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If a team has a 2/3rd chance of winning any game, what is the probability that it wins at most 4 out of 5 games? The answer I got was

The chances that they win no games = 1/243
The chances that they win one game = 10/243
The chances that they win two games = 40/243
The chances that they win three games = 80/243
The chances that they win four games = 80/243

For a total probability of 211/243 that they win at least 4 out of 5 games.

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  • $\begingroup$ It is too much work. Also, there is a slip at the end, you should have written "at most" not "at least" in the last line. $\endgroup$ – André Nicolas Nov 26 '13 at 20:51
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That is correct, but you could just use complementary counting. The probability that the team will win all five games is $ \left( \dfrac {2}{3} \right)^5 $. Thus, the probability that the team will not win all $5$ games is $$ 1 - \left( \dfrac {2}{3} \right)^5 = 1 - \dfrac {32}{243} = \boxed {\dfrac {211}{243}}, $$which is what you got.

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Yes, this is correct. Note, though, that it’s easier to compute the probability that they win all five games and subtract this from $1$:

$$1-\left(\frac23\right)^5=1-\frac{32}{243}=\frac{211}{243}\;.$$

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  • $\begingroup$ What! 56 seconds apart. :( I'll upvote once my vote lock is off. $\endgroup$ – Ahaan S. Rungta Nov 26 '13 at 20:41
  • $\begingroup$ @Ahaan: It happens surprisingly often. I’ve seen essentially identical answers come up within a couple of seconds of each other. $\endgroup$ – Brian M. Scott Nov 26 '13 at 20:44
  • $\begingroup$ Upvoted. $\mathrm{}\\$ $\endgroup$ – Ahaan S. Rungta Nov 27 '13 at 13:07

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