# Question: Green or Ball

## Comment on Green or Ball

### I know the probability of a

I know the probability of a green ball is given, but why can't you multiply the two probabilities to get P ( green and ball) = 1/6. Is it because you have the blue and red objects in play. ### P(green object) = 1/4 and P

P(green object) = 1/4 and P(ball) = 2/3
However, P(green and ball) = (1/4)(2/3) = 1/6 ONLY IF these two events (being green and being a ball) are independent. In this case, they clearly aren't independent since we're told that P(green and ball) = 1/5 (and not 1/6)
For more on independence see: https://www.greenlighttestprep.com/module/gre-probability/video/755 and https://www.greenlighttestprep.com/module/gre-probability/video/756

### A bit of a confusing question

A bit of a confusing question. When I read it, I assumed that selecting an object that was a ball AND a green since a green ball satisfies the condition that you selected an object that was a green or a ball.

The question could be better clarified by asking, "What is the probability of selecting an object that is green or a ball, but not both?" ### My calculations for P(green

My calculations for P(green or ball) INCLUDE the scenario that the object is green AND a ball.

I think the "or" probability formula is throwing you off, since it subtracts "P(green and ball)" at the end. However, this does not mean that we are saying that an object that is both a ball and green does not meet the conditions. We subtract "P(green and ball)" at the end in order to remove duplication in our calculations.

For more on why we need to subtract PA AND B), watch this earlier video: https://www.greenlighttestprep.com/module/gre-probability/video/748

### In this question, why can't

In this question, why can't we determine the probability of an object being green using the following formula:

P(green ball) = P(green)*P(ball)
1/5 = P(green)*2/3
P(green) = 3/10 ### That formula would be used if

That formula would be used if we were looking for the probability that the selected item is green AND is a ball. Also, the this approach would be valid ONLY IF the two events were independent (which they are not).

### The probability of selecting

The probability of selecting a green ball is already given as 1/5, why should we use the complement rule to find for p(green). this is a little confusing to me. ### The box contains balls AND

The box contains balls AND chips. Some BALLS are green and some CHIPS are green.

P(green) refers to the probability that we select EITHER a green ball OR a green chip.

So, knowing that P(selecting a green ball) = 1/5 is not enough to determine P(green). We must also consider P(selecting a green CHIP)

Does that help?

Cheers,
Brent

### Thanks. It is clear after

Thanks. It is clear after going over. Thanks again

### can a chip be colored to be

can a chip be colored to be "Red" in this box or only balls could be Red among the objects? ### The question tells us that

The question tells us that each OBJECT is red, blue or green. Since each object can be a ball or a chip, it's possible to have red balls AND red chips.

Cheers,
Brent