Question

ex. in a bag there are 3 red marbles, 2 blue marbles, and 4 green marbles. we select one marble and then another.
a. find p(g and b) if we do not replace the first marble.
b. find p(g and b) if we replace the first marble.
c. find p(r and r) if we do not replace the first marble.

Explanation:

Step1: Calculate total number of marbles

Total marbles = 3 (red) + 2 (blue) + 4 (green)=9

Step2: Calculate probability for part a (without replacement)

The probability of first picking a green marble $P(G_1)=\frac{4}{9}$. After picking a green marble, there are 8 marbles left. The probability of then picking a blue marble $P(B_2)=\frac{2}{8}$. By the multiplication - rule for dependent events, $P(G\text{ and }B)=P(G_1)\times P(B_2)=\frac{4}{9}\times\frac{2}{8}=\frac{8}{72}=\frac{1}{9}$

Step3: Calculate probability for part b (with replacement)

The probability of first picking a green marble $P(G_1)=\frac{4}{9}$. Since we replace the first marble, the total number of marbles is still 9 for the second pick. The probability of then picking a blue marble $P(B_2)=\frac{2}{9}$. By the multiplication - rule for independent events, $P(G\text{ and }B)=P(G_1)\times P(B_2)=\frac{4}{9}\times\frac{2}{9}=\frac{8}{81}$

Step4: Calculate probability for part c (without replacement)

The probability of first picking a red marble $P(R_1)=\frac{3}{9}$. After picking a red marble, there are 8 marbles left and 2 red marbles left. The probability of then picking a red marble $P(R_2)=\frac{2}{8}$. By the multiplication - rule for dependent events, $P(R\text{ and }R)=P(R_1)\times P(R_2)=\frac{3}{9}\times\frac{2}{8}=\frac{6}{72}=\frac{1}{12}$

Answer:

a. $\frac{1}{9}$
b. $\frac{8}{81}$
c. $\frac{1}{12}$