A team wins with a probability of $0.4$. Suppose that it plays $9$ matches.
(a) What is the probability that it will win exactly $5$ matches?
(b) What is the probability that this team will win $4$ matches before losing $5$ matches? [Hint: one possible way is to condition on the result of the first match and build a generic iterative expression for the probability of $n$ successes before $m$ failures.]
The answer to the part (a) solved by random variables looks like as follows, is it the correct approach?
$P(A=5) = ((1-0.4)^4) \cdot 0.4^5$
What could be the way of solving the part b?
P(Win) = 0.4 P(Lose) = 1-0.4 = 0.6 Assuming that only two events win and lose is possible. Number of ways of choosing 5 matches out of 9 matchers = $\binom{9}{5}$. Probability that it will win exactly 5 mtaches = $\binom{9}{5}$ $0.4^5$ $0.6^4$. Probability that this team will win 4 matches before losing 5 matches = $0.4^4$ $0.6^5$. Since there is only one way of winning 4 matches before losing 5 matches.