Question

A bag of contains 4 blue marbles, 6 green marbles and 3 yellow marbles. If two marbles are drawn at random from the bag, whats the probability of:

First drawing a green marble, and then drawing a yellow marble? Without Replacement


18/156


19/156


10/156


11/156

Answer 1

To calculate the probability of first drawing a green marble and then drawing a yellow marble without replacement, we can follow these steps:

1. **Count the total number of marbles**:
- Blue marbles: 4
- Green marbles: 6
- Yellow marbles: 3
- Total marbles = 4 + 6 + 3 = 13

2. **Probability of drawing a green marble first**:
- Number of green marbles = 6
- Total marbles = 13
- Probability of drawing a green marble \( P(G) = \frac{6}{13} \)

3. **Probability of drawing a yellow marble second**:
- After drawing a green marble, there will be 12 marbles left (13 - 1).
- Number of yellow marbles still = 3
- Probability of drawing a yellow marble \( P(Y|G) = \frac{3}{12} = \frac{1}{4} \)

4. **Combine the probabilities**:
- The probability of both events happening in order is:
\[
P(G) \times P(Y|G) = \frac{6}{13} \times \frac{1}{4} = \frac{6}{52} = \frac{3}{26}
\]

5. **Convert to a common denominator for comparison**:
- To express \(\frac{3}{26}\) with a denominator of 156:
\[
\frac{3}{26} = \frac{3 \times 6}{26 \times 6} = \frac{18}{156}
\]

Thus, the final probability of first drawing a green marble and then drawing a yellow marble is \(\frac{18}{156}\).

The correct answer is:

**18/156**