STA1501
ASSIGNMENT 2
2023
, PART A
Solution:
A = {2,4,6}
3
P(A) = = 0.5, then
6
P(Ac ) = 1 − P(A)
= 1 − 0.5
= 0.5 (TRUE)
Solution:
Assuming the events are independent
4 2
P(A) = =
6 3
4 2
P(B) = =
6 3
P(A and B) = P(A) ∙ P(B)
2 2
= ×
3 3
, = 0.44 (FALSE)
Solution:
P(A) = Probability that the selected woman did give birth before marriage
P(Ac ) = Probability that the selected woman did not give birth before marriage
P(A) = 0.73
P(Ac ) = 1 − P(A)
= 1 − 0.73
= 0.27
Probability that the selected woman did not give birth before marriage = 0.27
ANSWER: (FALSE)
Solution:
ANSWER: TRUE
Solution:
For mutually exclusive events P(A and B) = 0
ASSIGNMENT 2
2023
, PART A
Solution:
A = {2,4,6}
3
P(A) = = 0.5, then
6
P(Ac ) = 1 − P(A)
= 1 − 0.5
= 0.5 (TRUE)
Solution:
Assuming the events are independent
4 2
P(A) = =
6 3
4 2
P(B) = =
6 3
P(A and B) = P(A) ∙ P(B)
2 2
= ×
3 3
, = 0.44 (FALSE)
Solution:
P(A) = Probability that the selected woman did give birth before marriage
P(Ac ) = Probability that the selected woman did not give birth before marriage
P(A) = 0.73
P(Ac ) = 1 − P(A)
= 1 − 0.73
= 0.27
Probability that the selected woman did not give birth before marriage = 0.27
ANSWER: (FALSE)
Solution:
ANSWER: TRUE
Solution:
For mutually exclusive events P(A and B) = 0
