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A quiz consists of 7 multiple choice questions, each with 4 possible answers. To pass the quiz, it is you must get at least 3 questions correct. An unprepared student can do nothing except guess answers at random. What is the probability that the student passes?

I'm thinking you use the distributive property which gives you ${7 \choose 3} (0.25)^3 (0.75)^4$ but I'm not sure where to go from here.

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While your approach is largely correct, it needs to be adapted to accommodate for the possibility that the students guesses correctly more than three times; you computed the probability for the student to give precisely three correct answers.

Thus, simply compute the probabilities for 4,5,6,7 as well, and then add. Alternatively, you can compute the probabilities for 0,1,2 correct answers, and subtract the total from 1, giving the same result.

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If by (3 7) you mean ${7 \choose 3}$, the ways to select $3$ items from $7$, you have calculated the chance to get exactly $3$ correct, not the chance to get at least $3$.

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  • $\begingroup$ Then how would you calculate at least 3? $\endgroup$ – jrquick Oct 9 '12 at 17:32
  • $\begingroup$ @JeremyQuick: I would also calculate $4,5,6,$ and $7$ and add. Alternately, I would calculate $0,1,$ and $2$ and subtract from $1$ as there are fewer cases. $\endgroup$ – Ross Millikan Oct 9 '12 at 17:34

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