The probability is zero, because there are no red balls in the bag.
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
Slightly under 50% - (.4925, to be exact).----------------------------------------------------------------------------------------------------Another opinion.The probability that 3 balls are red is;P(3R,2W) =(5/11)(4/10)(3/9)(6/8)(5/7)[5!/(3!2!)] =0.324675...The probability that 4 balls are red is;P(4R,1W) =(5/11)(4/10)(3/9)(2/8)(6/7)[5!/(4!1!)] =0.064935...The probability that 5 balls are red is;P(5R) =5!6!/11! =0.0021645...The probability that at least 3 will be red balls is;P(atleast 3R) =P(3R,2W) + P(4R,1W) + P(5R) ~ 0.3918 ~ 39.2%
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.
Probability of first green: 5/7Probability of second green: 5/7Probability of a red: 2/7Probability of all 3 consecutive successes: (5/7) x (5/7) x (2/7) = 50/343 = 14.58% (rounded)
The probability is (6/14)*(5/13) = 30/182 = 0.1648 approx.
The probability is: 55/200 or 11/40
If I have 3 red balls 3green balls 2 white balls & 5black all in one sack what is the probability of finding a white ball?
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
2/7 = 28.57%
3 out of 13 or 3/13
17 out of 21
Yes, it certainly can if there is only one possible outcome. For instance, the probability of drawing a red ball from a bag containing nothing but red balls is equal to one.
The probability is 1 if you draw three balls without replacement. If only one draw, it is 3/5.
Slightly under 50% - (.4925, to be exact).----------------------------------------------------------------------------------------------------Another opinion.The probability that 3 balls are red is;P(3R,2W) =(5/11)(4/10)(3/9)(6/8)(5/7)[5!/(3!2!)] =0.324675...The probability that 4 balls are red is;P(4R,1W) =(5/11)(4/10)(3/9)(2/8)(6/7)[5!/(4!1!)] =0.064935...The probability that 5 balls are red is;P(5R) =5!6!/11! =0.0021645...The probability that at least 3 will be red balls is;P(atleast 3R) =P(3R,2W) + P(4R,1W) + P(5R) ~ 0.3918 ~ 39.2%
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.
Probability of first green: 5/7Probability of second green: 5/7Probability of a red: 2/7Probability of all 3 consecutive successes: (5/7) x (5/7) x (2/7) = 50/343 = 14.58% (rounded)
The probability that it contains exactly 3 balls is 6/45 = 0.133... recurring.