##### Independece (Probability) | MIT Assignment

You have a fair five-sided die. The sides of the die are numbered from 1 to 5. Each die roll is independent of all others, and all faces are equally likely to come out on top when the die is rolled. Suppose you roll the die twice.

(A)

Let event A to be “the total of two rolls is 10”, event B be “at least one roll resulted in 5”, and event C be “at least one roll resulted in 1”.

- Is event A independent of event B?
- Is event A independent of event C?

(B)

Let event D be “the total of two rolls is 7”, event E be “the difference between the two roll outcomes is exactly 1”, and event F be “the second roll resulted in a higher number than the first roll”.

- Are events E and F independent?
- Are events E and F independent given event D?

**Answer**

__(A)__

(1) *NO*

Consider the sample space(visualized below) when the five-sided die is rolled twice :

Overall, there are 25 different outcomes in the sample space. For a total of 10, we should get a 5 on both rolls.

Therefore A ⊂ B, and

P(B|A) = \(\frac{P(A \cap B)}{P(A)} = \frac{ P(A)}{P(A)} = 1\)

We observe that to get at least one 5 showing, we can have 5 on the first roll, 5 on the second roll, or 5 on both rolls, which corresponds to 9 distinct outcomes in the sample space. Therefore

\(P(B) = \frac{9}{25} \neq P(B | A)\)

Hence, A and B are not independent.

Ans NO.

(2) *NO*

Given event A, we know that both roll outcomes must be 5. Therefore, we could not have event C occur, which would require at least one 1 showing. Formally, there are 9 outcomes in C, and

\(P(C) = \frac{9}{25} \)

but, \(P(C | A) = 0 \neq P(C)\)

Hence, Ans NO.

__(B)__

(1) *NO*

Out of the total 25 outcomes, 5 outcomes correspond to equal numbers in the two rolls. In half of the remaining 20 outcomes, the second number is higher than the first one. In the other half, the first number is higher than the second. Therefore,

\(P(F) = \frac{10}{25}\)

There are eight outcomes that belong to event E:

E = {(1, 2),(2, 3),(3, 4),(4, 5),(2, 1),(3, 2),(4, 3),(5, 4)}.

To find P(F|E), we need to compute the proportion of outcomes in E for which the second number is higher than the first one:

\(P(F | E) = \frac{1}{2} \neq P(F) \)

Hence, Ans is NO.

(2) *YES,*

Conditioning on event D reduces the sample space to just four outcomes

{(2, 5),(3, 4),(4, 3),(5, 2)}

which are all equally likely. It is easy to see that

\(P(E|D) = \frac{2}{4} = \frac{1}{2} \\ P(F | D) = \frac{2}{4} = \frac{1}{2} \\ P(E \cap F | D) = \frac{1}{4} = P(E|D)P(F|D)\)

hence, Ans YES.