A & B play a game in which A’s chance of winning is 3/5. In a series of 6 games, approximately what…
The win count follows a binomial distribution with \(n=6\) and \(p=3/5\). Thus \(P(X\ge4)\) \(=15(3/5)^4(2/5)^2+6(3/5)^5(2/5)+(3/5)^6\) \(=1701/3125\) \(=0.54432\). This is…
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A & B play a game in which A’s chance of winning is 3/5. In a series of 6 games, approximately what percentage chance does A have of winning at least 4 games, rounded to the nearest whole percent?
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Correct answer A
The win count follows a binomial distribution with and . Thus . This is 54.432%, which rounds to 54%.
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Tamil Nadu SET (TNSET)