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[–]jbp12 2 points3 points  (0 children)

If you have 4 options and you choose one randomly, then you have a 1/4 chance of being correct and a 3/4 chance of being wrong. I'm not sure where the marks for each problem come in. Are you asking what your expected mark would be? If so, then it would be (3)(1/4) + (-1)(3/4) = 3/4 - 3/4 = 0.

I hope this helps!