Question

Consider an experiment where two 6-sided dice are rolled. We can describe the ordered sample space as below where the first coordinate of the ordered pair represents the first die and the second coordinate represents the second die. If the dice are ‘fair’, then all 36 of these possible outcomes are equally likely. 28. Describe the event E that the sum of the two dice is 5. 29. Find P(E). 30. Let F be the event that the number of the first die is exactly 1 more than the number on the second die. Find P(F|E)

Answer 1

Answer:

• $E$ = { (4,1) , (3,2) , (2,3) , (1,4) }

• $P(E)=\frac{1}{9}$

• $P(F|E)=\frac{1}{4}$

Step-by-step explanation:

Let's start writing the sample space for this experiment :

$S=$ { (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 $E$ ⇒

$E$ : '' The sum of the two dice is 5 ''

We can describe the event by listing all the favorables cases from $S$ ⇒

$E$ = { (4,1) , (3,2) , (2,3) , (1,4) }

In order to calculate $P(E)$ we are going to divide all the cases favorables to $E$ over the total cases from $S$. We can do this because all 36 of these possible outcomes from $S$ are equally likely. ⇒

$P(E)=\frac{4}{36}=\frac{1}{9}$ ⇒

$P(E)=\frac{1}{9}$

Finally we are going to define the event $F$ ⇒

$F$ : '' The number of the first die is exactly 1 more than the number on the second die ''

⇒

$F$ = { (2,1) , (3,2) , (4,3) , (5,4) , (6,5) }

Now given two events A and B ⇒

P ( A ∩ B ) = $P(A,B)$

We define the conditional probability as

$P(A|B)=\frac{P(A,B)}{P(B)}$ with $P(B)>0$

We need to find $P(F|E)$ therefore we can apply the conditional probability equation :

$P(F|E)=\frac{P(F,E)}{P(E)}$   (I)

We calculate $P(E)=\frac{1}{9}$ at the beginning of the question. We only need $P(F,E)$.

Looking at the sets $E$ and $F$ we find that (3,2) is the unique result which is in both sets. Therefore is 1 result over the 36 possible results. ⇒

$P(F,E)=\frac{1}{36}$

Replacing both probabilities calculated in (I) :

$P(F|E)=\frac{P(F,E)}{P(E)}=\frac{\frac{1}{36}}{\frac{1}{9}}=\frac{1}{4}=0.25$

We find out that $P(F|E)=\frac{1}{4}=0.25$