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Consider a multiple choice exam with four options per question. Suppose that the probability is 0.6 that you know (with certainty) the answer to a randomly selected question. The probability is 0.3 that you eliminate (with certainty) 2 options and guess randomly between 2 options. The probability is 0.1 that you guess randomly among all 4 options.

(a) Find the probability that you answer a randomly selected question correctly.

(b) Find the probability that you just randomly guessed among the 4 options given that you answered a question correctly.

This is what I have so far :

$$P(A)=0.6$$ $$P(B)=0.3$$ $$P(C)=0.1$$ $$P(A)+P(B)+P(C) = 0.6+0.3*0.5+0.1*0.25=0.775$$

Using Bayes Rule : $$P(C|Correct)=\frac{P(Correct|C)P(C)}{P(Correct|C)P(C)+P(Correct|B^c)P(B^c)}$$

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  • $\begingroup$ If "B^c" means the complement of B, I think you actually meant to use the complement of C; and anyway you probably should split whatever it is into two cases. $\endgroup$ – David K Apr 14 '15 at 6:51
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If you use Bayes's Rule, you have

$$ P(C \mid \mbox{correct}) = \frac{P(\mbox{correct} \mid C) P(C)}{P(\mbox{correct})} = \frac{(1/4)(1/10)}{31/40} = 1/31 $$

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(a) Law of Total Probability: $$P(Right) = P(R) = P(R\cap A) + P(R\cap B) + P(R \cap C).$$ Then $P(R \cap B) = P(B)P(R|B) = (.3)(1/2),$ and similarly for the other two terms.

(b) Bayes' Theorem: $P(C|R) = P(C \cap R)/P(R),$ where the denominator is the answer to (a) and the numerator is one of the terms in (a).

With this much help, you should be able to get the answer, or to understand formulas in your book to get the answer.

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