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?
Probability of Probability
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0For $n$ games, we need $(0.6)^n>0.8\implies n<1$, but $n\ge 1$, so no solution. – 2012-10-04
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2Is 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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5@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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0I don't see how the title relates to the question. – 2012-10-04