A Bag Of Marbles Contains 120 Red And Yellow Marbles

Asked by lyla on april 18 2014.

A bag of marbles contains 120 red and yellow marbles.

A bag contains 4 yellow 2 red and 6 green marbles. Asked 03 10 20 a bag contains yellow marbles and red marbles 16 in total. There are 114 red marbles in the bag. If lisa draws a random marble from the bag what is the probability that it will be a red green or blue marble.

A bag of marbles contains 8 red 4 green 3 blue and 2 yellow marbles. A bag contains 9 red marbles 8 white marbles and 6 blue marbles. The first is replaced before the second is drawn. There are 3 red marbles and 7 black ma.

A bag contains 70. A handful of 5 marbles is grabbed from the bag. The number of yellow marbles is 1 more than 2 times the number of red marbles. Thus the probability of choosing the second red marble is 4 11.

If there are only red and blue marbles and out of the 120 in total there are 80 blue marbles then there must be 120 80 40 red marbles. 2 put value of x in 1 we get put y 6 in 2 we get hence there are 114 red marbles in the bag. Y number of black marbles. 1 there are 19 red marbles every for black marble.

A bag contains nine red marbles four green marbles three blue marbles and two yellow marbles. Calculate the expected value of the random variable. The probability of choosing first red marble is 5 12 because there is not any replacement now 4 red marbles are remained and totally 11 marbles. A random variable assigns the number of red marbles to each outcome.

The ratio between blue and red marbles is therefore. The number of different handfuls of 5 marbles with exactly 2 green and no yellow marbles is. Two marbles are drawn. A bag contains 120 marbles.

The probability of consecutively choosing two red marbles and a green marble without replacement. Let x number of red marbles. P 3 10 7 9 2 8 p 7 120 since we were ask to find the probability of chosing 1 red 1 black and 1red marble without replacement after each draw we can use the fundamental method in finding the probability.