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The probability of winning a game is 0.6 and losing it is 0.4. Then how many games should one play, so that overall probability of winning that many games exceeds 0.8?

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    For $n$ games, we need $(0.6)^n>0.8\implies n<1$, but $n\ge 1$, so no solution.2012-10-04
  • 2
    Is the problem that you want to win $n$ games and are asking how many games $m$ should be played to have 0.8 chance of winning $n$ of them?2012-10-04
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    @labbhattacharjee: in your reading, which *is* what the question asks, we clearly need $n=0$. Then we have greater than 80% chance of winning no games.2012-10-04
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    I don't see how the title relates to the question.2012-10-04

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