23 and 0.99 times 1977 (calculated using complements(1-p))