From a box containing 7 blue, 5 red, 3 green, and 9 yellow marbles, 2 are drawn one at a time without replacing the first marble before the second marble is drawn. Find the probability that one is blue and one is yellow.
Blue then yellow:
P(BY)=7/24+9/23
Yellow then blue:
P(YB)=9/24+7/23
P(YB∨BY)=P(YB)+P(BY)
To find the probability that one marble is blue and one marble is yellow, we need to calculate the total number of ways to draw two marbles without replacement and count the number of ways to draw one blue marble and one yellow marble.
Step 1: Find the total number of ways to draw two marbles without replacement.
The total number of marbles in the box is 7 + 5 + 3 + 9 = 24. So, there are 24 marbles to choose from for the first draw, and 23 marbles remaining for the second draw (after the first marble is drawn without replacement). Thus, the total number of ways to draw two marbles without replacement is 24 * 23.
Step 2: Find the number of ways to draw one blue and one yellow marble.
Since we want one blue and one yellow marble, we need to calculate the number of ways to choose one blue marble from the 7 available and one yellow marble from the 9 available. The number of ways to choose one blue marble is 7, and the number of ways to choose one yellow marble is 9. Therefore, the number of ways to draw one blue and one yellow marble is 7 * 9.
Step 3: Calculate the probability.
The probability of an event occurring is given by the number of favorable outcomes divided by the number of possible outcomes. In this case, the number of favorable outcomes is the number of ways to draw one blue and one yellow marble (7 * 9) and the number of possible outcomes is the total number of ways to draw two marbles without replacement (24 * 23). Therefore, the probability is (7 * 9) / (24 * 23).
Now, let's calculate the probability:
(7 * 9) / (24 * 23) = 63 / 552 = 0.1141304348
So, the probability that one marble is blue and one marble is yellow is approximately 0.1141304348 or 11.41%.