So i could pick that green marble or that green marble.
There are 9 red and 6 green marbles.
Write the probability as a fraction in simplest form a decimal and a percent.
Jun 21 2018 1 3 explanation.
Finally multiply the probabilities and reduce to get 3 95 share.
So this is all the possible outcomes.
What is the probability of that child getting a green marble.
In the bag there are 6 green marbles and 12 red marbles.
Be careful letting children play with marbles they may swallow.
There are 9 red and 6 green marbles in a bag.
You wish to know t probability that the single chosen green or blue marble.
Indeed we have these equations where r is the number of red marbles g is the number.
In a jar of red green and blue marbles all but 6 are red marbles all but 8 are green and all but 4 are blue.
So i could pick that red marble or that red marble.
At this moment the bag becomes the same as it is before a green marble is drawn i e.
There are 9 red and 6 green marbles in a bag if a child reaches the bag randomly and picks 6.
There are 35 marbles in a bag.
There are still 6 green marbles and 9 red ones.
A child reaches in the bag and randomly takes one marble.
There are a total of 14 green and blue marbles so the probability is 14 25 or 56.
There s one blue marble.
If we are too draw another green marble the same probability 6 15 will be resulted.
Remember there is one less marble in the jar and that you assume the marble picked was green.
Solution from the condition we can determine how many marbles of each color were there in the jar.
What is the probability of drawing a green marble from the bag.
And then there s one blue marble in the bag.
The total number of marbles in the box is 25 that is 9 green 5 blue and 11 red.
There s two red marbles in the bag.
There s two green marbles in the bag.
There are 6 red 4 green 5 blue and 5 yellow marbles in a jar.
These are clearly all yellow.
What percent of the marbles aren t blue.
9 blue marbles 8 green marbles 4 red marbles 8 white marbles and 6 yellow marbles.
1 answer jim g.
As both requirements have to be fulfilled the required probability is the multiplication of both 6 15 6 15.