Question

Consider a random variable X such that represents the throw of a fair die.

Let Y be a random variable where P(Y=1) = (Your birth month)/15; P(Y=2), P(Y=3), P(Y=4), P(Y=5) and P(Y=6) are equally probable. The sample space for each variable is {1,..., 6}

Fill in columns 2 and 3 in the following table where each entry is the probability of the random variable having the indicated outcome.

Table:
n | P(X=n) | P(Y=n)
1
2
3
4
5
6

(Note: Birth month = 3)

Answer 1

go on mathway.com maybe it could help you w/ your question? :) Let me know if it helped

Answer 2

It didn't help :(
But it's a very interesting site!

Answer 3

What's the probability that you roll a 1 with a die if there are 6 possible faces?

Answer 4

1/6

Answer 5

Right. This is the same as saying \(P(X=1)=\dfrac{1}{6}\). Any roll has an equal outcome, so \(P(X=n)=\dfrac{1}{6}\) for all \(n\).

Answer 6

Yup :) I understand that

I'm mostly confused about the whole P(Y=1) = (Your birth month)/15.
Does this mean P(Y=1) =3/15 then the whole thing over 5?

Answer 7

***Does this mean P(Y=1) =3/15***
this part is true. or more simply, P(Y=1)= 1/ 5

to find P(Y= n) where n > 1, we use this info:
1. sum of all the P's must be 1
P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) + P(Y=6) =1

2. P(Y=2), P(Y=3), P(Y=4), P(Y=5) and P(Y=6) are equally probable.

if we call P(Y=2) =x (i.e. the unknown probability)
then we can say, using 1. ,
P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) + P(Y=6) =1
1 / 5 + 5x = 1
or
0.2 + 5x = 1
add -0.2 to both sides

5x = 1-0.2 or
5x = 0.8
x = 0.8/5 = 0.16
now you can fill in the P(Y) column