I believe it would be 5 red marbles of 14 total marbles (5/14)
There are a total of 25 Marbles The chances are 3 out of 25 drawing a Red marble. 3/25 = 12% chance of drawing a red marble
3 in 10
24 red marbles
Suppose probability of drawing a red marble is p. Then p = 2*(1 - p) that is p = 2 - 2p or p = 2/3 So 2/3 of the 24 marbles are red 24*(2/3) = 16 red marbles.
12
Since the box contains 16 marbles, seven of them white, then the probability of drawing one white marble is 7/16. If you replace the marble and draw again, the probability of drawing another white marble is still 7/16. The net probability of drawing two white marbles, while replacing the first, is 49/256.
There are a total of 25 Marbles The chances are 3 out of 25 drawing a Red marble. 3/25 = 12% chance of drawing a red marble
Without replacement: P(one green and one blue) = P(drawing green then blue) + P(drawing blue then green) = (6/23)(9/22) + (9/23)(6/22) = 104/506 = 52/253 With replacement: P(one green and one blue) = P(green)*P(blue) = (6/23)(9/23) = 54/529
It is (1/2)5 = 1/32
5/15 = 1/3 = 33 and 1/3 percent
Probability of drawing a red marble = 4/16 = 1/4 Probability of drawing not a red marble = 1 - 1/4 = 3/4
There would be a 7/19 or 36.84% chance of drawing a blue marble from the bag.
4/8 or 1/2(probability of first draw) * 3/8(probability of second draw which is 12/64 or 3/16 of the given scenario.
The probability of drawing two reds, with replacement, is the same as the probability of drawing a red, times itself. So: P(drawing two reds) = P(drawing a red)2 = (12/(2 + 12 + 6))2 = (12/20)2 = (3/5)2 = 9/25
56 marbles
700 bags if each bag contains 8 marbles is a total of 5600 marbles.
25%