# Conditional Probability

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

• 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. Apr 14, 2015 at 6:51

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