MHB Calculating Likelihood of Proposal Passage in Council Vote of 9 Members

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In a council vote with nine members, each having a 60% chance of voting in favor, at least five votes are needed for the proposal to pass. The initial calculation incorrectly estimated the likelihood of passage at 0.177 percent by simply multiplying probabilities for individual outcomes. The correct approach involves using the binomial probability formula, which accounts for the number of ways to achieve each possible outcome. By summing the probabilities for five to nine favorable votes, the accurate likelihood of the proposal passing can be determined. This method provides a comprehensive calculation of the proposal's chances in the council vote.
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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?
 
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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:

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

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 $k=9$:

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