Question

You have a bag of gummy worms. 12 are green, 3 are blue, 6 are red and 2 are orange. What is the probability that you will reach into the bag and pull a red worm, eat it, and then pull a green worm?answer choices:18/23 36/253 1/529. 72/529

Answer 1

To answer this question, we can proceed as follows:

1. Count the total of gummy worms into the bag:

12 green + 3 blue + 6 red + 2 orange = 23 gummy worms.

2. In the first event (pull a red worm), we have that the probability of it is:

`P(R)=(6)/(23)`

Since we have 6 red gummy worms from a total of 23.

3. Now, the probability of the second event is a little different. You have eaten one of the red gummy worms. Then, we have a total of 23 - 1 = 22 gummy worms.

The probability, now, of getting a green worm is (there are 12 green gummy worms):

`P(G)=(12)/(22)\Rightarrow P(G)=(6)/(11)`

4. Finally, the probability of these two events is, applying the multiplication rule:

`P(R\cap G)=(6)/(23)\cdot(6)/(11)\Rightarrow P(R\cap G)=(36)/(253)`

Then, the answer is 36/253 (second option.)

This is a case of conditional probability. We see that the probability of the second event was influenced by the first event.