A bag contains 4 red, 3 yellow, 3 blue, and 2 white marbles, for a total of 12 marbles. If Lucia randomly picks two marbles from the bag without putting the first marble back, what is the probability that the first marble is red and the second is white?

1 answer

To find the probability that the first marble Lucia picks is red and the second marble is white, we need to follow these steps:

  1. Calculate the total number of marbles: \[ 4 \text{ (red)} + 3 \text{ (yellow)} + 3 \text{ (blue)} + 2 \text{ (white)} = 12 \text{ marbles} \]

  2. Find the probability of picking a red marble first: The number of red marbles is 4 out of a total of 12 marbles. Thus, the probability of picking a red marble first is: \[ P(\text{Red first}) = \frac{4}{12} = \frac{1}{3} \]

  3. Calculate the probability of picking a white marble second after a red marble has been picked: After picking one red marble, there are now:

    • Total marbles: \( 12 - 1 = 11 \)
    • Red marbles remaining: \( 4 - 1 = 3 \)
    • Yellow marbles remaining: 3
    • Blue marbles remaining: 3
    • White marbles remaining: 2

    The probability of picking a white marble second is then the number of white marbles divided by the new total: \[ P(\text{White second | Red first}) = \frac{2}{11} \]

  4. Combine the probabilities: The probability of both events happening (picking a red marble first and a white marble second) is the product of the two probabilities calculated: \[ P(\text{Red first and White second}) = P(\text{Red first}) \times P(\text{White second | Red first}) = \frac{1}{3} \times \frac{2}{11} \] \[ = \frac{2}{33} \]

Thus, the probability that the first marble is red and the second one is white is: \[ \boxed{\frac{2}{33}} \]