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You roll two six-sided dice, one then the other.

A = {The first roll is a 1} B = {The sum of the two dice is 4}

What is the value of $P(A^c | B)$?

$A^c$ = $5/6$

B =$1/12$

I thought that the formula would be:

$$1/12 * 5/6 \div 1/12$$

but the result is 10, which is wrong

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  • $\begingroup$ Did you mean $P(A^C) = \frac{5}{6}$? $\endgroup$
    – N. F. Taussig
    Jan 31, 2016 at 17:16
  • $\begingroup$ Yes. sorry I'll edit that. $\endgroup$
    – user_123945839432
    Jan 31, 2016 at 17:17
  • $\begingroup$ what formula are you using? Bayes rule? $\endgroup$
    – AlanSTACK
    Jan 31, 2016 at 17:21
  • $\begingroup$ P(B∣A)=P(A∩B)P(A) $\endgroup$
    – user_123945839432
    Jan 31, 2016 at 17:23
  • $\begingroup$ Btw, the correct answer to the wrong formula you used is $\frac1{12}$. $\endgroup$
    – true blue anil
    Jan 31, 2016 at 17:39

1 Answer 1

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You have correctly found that $P(A^C) = \frac{5}{6}$ and that $P(B) = \frac{1}{12}$. The conditional probability that event $A^C$ occurs given that event $B$ occurs is $$P(A^C \mid B) = \frac{P(A^C \cap B)}{P(B)}$$ There are three ways for event $B$ to occur: (1, 3), (2, 2), (3, 1). Of these, two do not involve rolling a 1 on the first roll. Since there are $36$ ordered pairs of rolls in the sample space, $$P(A^C \cap B) = \frac{2}{36} = \frac{1}{18}$$
Hence, $$P(A^C \mid B) = \frac{P(A^C \cap B)}{P(B)} = \frac{\frac{1}{18}}{\frac{1}{12}} = \frac{2}{3}$$

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