a bag has 6 blue marbles, 4 red marbles, and 2 green marbles. two marbles are randomly chosen without…
a bag has 6 blue marbles, 4 red marbles, and 2 green marbles. two marbles are randomly chosen without replacement. what is the probability that both marbles picked are blue?\na. $\frac{1}{6}$\nb. $\frac{2}{24}$\nc. $\frac{5}{22}$\nd. $\frac{1}{4}$\ne. $\frac{1}{3}$
Answer
Explanation:
Step1: Calculate total marbles
Total marbles = 6 + 4+ 2 = 12
Step2: Calculate first - blue marble probability
The probability of picking the first blue marble is $\frac{6}{12}=\frac{1}{2}$
Step3: Calculate second - blue marble probability
Since one blue marble is already picked without replacement, there are 5 blue marbles left and 11 marbles in total. So the probability of picking the second blue marble is $\frac{5}{11}$
Step4: Calculate combined probability
The probability of both events happening is the product of their probabilities. So $P=\frac{1}{2}\times\frac{5}{11}=\frac{5}{22}$
Answer:
C. $\frac{5}{22}$