Question

If you draw 1 marble and then leave it out of the bag, then draw a second marble, what is the probability of picking a blue marble then a red marble?

a. P(B) = 1/5, P(R) = 1/5
b. P(B) = 4/10, P(R) = 3/9
c. P(B) = 3/10, P(R) = 4/9
d. P(B) = 2/5, P(R) = 2/5

Answer 1

The probability of picking a blue marble then a red marble without replacement can be calculated by multiplying the individual probabilities of each event. The probability of picking a blue marble is 4/7 and the probability of picking a red marble is 1/2, resulting in a final probability of 2/7.

Step-by-step explanation:

The probability of picking a blue marble and then picking a red marble without replacement can be determined by multiplying the individual probabilities of each event.

First, we draw a blue marble. There are 4 blue marbles out of a total of 7 marbles remaining in the bag, so the probability of drawing a blue marble is 4/7.

Next, we draw a red marble. There are now 3 red marbles out of a total of 6 marbles remaining in the bag, so the probability of drawing a red marble is 3/6, which simplifies to 1/2.

Multiplying these probabilities together, we get (4/7) * (1/2) = 4/14 = 2/7.

Therefore, the probability of picking a blue marble then a red marble is P(B) = 2/7, P(R) = 1/2, which corresponds to option d.