There are 6 red marbles 4 blue marbles to yellow marbles. Sabrina picks 2 marbles without putting the first one back what is the probability that she picks a not yellow marble and then a blue marble.

P = 3/11

Step-by-step explanation:

There are:

6 red marbles

4 blue marbles

2 yellow marbles.

So there are a total of 12 marbles in the bag, such that each one has the same probability of being randomly picked.

We want to find the probability that the first pick is not yellow (so it can be either blue or red), and the second pick is a blue marble.

Now there are two cases.

The first marble is red and the second blue

the first marble is blue and the second blue

We need to find the probability in each case.

The probability that the first marble is red is equal to the quotient between the number of red marbles (6) and the total number of marbles (12)

p = 6/12

Now, for the second draw, we need a blue marble, the probability in this case is the quotient between the number of blue marbles (4) and the total number of marbles in the bag (now there are 11, because one is already taken)

q = 4/11

The joint probability is equal to the product of the individual probabilities, then:

P₁ = p*q = (6/12)*(4/11) = 2/11

And for the other case, we draw two blue marbles:

For the first draw, the probability is:

p = 4/12

And for the second draw, now there are 3 blue marbles and 11 total marbles, then the probability is:

q = 3/11

The joint probability is:

P₂ = (4/12)*(3/11) = 1/11

The total probability is then:

P = P₁ + P₂ = 2/11 + 1/11 = 3/11

is there any photos or whatever?