Question

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?

Answer options with 5 options
A.
StartFraction 1 over 6 EndFraction

B.
StartFraction 2 over 24 EndFraction

C.
StartFraction 5 over 22 EndFraction

D.
StartFraction 1 over 4 EndFraction

E.
StartFraction 1 over 3 EndFraction

Answer 1

To find the probability that both marbles picked are blue, we can follow these steps:

1. Count the total number of marbles:
- Blue marbles = 6
- Red marbles = 4
- Green marbles = 2
- Total marbles = 6 + 4 + 2 = 12

2. Calculate the total number of ways to choose 2 marbles from 12 marbles:
- The number of ways to choose 2 marbles from 12 can be calculated using the combination formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).
- Here, \(n = 12\) and \(k = 2\):

\[
\binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66
\]

3. Calculate the number of ways to choose 2 blue marbles from 6 blue marbles:
- Similarly, we use the combination formula with \(n = 6\) and \(k = 2\):

\[
\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
\]

4. Now, we can find the probability that both marbles picked are blue:
\[
P(\text{both blue}) = \frac{\text{Number of ways to choose 2 blue marbles}}{\text{Total ways to choose 2 marbles}} = \frac{15}{66}
\]

5. Simplifying the fraction \(\frac{15}{66}\):
- The greatest common divisor (GCD) of 15 and 66 is 3:
\[
\frac{15 \div 3}{66 \div 3} = \frac{5}{22}
\]

Therefore, the probability that both marbles picked are blue is \(\frac{5}{22}\).

The correct option is **C. \( \frac{5}{22} \)**.