Answer:
![\frac{1}{6} ,\frac{1}{6} ,\frac{1}{36}\,,\,independent](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%7D%20%2C%5Cfrac%7B1%7D%7B6%7D%20%2C%5Cfrac%7B1%7D%7B36%7D%5C%2C%2C%5C%2Cindependent)
Step-by-step explanation:
Given: Let A be the event that the first die lands on 2 and B be the event that the second die lands on 2.
To find:
P(A), the probability that the first die lands on 2
P(B), the probability that the second die lands on 2
P(A and B): the probability that the first die lands on 2 and the second die lands on 2
Solution:
Probability refers to chances of occurrence of some event.
Probability = number of favourable outcomes/total number of outcomes
Sample space = ![\left \{ 1,2,3,4,5,6 \right \}](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%201%2C2%2C3%2C4%2C5%2C6%20%5Cright%20%5C%7D)
Total number of outcomes = 6
For P(A):
Number of favourable outcomes = 1
So,
![P(A)=\frac{1}{6}](https://tex.z-dn.net/?f=P%28A%29%3D%5Cfrac%7B1%7D%7B6%7D)
For P(B):
Number of favourable outcomes = 1
So,
![P(B)=\frac{1}{6}](https://tex.z-dn.net/?f=P%28B%29%3D%5Cfrac%7B1%7D%7B6%7D)
P(A and B) = ![P(A)P(B)=\left ( \frac{1}{6} \right )\left ( \frac{1}{6} \right )=\frac{1}{36}](https://tex.z-dn.net/?f=P%28A%29P%28B%29%3D%5Cleft%20%28%20%5Cfrac%7B1%7D%7B6%7D%20%5Cright%20%29%5Cleft%20%28%20%5Cfrac%7B1%7D%7B6%7D%20%5Cright%20%29%3D%5Cfrac%7B1%7D%7B36%7D)
Yes, A and B are independent events as happening of each of the event does not depend on the other.