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$Old$ 11-16-2008, 02:55 PM

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$Default$ Discrete Probability Distributions

A multiple-choice examination has 15 questions. Each question has four possible answers of which only one is correct. The probability that by just guessing a student will get exactly 7 correct is?

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$A_ n = \int D\mu \int D[X]$ $exp\left(-\frac{1}{4\pi \alpha} \int \partial_z X_{\mu}(z,\bar{z})\partial_{\bar{z}}X^{\mu}(z,\bar{z})dz + i\sum_{i=1}^N k_{i \mu}X^{\mu}(z_i,\bar{z_i})\right)$

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$Old$ 11-16-2008, 03:07 PM

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${15 \choose 7}(s)^7(1-s)^{15-7}$
Where $s$ is the probability of success.

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$Old$ 11-16-2008, 03:12 PM

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Thanks, and I assume that s would be 0.25 because of four possible choices and one correct answer.

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$A_ n = \int D\mu \int D[X]$ $exp\left(-\frac{1}{4\pi \alpha} \int \partial_z X_{\mu}(z,\bar{z})\partial_{\bar{z}}X^{\mu}(z,\bar{z})dz + i\sum_{i=1}^N k_{i \mu}X^{\mu}(z_i,\bar{z_i})\right)$

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