Answer:
1. 1/6
2. 1/6
Step-by-step explanation:
Let A be the event that the sum of the two die is 6 and B be an event that the green die is either 4 or 1.
The conditional probability will be given by P (A/B) = P (A∩B)/ P (B).
Now the total sample space consists of 36 outcomes .
And to find (A∩B) we need to find the outcomes in which green die is either 4 or 1 and the sum of the two die is 6.
So when green is 1 red must be 5
So when green is 4 red must be 2
So there are two ways in which green die is either 4 or 1 and the sum of the two die is 6.
Therefore the probability of (A∩B)= P (A∩B)= 2/36= 1/18
Now we find the probability of green die having 4 or 1
So when green is 4 red can have all the numbers from 1- 6
And when green is 1 red can have all the numbers from 1- 6
The total number would be 12 .
So probability of green die having 1 or 4 is given by = P (B)= 12/36
Now the conditional probability = P (A/B) = P (A∩B)/ P (B)=1/18/ 1/3
= 3/18= 1/6
2. Similarly we find the conditional probability of the two die when the red one is 6, given that the sum is 11.
When red is 6 the green must be 5 to get 11. So the probability
=P (A∩B)= 1/36
Now we find the probability of red die having 6 =P(B)= 6/36
Now the conditional probability = P (A∩B)/P(B) = 1/36/ 6/36= 1/6