Twenty marbles are in a bag. There are 5 green marbles, 9 blue marbles, and the rest are red. Two marbles are drawn at random without placement. Find the probability that neither is blue
To find the probability that neither of the marbles drawn is blue, we first need to determine the total number of marbles that are not blue.
Given that there are 5 green marbles and 9 blue marbles, the number of red marbles can be calculated as follows:
Total number of marbles - Number of green marbles - Number of blue marbles = 20 - 5 - 9 = 6
Therefore, there are 6 red marbles in the bag.
Next, we need to calculate the total number of ways to draw 2 marbles out of the 20 marbles without replacement. This can be done using a combination formula, denoted as "nCr" or "C(n, r)", which calculates the number of combinations of choosing "r" items from a set of "n" items. In this case:
Number of ways to choose 2 marbles out of 20 = C(20, 2) = (20!)/(2!(20-2)!) = (20!)/(2!18!) = ((20)(19))/((2)(1)) = 190
Now, to calculate the probability that neither of the marbles drawn is blue, we need to determine the number of ways to choose 2 marbles such that both are not blue. In this case, both marbles can be either green or red.
Number of ways to choose 2 marbles that are not blue = C(5, 2) + C(6, 2) = [(5!)/(2!(5-2)!)] + [(6!)/(2!(6-2)!)] = (10) + (15) = 25
Finally, the probability can be calculated by dividing the number of ways to choose 2 marbles that are not blue by the total number of ways to choose 2 marbles:
Probability = (Number of ways to choose 2 marbles that are not blue)/(Total number of ways to choose 2 marbles)
= 25/190
≈ 0.1316
Therefore, the probability that neither of the marbles drawn is blue is approximately 0.1316 or 13.16%.