Question

jerod has a bag of 20 marbles containing 12 red, 6 green, and 2 blue marbles. jerod selects two marbles at random without replacement. what is the probability that jerod selects two red marbles? a $\frac{9}{28}$ b $\frac{1}{10}$ c $\frac{33}{95}$ d $\frac{1}{190}$

Explanation:

Step1: Calculate first - draw probability

The probability of selecting a red marble on the first draw is the number of red marbles divided by the total number of marbles. There are 12 red marbles and 20 total marbles, so the probability $P_1=\frac{12}{20}=\frac{3}{5}$.

Step2: Calculate second - draw probability

Since we are drawing without replacement, after the first red - marble draw, there are 11 red marbles left and 19 total marbles left. So the probability of selecting a red marble on the second draw given that a red marble was selected on the first draw is $P_2 = \frac{11}{19}$.

Step3: Calculate combined probability

The probability of both events (selecting a red marble on the first draw and a red marble on the second draw) occurring is the product of the probabilities of each event. So $P = P_1\times P_2=\frac{3}{5}\times\frac{11}{19}=\frac{33}{95}$.

Answer:

C. $\frac{33}{95}$