Some clarifications and a question on basic probability.

Question

I have a few questions and some clarifications.

CLARIFICATIONS:

1. Assume we roll 2 four sided dice. What is P({sum of the rolls is even})? I answered the question correctly

I: Odd + Odd = Even
J: Even + Even = Even
K: Odd + Even = Odd

I answered the question with this line of thinking P(I) + P(J):

P(I) = (2 / 4) * (2 / 4) P(J) = (2 / 4) * (2 / 4) = 1 / 2

*Is this a valid concept of solving the problem?

2. Assume we roll 2 four sided dice. What is P({first roll larger than second roll})?

P(1,<2,3,4>) + P(2, <3,4>) + P(3, 4)

(1 / 4 * 3 / 4) + (1 / 4 * 2 / 4) + (1 / 4 * 1 / 4) = 3 / 8

*Is this a valid concept of solving the problem? Are there better solutions?

QUESTION:

Say you're in a classroom with 29 other students (30 total students). What is the calculated probability that two students share the same birthdate? For simplicity, assume there are 365 days in a year and all birthdates are equally probable. I need help on this one.

Answer 1

Your approach is correct. Your notation could use some work.

You are partitioning the event into a series of disjoint events, whose individual probabilities you can calculate and then sum using the Law of Total Probability. That is exactly the approach you should use.

1. Assume we roll 2 four sided dice. What is P({sum of the rolls is even})? I answered the question correctly

Let $X$ be the result of one die and $Y$ be the result of the other.

$$\begin{align} \mathsf P(\mathrm{Even}(X+Y)) & = \mathsf P(\mathrm{Even}(X))\;\mathsf P(\mathrm{Even}(Y))+\mathsf P(\mathrm{Odd}(X))\;\mathsf P(\mathrm{Odd}(Y)) \\[2ex] & = \frac 2 4 \frac 2 4 + \frac 2 4\frac 2 4 \\[2ex] & = \frac 1 2 \end{align}$$

Alternatively you can just notice that whatever result of the first dice, half of the results of the second dice will give an even sum.

2. Assume we roll 2 four sided dice. What is P({first roll larger than second roll})?

$$\begin{align} \mathsf P(X > Y) & = \sum_{x=2}^4 \mathsf P(X=x) \mathsf P(Y< x) \\[2ex] & = \mathsf P(X=2)\mathsf P(Y<2) + \mathsf P(X=3)\mathsf P(Y<3) + \mathsf P(X=4)\mathsf P(Y< 4) \\[2ex] & = \frac 1 4 \left(\frac 1 4 + \frac 2 4 + \frac 3 4\right) \\[1ex] & = \frac 3 8 \end{align}$$

Say you're in a classroom with 29 other students (30 total students). What is the calculated probability that two students share the same birthdate? For simplicity, assume there are 365 days in a year and all birthdates are equally probable. I need help on this one.

In this case you need to use the Rule of Complements.   The sum of the probabilities of complementary events is $1$.   The event that "at least two students share a birthdate" is the complement of the event that "all students have unique birthdates".   Can you calculate the probability that nobody shares a birthdate?