Question

study help:::
* dont need answer need to learn how to solve for this type of problems...
Six differently colored balls (red, blue,
green, orange, purple, and white) are
placed in a basket. Without looking, three
balls are removed. What is the total
number of combinations that include a
red ball?

Answer 1

I'll help in one sec.

Answer 2

Ok so basically your determining probabilities, right?

Answer 3

yes

Answer 4

If we ignored the fact that one of the balls has to be red, how many combinations would there be of three balls?

Answer 5

two

Answer 6

There would actually be quite a few more. We could have red, green and blue. Or maybe red, blue, and orange. Perhaps I have red, blue, and white. To find the total number of combinations, let's start with one.

How many combinations of a single ball are there?

Answer 7

six

Answer 8

Exactly. Now, how many combinations of 2 balls? Remember that there should be more than three combinations.

Answer 9

hmm 9?

Answer 10

Look at it this way. You have 6 options for the first ball you can choose. Now, since you've already chosen one, you only have 5 balls left to choose from. So you should have \[6\cdot5=30\]Ways to choose to balls.

Does this make sense?

Answer 11

yeah im getting multiply by watever u have left if there were more numbers id multiply them all

Answer 12

There's one more thing I forgot to mention though. The way we calculated this, if we chose a red ball and then the blue ball, we would have a different set than if we chose the blue ball then the red ball. So really, we want \(6\cdot5\) divided by the number of ways we can order the two balls. since there's only two balls, they can only be ordered in two ways. Thus, we actually want \[{6\cdot5 \over 2}=15\]

Sorry about the confusion. By the way, how much have you talked about combinations in your class so far?

Answer 13

i study online and theres only quizzes about this type of problems got a tst tomorrow and im sure this type of problems will be there and i got you on watch u just said about dividing

Answer 14

what*

Answer 15

Basically, what I did there by dividing by 2, is using the formula\[{n! \over k!(n-k)!}\]With \(n=6\) and \(k=2\). This formula gives you the number of combinations of \(k\) things from an overall group of \(n\) things.

Have you seen this formula before?

Answer 16

yes

Answer 17

This is what you want to use for combinations.

Using this formula, can you solve for the case of choosing 3 balls out of 6 total?

Answer 18

n=6 k=3? i mean using these?

Answer 19

Exactly.

Answer 20

6/9

Answer 21

Remember that \[6! =6\cdot5\cdot4\cdot3\cdot2\cdot1=720\]And that \[3! \cdot(6-3)!=3!\cdot3!=3\cdot2\cdot1 \cdot 3\cdot2\cdot1=36\]

Answer 22

ok il write that down

Answer 23

So in the end, you should get \(720/36 =20\) combinations with 3 balls from 6 original balls.

Answer 24

Now we have to get back to the original problem of choosing three balls from six if one of those three must be red.

Let's say you've already chosen the red ball. That leaves us with 5 balls left to choose from, and only two spots left. Thus, we're asking how many combinations of 2 balls can be chosen from a set of 5 original balls. This problem is solved by the formula I wrote above.

Answer 25

thank u kinggeorge for bearing with me