there are 5 discs, 6 jump ropes, 3 balls, and 12 pieces of sidewalk chalk in a bin. If two items are drawn at random without replacement, what is the probability that both items removed are not jump ropes?

0.584

Step-by-step explanation:

We have:

5 discs

6 jump ropes

3 balls

12 pieces of sidewalk.

5 + 6 + 3 + 12 = 26

If all of them have exactly the same probability of being removed, then:

in the first selection, we do not want to remove a jump rope, so we can remove one disc, one ball or one piece of sidewalk.

The total number of those objects is:

5 + 3 + 12 = 20.

Then the probability of removing one of those objects is:

P1 = 20/26 = 0.769

Now in the second selection, we have the same situation, but now we have 25 objects in total, and because in the previous selection we removed one ball, or one disc, or one piece of sidewalk, the total number of these things now is 19.

So the probability of removing another object of that set is:

P2 = 19/25 = 0.76

The joint probability is equal to the product of the individual probabilities, so we have:

P = P1*P2 = 0.769*0.76 = 0.584

the answers c