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