Final answer:
The probability of drawing one green marble and one red marble from an urn with 7 green and 2 red marbles, without replacement, is calculated by adding the probabilities of both possible sequences of drawing.
The calculation shows that the probability is 7/18, which corresponds to answer choice A.
Step-by-step explanation:
The question asks to find the probability of drawing one green marble and one red marble from an urn containing 7 green and 2 red marbles, when two marbles are drawn without replacement. To solve this, we can use the concept of conditional probability and combinations.
To calculate the probability of getting one green and one red marble, we consider the two scenarios:
- Drawing a green marble first and then a red marble.
- Drawing a red marble first and then a green marble.
The probability for the first scenario is (7/9) * (2/8), since there are 7 green marbles out of 9 total initially and then 2 red marbles out of the remaining 8. The probability for the second scenario is (2/9) * (7/8) for similar reasons. We then add these probabilities together to get the total probability of one marble of each color:
(7/9) * (2/8) + (2/9) * (7/8) = 14/72 + 14/72 = 28/72 = 7/18
The correct answer is 7/18, which is choice A. Therefore, the probability of drawing one marble of each color is 7/18 when drawing two marbles without replacement.