a. The joint probability table is:
B Bc Total
A 0.2 0.12 0.32
Ac 0.5 0.18 0.68
Total 0.7 0.3 1
b. The probability that A occurs is 0.32.
c. The probability that A and B occur is 0.2.
d. Given that B has occurred, the probability that A occurs is 0.2/0.3 = 0.6667.
e. Given that Ac has occurred, the probability that B occurs is 0.12/0.68 ≈ 0.1765.
f. No because P(A∩B) ≠ 0.
g. No because P(A|B) ≠ P(A).
Consider the following contingency table.
B Bc
A 20 12
Ac 50 18
a. Convert the contingency table into a joint probability table. (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
b. What is the probability that A occurs? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
c. What is the probability that A and B occur? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
d. Given that B has occurred, what is the probability that A occurs? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
e. Given that Ac has occurred, what is the probability that B occurs? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
f. Are A and B mutually exclusive events?
multiple choice 1
Yes because P(A | B) ≠ P(A).
Yes because P(A ∩ B) ≠ 0.
No because P(A | B) ≠ P(A).
No because P(A ∩ B) ≠ 0.
g. Are A and B independent events?
multiple choice 2
Yes because P(A | B) ≠ P(A).
Yes because P(A ∩ B) ≠ 0.
No because P(A | B) ≠ P(A).
No because P(A ∩ B) ≠ 0.
1 answer
1 answer