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

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• CGandC
In summary: If you're going to ask for help on this, make sure you've got the complete problem and everything you need to solve it.In summary, the probability of the first ball being white is ## \frac{1}{4} ##.
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.

## 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!

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

CGandC
FactChecker said:
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!

Last edited:
CGandC said:
## 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

## 1. What is the probability of selecting a white ball from a box of 120 balls?

The probability of selecting a white ball from a box of 120 balls depends on the number of white balls in the box. If there are 20 white balls in the box, the probability would be 20/120 or 1/6. This means that for every 6 balls selected, one of them is likely to be a white ball.

## 2. How do you calculate the probability of selecting a white ball from a box of 120 balls?

To calculate the probability, you need to know the total number of balls in the box and the number of white balls. Then, you can use the formula P(A) = n(A)/n(S), where P(A) is the probability of selecting a white ball, n(A) is the number of white balls, and n(S) is the total number of balls in the box.

## 3. What is the probability of selecting a white ball if there are 30 white balls in the box?

If there are 30 white balls in the box, the probability of selecting a white ball would be 30/120 or 1/4. This means that for every 4 balls selected, one of them is likely to be a white ball.

## 4. Is the probability of selecting a white ball from a box of 120 balls affected by the color of the other balls?

No, the probability of selecting a white ball from a box of 120 balls is not affected by the color of the other balls. As long as the total number of balls and the number of white balls remain the same, the probability will remain the same.

## 5. How can the probability of selecting a white ball from a box of 120 balls be increased?

The probability of selecting a white ball from a box of 120 balls can be increased by increasing the number of white balls in the box. This will increase the chances of selecting a white ball. Alternatively, you can also decrease the total number of balls in the box, which will also increase the probability of selecting a white ball.

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