If you draw enough balls, without replacement, the probability is 1.
The answer depends on
Unfortunately, your question gives no information on these matters.
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
The probability is 0.
The probability is zero, because there are no red balls in the bag.
The probability of choosing a red or black card from a standard deck of 52 cards is 52 in 52, or 1 in 1. In other words, it will happen no matter what.
3/12*3/11 = 9/132, or 6.818%.
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
2
11/18 x 10/17 = .359
A laundry bag contains 160 black socks and 300 red socks. 1940 black socks must be added so that the probability of choosing a black sock is 7 of 8.
The probability of a black ball in bag 1 is 4/12 = 1/3. If you add 3 black balls to bag 2, it will contain 5 black balls out of 15: the probability of a black ball being 3/15 = 1/3.
The probability is 0.
The probability, or probility, even, is 0 since tere can be no such thing as "choosing red card of the black".
The probability is zero, because there are no red balls in the bag.
The probability is: 55/200 or 11/40
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.
The probability of choosing a red or black card from a standard deck of 52 cards is 52 in 52, or 1 in 1. In other words, it will happen no matter what.
3/12*3/11 = 9/132, or 6.818%.