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There are two events, A & B, which revolve around the rolling of two normal dies. The events are as follows:

A = {The first roll of a dice lands up as a 1}

B = {The sum of the two dice lands up as 4}

So far, I have the following probabilities set up:

$P(A) = \frac{1}{6}, P(B) = \frac{1}{12}, P(A^c) = \frac{5}{6}, P(B^c) = \frac{11}{12}$

However, now I am required to find $P(A^c | B)$.

Work

I am attempting to use a mix of formulas here. Specifically, $$P(A^c | B) = \frac{P(A^c \cap B)}{P(B)}$$

The main issue comes down to the fact that I am entirely lost when it comes up to here. I have no idea on how to obtain this intersection. The best I can do at this point is use the formula:

$$P(A^c \cap B) = P(A^c \cup B) - P(A) - P(B)$$

However, I am unsure now of what to do because this just gives me that the intersection is $0$

How can I obtain $P(A^c | B)$?

  • 1
    The intersection $A^c\cap B$ is not zero. That would be the same as saying 'it can't happen that the first roll is not a 1 and the sum of the rolls is four" It can, as the possibility $(2,2)$ demonstrates. – 2017-02-06
  • 0
    Tip: You found the probabilities for events $A$ and $B$ from measuring(counting) the outcomes of the die rolls against the sample space. Do the same for the event "the first die shows *not* one, and the sum of the dice is four." – 2017-02-07

3 Answers 3

1

The figure below depicts all the possibilities.

enter image description here

  • Here, $A$ is yellow with one green box; number of boxes = $6$.

  • $B$ is blue with one green box; number of boxes = $3$.

  • $ A^c$ is framed in red; number of boxes = $30$.

  • $ A^c \cap B$ is blue; number of boxes = $2$.

Then

$$P( A^c \mid B)=\frac{P( A^c \cap B)}{P(B)}=\frac{\frac{2}{36}}{\frac{3}{36}}=\frac{2}{3}.$$

2

You need the probability that the first roll is not a $1$ given that the sum of the two dice is four. If the sum of the two dice is four, then it could be $(1,3),$ $(3,1)$ or $(2,2)$ with equal probability. Only $(2,2)$ and $(3,1)$ are in $A^c$ (they don't have one for the first die). So the probability is $2/3.$

1

There are several ways to do this.

First, your event $B$ is $(1,3)$, $(2,2)$ and $(3,1)$ on the first and the second dice respectively. This is how you calculated $\mathbb{P}[B]=\frac{3}{36}$. Now you know $B$ occurred and you are asking what is the probability that $1$ did not happen on the first dice. Since $(1,3)$, $(2,2)$ and $(3,1)$ have the same probability, it is just $\frac{2}{3}$.

Another, $A^{c}\cap B$ is event that the sum on the two dice is $4$ but $1$ did not occur on the first dice. That is, it is the event that the rolls were $(2,2)$ or $(3,1)$. Probability of those two is $\frac{1}{18}$. Hence the probability you are after is $\frac{1}{18}/\frac{1}{12}=\frac{12}{18}$.