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If A box contains three black balls and four gold balls Two balls are randomly drawn in succession from the box If there is no replacement what is the probability that both balls are black?
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
Suppose 5 balls are drawn simultaneously from an urn containing 5 red and 6 white balls. Find the probability that at least 3 will be red balls.?
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%
There are 2 red and 5 green balls in a bag If you randomly choose balls one at a time with replacement what is the probability of choosing 2 green balls and then 1 red ball?
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)
If you draw 5 balls with replacement from an urn containing white balls and black balls is the probability of getting 3 white balls always higher than if you draw the balls without replacement?
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.
What is the probability that a green ball is drawn from a red and blue balls?
The probability of drawing a green ball from a collection that only contains red and blue balls is 0%. Since there are no green balls present in the selection, it is impossible to draw one, making the probability zero.
There are 200 balls there are 100 white balls 55 red balls 45 black balls if a student randomly picks up a ball what's the the probability that he picks up a red ball?
The probability is: 55/200 or 11/40
What are questions you could ask about probability?
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?
If A box contains three black balls and four gold balls Two balls are randomly drawn in succession from the box If there is no replacement what is the probability that both balls are black?
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
What is the probability of getting a red balls from an urn containing 2 red 3 green and 2 white balls?
2/7 = 28.57%
What is the probability of picking a red ball out of a bag containing 3 red balls 4 blue balls and 6 green balls?
3 out of 13 or 3/13
One ball is drawn at random from a bag containing 17 red balls and 4 white balls what is the probability that the ball is red?
17 out of 21
Can the probability of an outcome be 1?
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.
What is the chance of drawing a red ball from a bag containing 3 red and 2 white balls?
The probability is 1 if you draw three balls without replacement. If only one draw, it is 3/5.
Suppose 5 balls are drawn simultaneously from an urn containing 5 red and 6 white balls. Find the probability that at least 3 will be red balls.?
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%
A box contains three red and two black balls Four balls are removed from the box one by one without replacement The probability of the ball remaining in the box being red is?
To find the probability of the remaining ball being red after removing four balls, we first consider the total combinations of removing four balls from the box containing three red and two black balls. The only possible scenario where a red ball remains is if at least one red ball is not removed. Given that there are three red balls, the probability of removing all four balls as red is impossible. Hence, the probability of at least one red ball remaining is certain, making the probability of the remaining ball being red effectively (1) or (100%), assuming that we only have a single ball left in the box.
There are 2 red and 5 green balls in a bag If you randomly choose balls one at a time with replacement what is the probability of choosing 2 green balls and then 1 red ball?
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)
If you draw 5 balls with replacement from an urn containing white balls and black balls is the probability of getting 3 white balls always higher than if you draw the balls without replacement?
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.
