We want to prove

which is akin to saying that, for any given
, we guarantee that

for all
exceeding some threshold
.
We want to end up with


which suggests that we guarantee that
is arbitrarily close to 1 if
. (Take the ceiling to ensure
is a natural number.)
Now for the proof: Let
be given. Then for
we have

QED
Answer: 2
Step-by-step explanation: because the x in the middle has to be the same
This problem fits the conditional probability formula very well. The formula is P(B|A) = P(B ∩ A)/P(A). If event A is winning the first game, and event B is winning the second, then P(B ∩ A) = 0.44, and P(A) = 0.6. So P(B|A) is obtained by dividing 0.44 by 0.6, which is about 0.733.