Answer:
= { (4,1) , (3,2) , (2,3) , (1,4) }
Step-by-step explanation:
Let's start writing the sample space for this experiment :
{ (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }
Let's also define the event
⇒
: '' The sum of the two dice is 5 ''
We can describe the event by listing all the favorables cases from
⇒
= { (4,1) , (3,2) , (2,3) , (1,4) }
In order to calculate
we are going to divide all the cases favorables to
over the total cases from
. We can do this because all 36 of these possible outcomes from
are equally likely. ⇒
⇒
![P(E)=\frac{1}{9}](https://tex.z-dn.net/?f=P%28E%29%3D%5Cfrac%7B1%7D%7B9%7D)
Finally we are going to define the event
⇒
: '' The number of the first die is exactly 1 more than the number on the second die ''
⇒
= { (2,1) , (3,2) , (4,3) , (5,4) , (6,5) }
Now given two events A and B ⇒
P ( A ∩ B ) = ![P(A,B)](https://tex.z-dn.net/?f=P%28A%2CB%29)
We define the conditional probability as
with ![P(B)>0](https://tex.z-dn.net/?f=P%28B%29%3E0)
We need to find
therefore we can apply the conditional probability equation :
(I)
We calculate
at the beginning of the question. We only need
.
Looking at the sets
and
we find that (3,2) is the unique result which is in both sets. Therefore is 1 result over the 36 possible results. ⇒
![P(F,E)=\frac{1}{36}](https://tex.z-dn.net/?f=P%28F%2CE%29%3D%5Cfrac%7B1%7D%7B36%7D)
Replacing both probabilities calculated in (I) :
![P(F|E)=\frac{P(F,E)}{P(E)}=\frac{\frac{1}{36}}{\frac{1}{9}}=\frac{1}{4}=0.25](https://tex.z-dn.net/?f=P%28F%7CE%29%3D%5Cfrac%7BP%28F%2CE%29%7D%7BP%28E%29%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B36%7D%7D%7B%5Cfrac%7B1%7D%7B9%7D%7D%3D%5Cfrac%7B1%7D%7B4%7D%3D0.25)
We find out that ![P(F|E)=\frac{1}{4}=0.25](https://tex.z-dn.net/?f=P%28F%7CE%29%3D%5Cfrac%7B1%7D%7B4%7D%3D0.25)