11/18 x 10/17 = .359
It is 0.6050
If the balls are selected at random, then the probability is 12/95.
with replacement: binominal distribution f(k;n,p) = f(0;5,5/12) without replacement: hypergeometric distribution f(k;N,m,n) = f(0;12,5,5)
The probability is 1 if you draw three balls without replacement. If only one draw, it is 3/5.
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The probability is (6/14)*(5/13) = 30/182 = 0.1648 approx.
If you draw enough balls, without replacement, the probability is 1.The answer depends onhow many balls are drawn, andwhether or not they are replaced.Unfortunately, your question gives no information on these matters.
It is 0.6050
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
If the balls are selected at random, then the probability is 12/95.
with replacement: binominal distribution f(k;n,p) = f(0;5,5/12) without replacement: hypergeometric distribution f(k;N,m,n) = f(0;12,5,5)
The probability of getting 3 white balls in a draw of 5 balls with replacement from an urn containing white balls and black balls is always greater than the same test without replacement, because the number of white balls decreases when you draw a white ball and do not replace it. The ratio of white to black with replacement is constant, and is always less than one, assuming there is at least one black ball. The ratio of white to black without replacement decreases each turn, and is still less than one, and is less than the previous ratio, unless the question asked about 2 white balls or less.
The probability the first ball will be red is 5/8. The probability that the first and second balls will be red is 5/8 x 4/7. The probability that the first, second, and third balls will be red is 5/8 x 4/7 x 3/6, or overall 60/336 (about 17.86%).
The probability is 1 if you draw three balls without replacement. If only one draw, it is 3/5.
The probability that it contains exactly 3 balls is 6/45 = 0.133... recurring.
If you take 4 balls and there are only 3 colors, there is no way you cannot get 2 of the same color. So 100%
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