Let us consider a sample space S in connection with a random experiment and let A be an event defined on the sample space S. That is, A ≤ S.
Then a real valued function P defined on S is known as a probability measure and P(A) is defined as the probability of A if P satisfies the following axioms :
(i) P(A) ≥ 0 for every A ⊆ S (subset)
(ii) P(S) = 1
(iii) For any sequence of mutually exclusive events A1, A2, A3,.................
P(A1UA2UA3.............) = P(A1) + P(A2) + P(A3) + .............
Problem 1 :
A number is selected from the first 25 natural numbers. What is the probability that it would be divisible by 4 or 7 ?
Solution :
Let A be the event that the number selected would be divisible by 4 and B, the event that the selected number would be divisible by 7.
Then AuB denotes the event that the number would be divisible by 4 or 7.
Next we note that
A = {4, 8, 12, 16, 20, 24} and B = {7, 14, 21}
whereas S = {1, 2, 3, ……... 25}.
Here, AnB = Null set.
The two events A and B are mutually exclusive and as such we have
P(AuB) = P(A) + P(B) -----(1)
Since
P(A) = n(A) / n(S) = 6/25
and
P(B) = n(B) / n(S) = 3/25
Thus from (1), we have
P(AuB) = 6/25 + 3/25
P(AuB) = 9/25
So, the probability that the selected number would be divisible by 4 or 7 is 9/25 or 0.36
Problem 2 :
A coin is tossed thrice. What is the probability of getting 2 or more heads ?
Solution :
If a coin is tossed three times, then we have the following sample space.
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
2 or more heads imply 2 or 3 heads.
Next we note that
If A and B denote the events of occurrence of 2 and 3 heads respectively, then we find that
A = { HHT, HTH, THH } and B = { HHH }
P(A) = n(A)/n(S) = 3/8
and
P(B) = n(B)/n(S) = 1/8
As A and B are mutually exclusive, the probability of getting 2 or more heads is
P(AuB) = P(A) + P(B)
P(AuB) = 3/8 + 1/8
P(AuB) = 0.50
So, the probability of getting 2 or more heads is 0.50.
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