Question:
Number X is randomly selected from the set of odd numbers and Y is randomly selected from the set of even numbers of the set {1, 2, 3, 4, 5, 6, 7}. Let Z = (X + Y). What is P (Z > 11) equal to?
Number X is randomly selected from the set of odd numbers and Y is randomly selected from the set of even numbers of the set {1, 2, 3, 4, 5, 6, 7}. Let Z = (X + Y). What is P (Z > 11) equal to?
Options:
- 1/6
- 0
- 1/4
- 1/12
- 1/6
- 0
- 1/4
- 1/12
Correct Answer: a)
Explanation:
‘ Gi√en that, X = Set of odd numbers from the set A.
1 Y = Set of e√en numbers from set A.
‘ LetsetA=1,2,3,4,56,7 and Z=X +Y
‘ Now, Z > 11 is only possibel when X =7 and Y = 6
1 .”. Fa√ourable case = 7, 6
‘ = n(E3)=1
. _nBs) _ 1
‘ SPEZ> 1D) = = 1p
‘ Gi√en that, X = Set of odd numbers from the set A.
1 Y = Set of e√en numbers from set A.
‘ LetsetA=1,2,3,4,56,7 and Z=X +Y
‘ Now, Z > 11 is only possibel when X =7 and Y = 6
1 .”. Fa√ourable case = 7, 6
‘ = n(E3)=1
. _nBs) _ 1
‘ SPEZ> 1D) = = 1p