Question

comparing the probabilities of independent and dependent events
there are 4 red marbles, 2 white marbles, and 2 black marbles in a bag. consider the following compound events.
what is the probability that maria will pick two red marbles if she returns the first marble to the bag?
$p((red))p((red)) = \left(\frac{4}{8}\
ight)\left(\frac{4}{8}\
ight)$
$=$
what is the probability that maria will pick up two red marbles in a row if she did not replace the first marble in the bag?
$p((red))p((red)) = \left(\frac{4}{8}\
ight)\left(\frac{3}{7}\
ight)$
$=$
1/8
3/14
4/12
3/8

Explanation:

Step1: Calculate probability with replacement

The total number of marbles is \(4 + 2+2=8\). The probability of picking a red marble on the first draw and replacing it is \(P(\text{red})=\frac{4}{8}\), and since the marble is replaced, the probability of picking a red marble on the second draw is also \(P(\text{red})=\frac{4}{8}\). The probability of both events is \(P = \frac{4}{8}\times\frac{4}{8}=\frac{16}{64}=\frac{1}{4}\).

Step2: Calculate probability without replacement

The probability of picking a red marble on the first draw is \(P_1(\text{red})=\frac{4}{8}\). Since the marble is not replaced, there are now 3 red marbles and 7 total marbles left. So the probability of picking a red marble on the second draw is \(P_2(\text{red})=\frac{3}{7}\). The probability of both events is \(P=\frac{4}{8}\times\frac{3}{7}=\frac{12}{56}=\frac{3}{14}\).

Answer:

The probability when replacing the first marble is \(\frac{1}{4}\).
The probability when not - replacing the first marble is \(\frac{3}{14}\).