An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles.
If two marbles are picked at random, what is the probability that either both are green or both are yellow?

Options

A. `5/91`
B. `1/35`
C. `1/3`
D. `4/105`

Answer

To solve the problem of finding the probability that either both marbles picked are green or both are yellow, we will follow these steps:

### Step 1: Determine the total number of marbles in the urn.
The urn contains:
- 6 red marbles
- 4 blue marbles
- 2 green marbles
- 3 yellow marbles

**Total number of marbles = 6 + 4 + 2 + 3 = 15.**

### Step 2: Calculate the total number of ways to pick 2 marbles from 15.
The number of ways to choose 2 marbles from 15 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).

So, the total ways to choose 2 marbles from 15 is:
\[ C(15, 2) = \frac{15!}{2!(15-2)!} = \frac{15 \times 14}{2 \times 1} = 105. \]

### Step 3: Calculate the probability of picking 2 green marbles.
There are 2 green marbles in the urn. The number of ways to choose 2 green marbles is:
\[ C(2, 2) = 1. \]

Thus, the probability of picking 2 green marbles is:
\[ P(\text{both green}) = \frac{C(2, 2)}{C(15, 2)} = \frac{1}{105}. \]

### Step 4: Calculate the probability of picking 2 yellow marbles.
There are 3 yellow marbles in the urn. The number of ways to choose 2 yellow marbles is:
\[ C(3, 2) = 3. \]

Thus, the probability of picking 2 yellow marbles is:
\[ P(\text{both yellow}) = \frac{C(3, 2)}{C(15, 2)} = \frac{3}{105} = \frac{1}{35}. \]

### Step 5: Combine the probabilities.
Since we want the probability that either both marbles are green or both are yellow, we add the two probabilities together:
\[ P(\text{both green or both yellow}) = P(\text{both green}) + P(\text{both yellow}) = \frac{1}{105} + \frac{3}{105} = \frac{4}{105}. \]

### Final Answer:
The probability that either both marbles picked are green or both are yellow is:
\[ \frac{4}{105}. \]

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