What percentage of games does a team have to win for there to be at least a 20% chance the team doesn't lose two games in a row over the course of the season. For the sake of the problem, assume that the team has a 90% chance of winning each game and that there are 50 games in a season.
I am new to probability and am trying to learn it this summer before I take a class that requires knowledge of it. I saw this problem in my textbook and I am not really sure how to approach it so any help would be appreciated!
if you lose $n_l$ games your season could be represented by a string of 50 characters consisting of $n_l$ "L"s and $(50-n_l)$ "W"s
out of the 50 positions choose $n_l$ of them to be "l"s and fill the rest with "W"s
There are $\binom{50}{n_l}$ equally probable strings
To count the number of these strings in which the "l"s are all isolated imagine that you begin forming your string with $(50 - n_l)$ "W"s and there are now $(51-n_l)$ places to position the $n_l$ "L"s
There are $\binom{51-n_l}{n_l}$ equally probable strings of this type. Note that $\binom nk =0$ whenever $k>n$ so there are zero strings of this type if $n_l>25$
We can now write the probability that a team goes through a season without back to back losses as a function of $n_l$
$$ P(n_l) = \dfrac{\binom{51-n_l}{n_l} }{\binom{50}{n_l} } $$
plugging in some values for $n_l$ ...
$$P(5)=0.65 : P(8) =0.27: P(9) = 0.18$$
So you must lose fewer than 9 games which means winning at least 42 games (84% of the games) will give you better than 20% chance of having no back to back losses.