
Could someone explain to my dumb head why we are seeking $P(X \leq2)$? Is it because that the "majority" of "five" is 3? And we want to find three correct transmissions?

Could someone explain to my dumb head why we are seeking $P(X \leq2)$? Is it because that the "majority" of "five" is 3? And we want to find three correct transmissions?
There are alltogether $5$ bits and you are interested in the total number of correct decisions for at least $3$ of them to say that a correct $1$ or correct $0$ is sent. It can also be interpreted that at most there are $k=2$ errors. this is exactly what $P(X\leq 2)$ is.