Question

A paper bag contains 3 blue marbles, 2 red marbles, 4 green marbles, and 1 white marble. The marbles are drawn from the bag, one after the other, and not put back in the bag after drawing. What is the probability of drawing one blue marble and one red marble (in either order)?

Answer 1

2/15

Explanation:

There are 10 marbles in total. After the first marble is drawn, there are 9 left.

P(blue then red) = (3/10) × (2/9) = 6/90 = 1/15

P(red then blue) = (2/10) × (3/9) = 6/90 = 1/15

Therefore, the probability of either is:

1/15 + 1/15 = 2/15

Answer 2

To find the probability of drawing one blue marble and one red marble (in either order) from the bag, we can use the concept of combinations.

First, let's determine the total number of marbles in the bag. There are 3 blue marbles, 2 red marbles, 4 green marbles, and 1 white marble, so the total number of marbles is 3 + 2 + 4 + 1 = 10.

Next, let's find the number of ways we can choose one blue marble and one red marble. We can choose one blue marble from the 3 blue marbles in C(3, 1) = 3 ways, and one red marble from the 2 red marbles in C(2, 1) = 2 ways. Since the order of drawing doesn't matter, we need to consider both possibilities: blue then red or red then blue.

The total number of ways to choose one blue marble and one red marble is 3 * 2 * 2 = 12.

Now, let's find the total number of possible outcomes. Since we are drawing marbles one after the other without replacement, the total number of possible outcomes is the number of ways to choose 2 marbles from the 10 marbles in the bag, which is C(10, 2) = 45.

Finally, we can calculate the probability by dividing the number of favorable outcomes (choosing one blue and one red marble) by the total number of possible outcomes:

Probability = Number of favorable outcomes / Total number of possible outcomes

= 12 / 45

= 4 / 15

So, the probability of drawing one blue marble and one red marble (in either order) from the bag is 4/15.