# Statistics, ping pong balls in bag probability question

• Calculator14
In summary, the conversation is discussing the probability of two balls being randomly selected from a bag of twelve colored ping-pong balls and being the same color. The solution involves finding the probability of each individual color and adding them together. The equation used is P(two same color) = P(red) + P(blue) + P(green). The attempt at a solution involves multiplying fractions, but the correct method is to add them together. The question is then posed, "what is P(both balls are blue)?"
Calculator14

## Homework Statement

Twelve colored ping-pong balls are placed into a shopping bag and well mixed. There are two red balls, six blue balls and four green balls. One ball is selected at random, its color noted and then it is set aside. A second ball is then randomly selected and its color noted. What is the probability that both balls are the same color?

## Homework Equations

P(two same color) = P(red) + P(blue) + P(green)

## The Attempt at a Solution

P(two same color) = (2/12 + 1/12) * (6/12 + 5/12) * (4/12 + 3/12) = .8125

Hi Calculator14!
Calculator14 said:
P(two same color) = P(red) + P(blue) + P(green)

## The Attempt at a Solution

P(two same color) = (2/12 + 1/12) * (6/12 + 5/12) * (4/12 + 3/12) = .8125

I don't understand

and eg what is 6/12+5/12 supposed to be?

Start again …

what is P(both balls are blue)?

Hi tiny-tim! I'm sorry, I am lost on how to start this equation. Would you be able to guide me a bit through this? Thank you for your help!

Hi Calculator14!

(just got up :zzz: …)

If you select two balls out of 12, what is the probability that both balls are blue?

I would approach this problem by first analyzing the given information and understanding the context. The problem is asking for the probability of selecting two balls of the same color from a bag containing 12 ping-pong balls of three different colors: red, blue, and green. The total number of possible outcomes is 12C2 (12 choose 2) which is equal to 66.

To find the probability of selecting two balls of the same color, we can use the formula P(two same color) = P(red) + P(blue) + P(green). This is because there are three possible outcomes for selecting two balls of the same color: both red, both blue, or both green.

To calculate the probability of selecting a red ball, we use the formula P(red) = 2/12 (2 red balls out of 12 total balls). Similarly, the probabilities of selecting a blue and green ball are 6/12 and 4/12, respectively.

Therefore, the probability of selecting two balls of the same color is P(two same color) = (2/12 * 1/11) + (6/12 * 5/11) + (4/12 * 3/11) = 0.2727 + 0.2727 + 0.0909 = 0.6363. This means that there is a 63.63% chance of selecting two balls of the same color from the bag.

It is important to note that this solution assumes that the balls are replaced after each selection, meaning that the total number of balls in the bag remains the same. If the balls are not replaced, then the probabilities would change for each subsequent selection.

As a scientist, it is also important to consider any potential limitations or assumptions made in the solution. For example, we are assuming that the balls are well-mixed and that there is an equal chance of selecting any ball from the bag. These assumptions may not hold true in real-world situations and could affect the accuracy of our solution.

## 1. What is the basic concept behind the ping pong balls in bag probability question?

The basic concept behind this question is that of probability. It involves calculating the likelihood of certain outcomes when drawing ping pong balls from a bag without replacement.

## 2. How do I approach solving this type of probability question?

When solving this type of question, it is important to first identify the total number of outcomes and the desired outcome. Then, use the formula P(desired outcome) = (number of desired outcomes)/(total number of outcomes) to calculate the probability.

## 3. What is the difference between with replacement and without replacement scenarios?

In a with replacement scenario, the selected ball is put back into the bag before the next selection, whereas in a without replacement scenario, the selected ball is not put back into the bag. This affects the total number of outcomes and therefore the probability of desired outcomes.

## 4. Can this type of question be solved using a tree diagram?

Yes, a tree diagram can be a helpful visual aid for solving this type of question. It can help to visually represent the different possible outcomes and their associated probabilities.

## 5. Are there any real-life applications of this type of probability question?

Yes, this type of question is commonly used in fields such as statistics, economics, and engineering to make predictions and inform decision-making based on probability calculations. It can also be applied in situations involving random sampling and risk analysis.

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