Question

a bag contains 9 marbles: 2 are green, 3 are red, and 4 are blue. bob 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.

Explanation:

Step1: Probability of first red marble

There are 3 red marbles out of 9 total. So, the probability of choosing a red marble first is $\frac{3}{9}$.

Step2: Probability of second red marble

After removing one red marble, there are 2 red marbles left and 8 total marbles. So, the probability of choosing a red marble second is $\frac{2}{8}$.

Step3: Multiply the probabilities

To find the probability of both events happening, multiply the two probabilities: $\frac{3}{9} \times \frac{2}{8} = \frac{6}{72}$. Simplify this fraction by dividing numerator and denominator by 6: $\frac{6 \div 6}{72 \div 6} = \frac{1}{12}$.

Answer:

$\frac{1}{12}$