The probability is B*G/(B+G+R)^2where
B = number of Blue marbles
G = number of Green marbles and
R = number of marbles of other colours.
probability of pulling out a purple marble = 20/85probability of NOT pulling out a purple marble = 1 - 20/85 = 65/85 = 13/17
your probability would be 13/13. you would have a 100 percent chance of getting a green marble
the probability is you'd get a green marble any other color is impossible. So, the probability is certain
It depends on how many yellow-green marbles there are, and on how many total marbles there are. There is insufficient information in the question to answer it. Please restate the question, giving this other information.
100%
probability of pulling out a purple marble = 20/85probability of NOT pulling out a purple marble = 1 - 20/85 = 65/85 = 13/17
your probability would be 13/13. you would have a 100 percent chance of getting a green marble
the probability is you'd get a green marble any other color is impossible. So, the probability is certain
5/(5+6+9) = 5/20 = 1/4 or 0.25
It depends on how many yellow-green marbles there are, and on how many total marbles there are. There is insufficient information in the question to answer it. Please restate the question, giving this other information.
The probability of choosing a blue marble is 5 in 15 or 1 in 3. The probability of then choosing a green marble is 5 in 14. (One is missing) Multiply the two probabilities and you get 5 in 42.(P = 0.1190... about 12%).
Let's consider first one bag: The probability to grab one green marble is Pg=1/5 The probability to grab one red marble is Pr=1-Pg=4/5 For the 4 bags, it's a binomial distribution: probability to get k green out of n bags, P(X=k)=nCk pk (1-p)n-k = nCk Pgk Prn-k Now, the probability to grab at least 2 green marbles from 4 bags is 1 - (Probability to get no green marble from 4 bags) - (Probability to get just one green marble from 4 bags) Probability to get no green marble is = 4C0 (4/5)4 = (4/5)4 (n=4, k=0, no green marble from each bag, 4C0 Pr4) Probability to get just one green marble is = 4C1 (1/5) (4/5)3 (n=4, k=1, one green marble from one bag and red marbles from the other ones, 4C1 Pg Pr3) Probability to grab at least 2 green marbles from 4 bags is 1-(4/5)4-4*(1/5)(4/5)3 = 0.1801
The probability of pulling a red or yellow marble out of a bag of 3 green 8 red 8 yellow and 3 black marbles is 16 out of 22, or about 0.73.
100%
This is the same as the probability of choosing either a red of a blue marble. There are 5+4 out of 15 ways of doing this. The probability is therefore 9/15 = 3/5.
If one marble is chosen at random, the probability is 6/(4+6+5) = 6/15 = 2/5
Probability of drawing a red marble = 4/16 = 1/4 Probability of drawing not a red marble = 1 - 1/4 = 3/4