How many marbles are selected at random from an urn?

Give your answer as a decimal number, with 3 decimal places. You can put this solution on YOUR website! An urn contains 4 red, 6 white, and 5 blue marbles. Three marbles are selected at random and without replacement. What is the probability that one of each color is selected?

What is the probability that 2 randomly chosen marbles are not pink?

To obtain the probability that is asked, simply compute 1 – (2/9) = 7/9. The probability that the 2 randomly chosen marbles are not both pink is 7/9.

What is the probability of drawing both aces without replacement?

Thus, for the first ace, there is a 4/52 probability and for the second there is a 3/51 probability. The probability of drawing both aces without replacement is thus 4/52*3/51, or approximately .005. In a bag, there are 10 red, 15 green, and 12 blue marbles.

What is the probability of drawing a red on the first?

On the first, you have 10/16 chances to draw a red. Supposing this red is not replaced, the chance of drawing a second red will be 9/15; therefore, the probability of A is (10/16) * (9/15) = 0.375. Event B is translated into 2 events: Blue + (White or Red) or (White or Red) + Blue.

What is the probability of getting the second black marble?

After you get a red, now we need to calculate the probability of getting a second black. There was no replacement, so you have 1 black in your hand and 3 in the urn. There is a 3 9 probability of getting the second black marble. 9 marbles in total because you have one in your hand.

What is the probability of drawing a black from the urn?

There are 4 blacks to possible draw from the urn, there are 10 total marbles. So the probability of drawing 1 black is 4 10 . After you get a red, now we need to calculate the probability of getting a second black. There was no replacement, so you have 1 black in your hand and 3 in the urn.

How many ways can you choose 3 marbles from 15 different marbles?

15C3 is the denominator (i.e. the number of ways of choosing 3 marbles out of 15 distinct ones). The sample space that comprises one green, one red, and one black marble is 5 x 4 x 6 combinations. In other words, the answer is 5 x 4 x 6 / 15C3 = 5 x 4 x 6 x 3! / (15 x 14 x 13) = 4 x 3! / 91 = 24 / 91.