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.
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).
a. To convert the contingency table into a joint probability table, we divide each cell by the total number of observations.
The total number of observations is the sum of all the cells in the table: 20 + 12 + 50 + 18 = 100.
The joint probability table is as follows:
B Bc
A 0.20 0.12
Ac 0.50 0.18
b. The probability that event A occurs is the sum of the probabilities in the A row of the joint probability table:
P(A) = 0.20 + 0.12 = 0.32
c. The probability that events A and B both occur is the value in the cell of the joint probability table where A and B intersect:
P(A ∩ B) = 0.20
d. Given that B has occurred, the probability that A occurs is calculated by dividing the joint probability of A and B by the probability of B:
P(A|B) = P(A ∩ B) / P(B) = 0.20 / (0.20 + 0.50) = 0.20 / 0.70 = 0.2857
e. Given that Ac has occurred, the probability that B occurs is calculated by dividing the joint probability of A' and B by the probability of A':
P(B|Ac) = P(A' ∩ B) / P(A') = 0.12 / (0.12 + 0.18) = 0.12 / 0.30 = 0.4000
f. Are A and B mutually exclusive events?
No because P(A ∩ B) ≠ 0.
g. Are A and B independent events?
No because P(A | B) ≠ P(A).
a. To convert the contingency table into a joint probability table, we need to divide each cell count by the total number of observations. In this case, the total number of observations is the sum of all the cell counts in the contingency table, which is 20 + 12 + 50 + 18 = 100.
The joint probability table will have the same structure as the contingency table, but each cell entry will represent the probability of the corresponding event occurring.
B Bc
A P(A∩B) / 100 P(A∩Bc) / 100
Ac P(Ac∩B) / 100 P(Ac∩Bc) / 100
Calculating the entries for the joint probability table:
P(A∩B) = 20 / 100 = 0.2
P(A∩Bc) = 12 / 100 = 0.12
P(Ac∩B) = 50 / 100 = 0.5
P(Ac∩Bc) = 18 / 100 = 0.18
The joint probability table becomes:
B Bc
A 0.2 0.12
Ac 0.5 0.18
b. The probability that A occurs is the sum of the probabilities in the "A" row of the joint probability table:
P(A) = P(A∩B) + P(A∩Bc) = 0.2 + 0.12 = 0.32
c. The probability that A and B occur is the entry in the cell corresponding to A and B in the joint probability table:
P(A∩B) = 0.2
d. Given that B has occurred, we want to find the probability that A occurs. This is calculated using conditional probability:
P(A | B) = P(A∩B) / P(B)
To find P(B), we sum the probabilities of the B column in the joint probability table:
P(B) = P(A∩B) + P(Ac∩B) = 0.2 + 0.5 = 0.7
Now we can calculate:
P(A | B) = 0.2 / 0.7 ≈ 0.2857
e. Given that Ac has occurred, we want to find the probability that B occurs. Again, we use conditional probability:
P(B | Ac) = P(Ac∩B) / P(Ac)
To find P(Ac), we sum the probabilities of the Ac row in the joint probability table:
P(Ac) = P(Ac∩B) + P(Ac∩Bc) = 0.5 + 0.18 = 0.68
Now we can calculate:
P(B | Ac) = 0.5 / 0.68 ≈ 0.7353
f. Are A and B mutually exclusive events?
The definition of mutually exclusive events is that P(A∩B) = 0. If P(A∩B) ≠ 0, then A and B are not mutually exclusive.
The correct answer is: No because P(A ∩ B) ≠ 0.
g. Are A and B independent events?
The definition of independent events is that P(A | B) = P(A) and P(B | A) = P(B). If P(A | B) ≠ P(A) or P(B | A) ≠ P(B), then A and B are not independent.
The correct answer is: No because P(A | B) ≠ P(A).