What is the probability that 2 red marbles are drawn from a jar that contains 4 red, 5 white, and 3 blue?

Answer 1

4/12 * 3/11 = 1/11

Answer 2

Start with basics

Probability = number of desirable outcomes / num of all possible outcomes.

I assume that the marbles are not replaced.

the jar contains 4R, 5W and 3B marbles. A total of 12 marbles.

The probability that the first drawn is a red is therfore = 4/12
(from defn on line one in this answer)

If it is not replaced, the probability that the second drawn is red =3/11 (three red marbles left, eleven in total)

So the probability that the first drawn AND the second drawn are red is 4/12 * 3/11 (the AND indicates we have to multiply)
=12/132

which is around 9 % ///

If we were to put the second marble and draw again the probability the second marble is red would be the same as the first = 4/12

And both would be 4/12 * 4/12 = 16/144 = 11% (this is logical because if we put back a red marble we'd expect the chance to improve)

Answer 3

Prob = Success/Total

So,

(Are we replacing the marbles?)

If we are replacing:

P = 4/12 = 1/3

Answer 4

chance of 1st red is 4/12
chance of another red is 3/11, because you just took one out.

multiply and get 12/132, which simplifies to 1/11

P(R,R) = 1/11

Answer 5

Yes, I think it is 1/11. Here is this one other line of reasoning:

There are 12x11 different outcomes. (because for your first marble you have 12 choices, and then for your second marble you have 11 choices)

Out of these 12x11 there are 12 cases that are both red. Thus 12/(12x11)=1/11

The 12 Red-Red pairs are:R1-R2, R1-R3, R1-R4, then R2-R1, R2-R3, R2-R4..., etcetera.

Answer 6

(4/12)(3/11) = (1/3)(3/11) = 3/33 = 1/11 or about 9.09%

Answer 7

1/6...