In: Statistics and Probability
Solution:
a) Probability of an event A is given by,
Where, n(A) is number of elements in A and n(S) is number of elements in sample space S.
We have, n(A) = 2 and n(S) = 8
Hence, P(A) = 2/8 = 1/4
Probability of an event B is given by,
Where, n(B) is number of elements in B and n(S) is number of elements in sample space S.
We have, n(B) = 4 and n(S) = 8
Hence, P(B) = 4/8 = 1/2
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Where,
n(A ∩ B) is number of elements which are available in both of A and B.
Since, only one element is available in both of A and B, therefore, n(A ∩ B) = 1.
n(S) = 8
P(A ∩ B) = 1/8
and we have, P(A) = 1/4, P(B) = 1/2
b)
n(A ∩ B) is number of elements which are available in both of A and B.
Since, only one element is available in both of A and B, therefore, n(A ∩ B) = 1.
n(S) = 8
P(A ∩ B) = 1/8
Already we have computed P(B) = 1/2 in part (a).
Two sets are said to be mutually exclusive if no element is common for both the sets.
A = {1,2,} B = {2,3,4,6} and C = {5,7}
(A ∪ B) = {1,2,3,4,6}
(A ∪ C) = {1,2,5,7}
Since, the two elements are common in both of (A ∪ B) and (A ∪ C), therefore (A ∪ B) and (A ∪ C) are not mutually exclusive.
Howet we don't need to obtain (A ∪ B) and (A ∪ C) to tell whether these two are mutually exclusive or not. Since, always the elements of A will be common in both of (A ∪ B) and (A ∪ C), therefore these two sets are not mutually exclusive.
c) The three sets A, B and C are said to be collectively exhaustive if,
(A ∪ B ∪ C) = S
We have, A = {1,2}; B = {2,3,4,6}; C = {5,7}; S = {1,2,3,4,5,6,7,9}
Hence, (A ∪ B ∪ C) = {1,2,3,4,5,6,7}
Since, therefore A, B and C are not collectively exhaustive.
Three sets A, B and C are said to be mutually exclusive if no elements is common in all of A, B and C.
Since, there is no element which is common in all of A, B and C, therefore A, B and C are mutually exclusive.
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