Question

part a
a bag contains 4 blue and 6 green marbles. two marbles are selected at random from the bag. if the first marble is replaced before the second marble is drawn, what is p(blue
second | green first)?
a $\frac{2}{10}$
b $\frac{4}{10}$
c $\frac{5}{10}$
d $\frac{6}{10}$
part b
if the first marble is not replaced before the second marble is drawn, what is p(blue second | green first)?

Explanation:

Step1: Calculate total number of marbles

Total marbles = 4 (blue) + 6 (green) = 10 marbles.

Step2: Analyze independence for Part A

Since the first marble is replaced, the two events are independent. The probability of getting a blue marble is \(P(\text{blue})=\frac{4}{10}\).

Step3: Analyze dependence for Part B

If the first marble is green (not replaced), then there are 9 marbles left. There are still 4 blue marbles. So \(P(\text{blue}|\text{green first})=\frac{4}{9}\). But since we are looking for the answer to Part A only (as per the options given in the problem), we focus on the independent case.

Answer:

A. \(\frac{4}{10}\)