Statistics, ping pong balls in bag probability question

[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

 

[tiny-tim]

Hi Calculator14!

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 …

why are you multiplying instead of adding?

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

Start again …

what is P(both balls are blue)? ​

 

[Calculator14]

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!

 

[tiny-tim]

Hi Calculator14!

(just got up :zzz: …)

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

 

[DeeNos]

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.