Question

Maria had a bag that contained 2 blue marbles, 3 green marbles, 4 red marbles, 5 yellow marbles, and 6 orange marbles. Maria removed one marble without looking and she recorded the result. She then removed a second marble without replacing the first marble.



What is the probability that Maria removed a red marble followed by a yellow marble?

Answer 1

To find the probability that Maria removed a red marble followed by a yellow marble, we can follow these steps:

1. **Count the total number of marbles**:
\[
2 \, (\text{blue}) + 3 \, (\text{green}) + 4 \, (\text{red}) + 5 \, (\text{yellow}) + 6 \, (\text{orange}) = 20 \, \text{marbles}
\]

2. **Calculate the probability of removing a red marble on the first draw**:
\[
P(\text{red first}) = \frac{\text{number of red marbles}}{\text{total number of marbles}} = \frac{4}{20} = \frac{1}{5}
\]

3. **Update the number of marbles after removing one red marble**: After removing a red marble, the total number of marbles is now 19 (since one red marble has been removed), and there are now 3 red marbles left.

4. **Calculate the probability of removing a yellow marble on the second draw**:
\[
P(\text{yellow second} | \text{red first}) = \frac{\text{number of yellow marbles}}{\text{total number of marbles remaining}} = \frac{5}{19}
\]

5. **Calculate the combined probability**:
To find the overall probability of drawing a red marble first and a yellow marble second, we multiply the probabilities from steps 2 and 4:
\[
P(\text{red first and yellow second}) = P(\text{red first}) \times P(\text{yellow second} | \text{red first}) = \frac{1}{5} \times \frac{5}{19} = \frac{5}{95} = \frac{1}{19}
\]

Thus, the probability that Maria removed a red marble followed by a yellow marble is \( \boxed{\frac{1}{19}} \).