Calculating Likelihood of Proposal Passage in Council Vote of 9 Members

[tmt1]

In a council vote for a proposal, there a nine council members. There is a 60% chance that each council person will vote in favour of the proposal. What is then likelihood that the proposal will pass?

So, this means that 5 or greater members must vote in favour.

These are independent events, so the odds that 5 vote in favour is $0.6 * 0.6 * 0.6 * 0.6 * 0.6 = 0.07776$,
that 6 vote in favour is $0.6 * 6 = 0.046$, that 7 vote in favour is $0.6 ^ 7 = 0.0279 $, that 8 vote in favour is $0.6 ^ 8 = 0.0167$ and that 9 vote in favour is $0.01$. Add all these up and I get 0.177 percent.

So this is incorrect, how do I fix my calculation to make it correct?

 

[MarkFL]

For each particular number of yes votes, let's say there are $k$ yes votes, and so $9-k$ no votes, we have to compute how many ways there are to choose $k$ from $9$ and then multiply that by the probability for one particular way to make that choice:

\(\displaystyle {9 \choose k}(0.6)^k(0.4)^{9-k}\)

And then, calling the event that it passes $X$, to find the probability of $X$ happening, we need to sum these up for $k=5$ to $k=9$:

\(\displaystyle P(X)=\sum_{k=5}^{9}\left({9 \choose k}(0.6)^k(0.4)^{9-k}\right)\)