There are 6 red, 4 green, and 8 white marbles in a bag. What's the probability of drawing with replacement first a red marble and then a green marble?
2 Answers
Explanation:
We start out with 18 marbles, 6 of which are red. On our first draw, the probability of getting a red marble is:
We put the red marble back and so are back to the original 18 marbles. We then want to draw a green marble. There are 4 green marbles, and so the probability of drawing a green one is:
So now the question is - Do we add these two fractions together or do we multiply?
One way to approach this is to see that each draw is an independent event and so we multiply the probabilities. But let's talk about this this way - if I put the green marble back and do a third draw and want a blue one (probability of
If it's more likely, we add the probabilities together. That'd be
As we add conditions, the probability that things will happen in certain way becomes smaller - and so we're multiplying (when we multiply a number by a fraction that reduces to less than one, the product gets smaller).
When we multiply the red draw and the green draw together, we get:
Explanation:
The probability of an event is equal to the number of times that event can happen over all possible events, and it looks like this:
In this set, we have 18 elements, and 6 of them are red. Using our formula, the probability for getting red first is
Since we put the red marble back, the probability of getting a green marble is the same on the first or second go. There are 4 green, so
The probability of picking two independent events is equal to
I hope that helps!