Answer:
a. P(AUB) = 0.74
b. P(A∩C) = 0.24
c. P(BUC) = 0.71
d. P(Bc) = 0.55
e. P(A∩B∩C) = 0.11
Step-by-step explanation:
n(U) = Total number of elements in the set = number of elements in the universal set = 38
A: (Outcome is an odd number (00 and 0 are considered neither odd nor even)]
An odd number referred to any integer, that is not a fraction, which is not possible to be divided exactly by 2. Therefore, we have:
A: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35
B: 2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33, 35
C: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18
a. Calculate the probability of AUB
AUB picks not more than one each of the elements in both A and B without repetition. Therefore, we have:
AUB = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 15, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 33, 35
n(AUB) = number of elements in AUB = 28
P(AUB) = probability of AUB = n(AUB) / n(U) = 28 / 38 = 0.736842105263158 = 0.74
b. Calculate the probability of A ∩ C
A∩C picks only the elements that are common to both A and C. Therefore, we have:
A∩C: 1, 3, 5, 7, 9, 11, 13, 15, 17
n(A∩C) = number of elements in A∩C = 9
P(A∩C) = probability of A∩C = n(A∩C) / n(U) = 9 / 38 = 0.236842105263158 = 0.24
c. Calculate the probability of BUC
BUC picks not more than one each of the elements in both B and C without repetition. Therefore, we have:
BUC: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 22, 24, 26, 28, 29, 31, 33, 35
n(BUC) = number of elements in BUC = 27
P(BUC) = probability of A∩C = n(BUC) / n(U) = 27 / 38 = 0.710526315789474 = 0.71
d. Calculate the probability of Bc
Bc indicates B component it represents all the elements in the universal set excluding the elements in B. Therefore, we have:
Bc: 00, 0, 1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36
n(Bc) = number of elements in Bc = 20
P(Bc) = probability of Bc = n(Bc) / n(U) = 20 / 38 = 0.526315789473684 = 0.55
e. Calculate the probability of A∩B∩C
A∩B∩C picks only the elements that are common to A, B and C. Therefore, we have:
A∩B∩C: 11, 13, 15, 17
n(Bc) = number of elements in A∩B∩C = 4
P(A∩B∩C) = probability of A∩B∩C = n(A∩B∩C) / n(U) = 4 / 38 = 0.105263157894737 = 0.11
Note: All the answers are rounded to 2 decimal places.
