# Question: Draw a Red Chip

## Comment on Draw a Red Chip

### I like this formula. Other

I like this formula. Other formula made me confused
My question is can I use this formula in all probability questions? ### Many student's prefer the

Many student's prefer the basic probability formula. Unfortunately, there are some questions where the formula won't help you. However, it will work for the majority of questions.

Also keep in mind that you will probably see 1 maybe 2 probability questions on test day (perhaps zero even).

For more on the various probability techniques, see https://www.greenlighttestprep.com/module/gre-probability/video/759

### I've heard that most of the

I've heard that most of the test questions are word problems and algebra?
is that correct? ### It depends on what you mean

It depends on what you mean by "Word Problems." Would you call the above video question a word problem?

If you check out the practice questions (and practice tests) in the Official Guide to the GRE General Test, you'll get a good idea of the breakdown of questions.

### Can't we directly assign

Can't we directly assign numbers and solve it? For instance I took red chips as 1 so green chips will be 2.. Total chips 3 and probability of red chip as 1/3. ### That's a totally valid

That's a totally valid approach in this case.
In fact, it works for any values we choose.
For example, if we say there are 3 red chips and 6 green chips, then P(red chip) = 3/9 = 1/3
Similarly, if we say there are 10 red chips and 20 green chips, then P(red chip) = 10/30 = 1/3
Etc,

However, that strategy comes with a word of caution.

If the question had read "The number of green chips is 3 GREATER than the number of red chips", then assigning values to the number of each chip will NOT work.

For example, it COULD be the case that there is 1 red chip and 4 green chips. In this case, P(red chip) = 1/4
It could also be the case that there are 4 red chips and 7 green chips. In this case, P(red chip) = 4/11
Or it could be the case that there are 50 red chips and 53 green chips. In this case, P(red chip) = 50/53
Etc.

Cheers,
Brent

### Can this be solved using

Can this be solved using compliments ### You bet.

You bet.

P(selected chip is red) = 1 - P(selected chip is NOT red)
P(selected chip is red) = 1 - P(selected chip is GREEN)

Let's find P(selected chip is GREEN)

Let R = # of red chips
So, 2R = # of green chips
and 3R = TOTAL number of chips

So, P(chip is green) = 2R/3R = 2/3

This means P(selected chip is red) = 1 - 2/3
= 1/3

Cheers,
Brent