The Hockey team for California is visiting Florida to play a series comprising five matches. In each match, assume that the California has a $70\%$ chance of winning. Further, assuming that the matches are independent of each other, what is the probability that:
a. The California will win $3$ matches in the series b. The California will win $4$ matches in the series
As per me, the solution to the above is:
- Probability of team wining $3$ matches: $${5 \choose 3} \left(\dfrac{3}{10}\times\dfrac{3}{10}\times\dfrac{7}{10}\times\dfrac{7}{10}\times\dfrac{7}{10}\right) = \dfrac{10* 3087}{100000} = 0.3087$$
- Probability of team wining 4 matches: $${5 \choose 4} \left(\dfrac{3}{10}\times\dfrac{7}{10}\times\dfrac{7}{10}\times\dfrac{7}{10}\times\dfrac{7}{10}\right) = \dfrac{5\times 7203}{100000} = 0.36015$$
It turns out that winning probability for $4$ matches is more than winning $3$ matches. That sounds little confusing, as for a layman chances for winning $3$ matches should be greater than $4$ matches.
Is the above solution even correct ? Am I doing something wrong here ?
As discussed in the comment, the question was apparently due to a failure to distinguish between at least $n$ matches and exactly $n$ matches. There is no contradiction between the fact that the probability to win at least $n$ matches is greater or equal than the probability to win at least $n+1$ matches and the fact that in some cases the probability to win exactly $n+1$ matches is greater than the probability to win exactly $n$ matches.