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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?

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Yes, when there are less than or equal to $2$ errors, you will receive the correct bit. – Andrew Jul 29 '12 at 1:20
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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.

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