A bag contains 2 white marbles, 4 blue marbles, and 5 red marbles. Three marbles are drawn from the bag. What is the probability that not all of them are red?

First find the prob that ALL of them ARE red, which is

(6/11)(5/10)(4/9) = 2/33

So the prob that not all of them are red
= 1-2/33
=31/33

There are 5 marbles in a box. One is black, one is yellow, one is red, one is purple and one is orange.

Please answer this

The answer is 5 marbles in a box. One is black, one is yellow, one is red, one is purple and one is orange.

It will be 5/5 because it 's only one each

To find the probability that not all of the three marbles drawn are red, we need to calculate the probability of drawing at least one non-red marble.

First, let's find the total number of ways to draw three marbles from the bag. This can be calculated using the combination formula:

C(n, r) = n! / (r!(n-r)!)

where "n" is the total number of marbles in the bag, and "r" is the number of marbles drawn.

In this case, n = 11 (2 white + 4 blue + 5 red), and r = 3. So, the total number of ways to draw three marbles is:

C(11, 3) = 11! / (3!(11-3)!) = 11! / (3!8!) = (11 * 10 * 9) / (3 * 2) = 165

Now, let's find the number of ways to draw three red marbles. We have a total of 5 red marbles in the bag, so the number of ways to draw three red marbles is:

C(5, 3) = 5! / (3!(5-3)!) = 5! / (3!2!) = 10

Finally, to find the probability that not all of the three marbles drawn are red, we subtract the number of ways to draw three red marbles from the total number of ways to draw three marbles, and divide by the total number of ways:

P(not all red) = (total ways - ways to draw 3 red marbles) / total ways
= (165 - 10) / 165
= 155 / 165
= 31 / 33

Therefore, the probability that not all of the three marbles drawn are red is 31/33, or approximately 0.939.