I Probability of White Ball in Box of 120 Balls: Solved!

[CGandC]

Problem:
In a box there are $120$ balls with $X$ of them being white and $120 - X$ being red for random variable $X$.
We know that $E[ X] = 30$. We are taking out $k$ balls randomly and with returning ( we return each ball we take out, so there is equal probability for each ball every time we take out a ball ), for $k \geq 2$.
Let $Y$ be the number of white balls in the sample that was taken out.
What is the probability of the first ball being white?

Note: This was the first question of an old exam and there were more questions that were based on it in the exam later on ( so it might be the case that we don't need to use $k$ here. ), none of them helped in inferring the reasoning for the answer to this problem.

The official answer was:
$E[ \frac{X}{120}] = \frac{1}{4}$, hence the probability of the first ball being white is $\frac{1}{4}$

Question:
I tried different things, among them being attempting to use binomial ,negative-binomial, hyper-geometric distributions in the problem, but I kept getting stuck because I don't know what $X$ is since it is a random variable.
Then I tried getting ahead with the following equations:
$E[ X] = \sum_{i=1}^{120} x P(X = x)$ , $E[ X] = \sum_{i=1}^{120} P(X \geq i)$ , but I was unable to proceed.

Do you have any explanation for the answer? I'm unable to retrace the steps necessary to arrive to it so I can't figure out how to arrive at the answer.

Thanks in advance for any help!

 

[FactChecker]

You should review the basic properties of expected values. What is the expected value of $aX$, where $a$ is a constant real number?

 

[CGandC]

You should review the basic properties of expected values. What is the expected value of $aX$, where $a$ is a constant real number?

$E[ aX] = aE[ X]$, I'm familiar with the basic properties of expected value butthese don't give a clue to why the answer is as is. All I can understand from this is that $E[X] = 120 \cdot p$ where p is the wanted probability, but there I don't understand the reasoning for why the sum on the right-hand side of the equation is as is.

Edit: I've asked this question on mathexchange and I was told to define an indicator random variable $X_i$ which says " the value of $X$ is $i$ " and then to use linearity of expectation $E[ X]=\sum E[X_i]$. Now I understand the solution, though the question itself was misleading and confusing.

Thanks for the help!

 

[statdad]

$E[ aX] = aE[ X]$, I'm familiar with the basic properties of expected value butthese don't give a clue to why the answer is as is. All I can understand from this is that $E[X] = 120 \cdot p$ where p is the wanted probability, but there I don't understand the reasoning for why the sum on the right-hand side of the equation is as is.

Edit: I've asked this question on mathexchange and I was told to define an indicator random variable $X_i$ which says " the value of $X$ is $i$ " and then to use linearity of expectation $E[ X]=\sum E[X_i]$. Now I understand the solution, though the question itself was misleading and confusing.

Thanks for the help!

First think about finding P(First ball is white | X = x] : this will be an expression that involves x, call it [for my illustration] f(x)

Then P(First ball is white) = E[First ball is white | [X = x]] = E(f(x))

This should lead you through it and see where the information about E(X) is useful.

finally: this is [in my opinion] a crappily worded question: You're told that k >=2 balls will be selected and Y is the number of white balls in the sample, but nothing about that is needed for solving the problem about P(First ball is white). I see no reason the business about Y and the sample needs to be in this question