The probability is 1/9.
Because you are replacing the marbles then it is an independent event. P(1st one is not green) = 1 - P(first green), equally P(2nd one is not green) = 1 - (second green), Thus it reads P(¬G ^ ¬G) = P(¬G) * P(¬G) = 15/20 * 15/20 = 225/400 = 9/16
If the two marbles are drawn without replacement, the probability is 16/33.
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1 out of 15 Probab. = Prob. of red x Prob. of blue Probab. = (3/10)x(2/9) = 5/90 = 1/15
8:6
Total marbles in the bag = 10Number of red ones = 3Probability of pulling a red one on the first draw = 3/10 = 0.3Total marbles remaining after the first draw = 9Number of green ones = 5Probability of pulling a green one after a red one has been withdrawn = 5/9Probability of both outcomes = (3/10) x (5/9) = (15/90) = 1/6 = (16 and 2/3) percent.
Because you are replacing the marbles then it is an independent event. P(1st one is not green) = 1 - P(first green), equally P(2nd one is not green) = 1 - (second green), Thus it reads P(¬G ^ ¬G) = P(¬G) * P(¬G) = 15/20 * 15/20 = 225/400 = 9/16
He has 10 green marbles.
He will have 13 blue marbles and 10 green marbles.
If the two marbles are drawn without replacement, the probability is 16/33.
10 Green marbles, 13 Blue marbles.
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.
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sure chance
There are at least 11 green marbles in the bag.
We can't answer that without knowing what else is in the bowl.
1 chance in 10.10 %