Calculating Probability of 2 Black Balls from 3-Ball Urn w/ Pascal Triangle

[firavia]

if we toss 4 coins and we want to know the probability of having 3 heads we can easily calculate that using pascal triangle .
I want to know if we have an Urn that contain 3 balls : 1 black , 1 red, 1 yellow.
and we want to draw 3 balls , but when we draw the first ball we put it back in the urn and thn we draw the 2nd ball which thn we put it back and finally we draw a third ball.

in this example we have 27 possible way of selecting 3 balls.

I want to know how can we calculate the probability of having at least 2 black ball , without using a tree diagram, probably wiht pascal triangle or any other formula with the explanation.

and you are very thankfull.

It is not a homework , I just want to know how this kind of problems works.

 

[firavia]

any help?

 

[Cant or Wont]

You could set it out as a binomial tree if you fancied? Reducing the outcomes to Black and Not Black - i know you said you wanted to do it without a tree diagram but binomial trees never get old. (Much like probability questions involving urns)

This may, or may not, help you get a feel for it. If not try considering the "Events" that need to occur for you to get "at least 2 blacks" and then attach probabilities to these situations.

 

[dacruick]

The probability of having at least 2 black balls means that you want to add the probability of getting 3 black balls (obviously 1 in 27) with the probability of having 2 black balls.

the other way to go is adding the probability of getting no black balls with the probability of getting 1 black ball, and subtracting that sum from 1.

 

[blue_raver22]

Thank you for your question. Calculating probability in this type of scenario can be done using the concept of combinations. In the urn example you provided, there are 3 balls and we want to draw 3 balls with replacement. This means that the order in which we draw the balls does not matter. We can use the combination formula, nCr = n!/(r!(n-r)!), to calculate the number of ways we can draw 3 balls from the urn.

In this case, n = 3 (since there are 3 balls) and r = 3 (since we want to draw 3 balls). Plugging these values into the formula, we get 3C3 = 3!/(3!(3-3)!) = 1.

This means that there is only one way to draw 3 balls from the urn with replacement. Now, let's consider the different outcomes that can occur when we draw 3 balls from the urn with replacement:

1. All 3 balls are black
2. 2 black balls and 1 non-black ball (either red or yellow)
3. 1 black ball and 2 non-black balls (either red or yellow)
4. All 3 balls are non-black

Out of these 4 possible outcomes, only 2 and 3 fulfill the condition of having at least 2 black balls. Using the combination formula again, we can calculate the number of ways these outcomes can occur:

2. 3C2 * 1C1 = 3
3. 3C1 * 1C2 = 3

Adding these together, we get a total of 6 ways that we can have at least 2 black balls when drawing 3 balls with replacement from the urn. Therefore, the probability of having at least 2 black balls can be calculated as 6/27 = 2/9.

I hope this explanation helps you understand how to approach these types of probability problems. Please keep in mind that this is just one method of solving the problem and there may be other ways to approach it. It is important to understand the concept of combinations and how to apply it in different scenarios. Thank you for your question and I wish you all the best in your studies.