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:

• = { (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. ⇒

⇒

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 ) =

We define the conditional probability as

with

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. ⇒

Replacing both probabilities calculated in (I) :

We find out that

Final answer:

When rolling two dice, there are 4 combinations that sum to 5. Hence, probability P(E) is 1/9. If considering the event F where the roll on the first die is 1 more than on the second die, it has 5 possible outcomes. So P(F) is 5/36. However, if event E has already happened, P(F|E) is 1/4.

Explanation:

The subject of this question is probability, which is part of Mathematics, specifically, it is a high school-level question. The event E described here is the scenario in which the sum of the numbers rolled on the two dice equals 5. There are 4 possibilities for this event: (1,4), (2,3), (3,2), and (4,1). As there are 36 possible outcomes when rolling two dice, the probability P(E) is 4/36 = 1/9.

Now considering event F where the number on the first die is exactly 1 more than the number on the second die, we have five possible pairs: (2,1), (3,2), (4,3), (5,4), (6,5). So the P(F) is 5/36. However, we're asked to find P(F|E), the probability of event F given that event E has occurred. Looking at the pairs that fit both conditions, we see that there is only one pair: (3,2). Therefore, P(F|E) is 1/4.

Learn more about Probability here:

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