A bag contains 9 marbles: 3 are green, 4 are red, and 2 are blue. Deshaun chooses a marble at random, and without putting it back, chooses another one at random. What is the probability that both marbles he chooses are red? Write your answer as a fraction in simplest form.
Help me pleaseeee!! I am stuck and in need of help?
there is a four in nine chance that the 1st marble is red
... and a three in eight chance that the 2nd marble is red
(4 / 9) * (3 / 8) = ?
To find the probability of choosing two red marbles, we first need to determine the total number of possible outcomes and the number of favorable outcomes.
The total number of marbles in the bag is 9, so for the first pick, Deshaun has 9 choices.
After Deshaun selects the first marble, there will be a total of 8 marbles left in the bag. Since Deshaun does not replace the first marble, for the second pick, he will have 8 choices.
The number of favorable outcomes, in this case, is the number of ways to choose 2 red marbles out of the 4 red marbles. We can calculate this using combinations.
The number of ways to choose 2 items from a set of 4 is equal to 4C2, which can be calculated as:
4C2 = 4! / ((4-2)! * 2!) = 6.
So, there are 6 ways to choose 2 red marbles out of the 4 red marbles.
The total number of possible outcomes is the number of ways to choose 2 marbles out of the 9 marbles, which can be calculated as:
9C2 = 9! / ((9-2)! * 2!) = 36.
Therefore, the probability of choosing two red marbles is 6/36, which simplifies to 1/6.
So, the probability that both marbles Deshaun chooses are red is 1/6.