A bag contains 6 red marbles nad 4 blue marbles if Dana draws one marble from the bag and then draws another without replacing the first, what is the probability that both marbles will be red?
To find the probability of drawing two red marbles without replacement, we need to consider the number of favorable outcomes (drawing two red marbles) and the number of possible outcomes (drawing any two marbles from the bag).
First, let's determine the number of possible outcomes. When drawing the first marble, there are a total of 10 marbles in the bag (6 red + 4 blue). After drawing the first marble, there will be 9 marbles left in the bag.
For the second draw, since the first marble was not replaced, there will be one less marble in the bag, resulting in a total of 9 marbles to choose from for the second draw.
Next, let's determine the number of favorable outcomes. Since we want both marbles to be red, the first draw must be red, leaving us with 5 red marbles. For the second draw, there will be 4 red marbles remaining.
To calculate the probability, divide the number of favorable outcomes by the number of possible outcomes:
Probability = Favorable outcomes / Possible outcomes
Probability = (Number of ways to choose 2 red marbles) / (Number of ways to choose any 2 marbles)
Probability = (6C1 * 5C1) / (10C1 * 9C1)
(Note: nCr represents combinations, which calculates the number of ways to choose r items from a set of n items without considering the order)
Evaluating the expression:
Probability = (6 * 5) / (10 * 9)
Probability = 30 / 90
Probability = 1 / 3
Therefore, the probability of drawing two red marbles is 1/3.