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Inverse Trigonometric Functions
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Differential Equations
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Three-dimensional Geometry
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Linear Programming
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Algebra
Calculus
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Financial Mathematics
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Linear Programming
Text
Let E and F be events of a sample space S of an experiment, then we have
Property : P(S|F) = P(F|F) = 1
We know that
P(S|F) = `(P(S ∩ F))/(P(F)) = (P(F))/(P(F)) = 1`
Also P(F|F) = `(P(F ∩ F))/(P(F)) = (P(F))/(P(F)) = 1`
Thus P(S|F) = P(F|F) = 1
Property : If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0, then
P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)
In particular, if A and B are disjoint events, then
P((A∪B)|F) = P(A|F) + P(B|F)
We have
P((A∪B)|F) = `(P[(A ∪ B) ∩ F]) /(P(F))`
= `(P[(A ∩ F )∪ (B ∩ F)])/ (P(F))`
(by distributive law of union of sets over intersection)
`= (P(A ∩ F) + P (B ∩ F) - P(A ∩ B ∩ F))/(P(F))`
`= (P(A ∩ F))/(P(F)) + (P (B ∩ F)) / (P(F)) - (P[(A ∩ B ∩ F)]) /(P(F))`
= P(A|F) + P(B|F) – P((A∩B)|F)
When A and B are disjoint events, then
P((A ∩ B)|F) = 0
⇒ P((A ∪ B)|F) = P(A|F) + P(B|F)
Property : P(E′|F) = 1 − P(E|F)
From first Property , we know that P(S|F) = 1
⇒ P(E ∪ E′|F) = 1 since S = E ∪ E′
⇒ P(E|F) + P (E′|F) = 1 since E and E′ are disjoint events
Thus, P(E′|F) = 1 − P(E|F)
Related QuestionsVIEW ALL [27]
Read the following passage and answer the questions given below.
 There are two antiaircraft guns, named as A and B. The probabilities that the shell fired from them hits an airplane are 0.3 and 0.2 respectively. Both of them fired one shell at an airplane at the same time. |
- What is the probability that the shell fired from exactly one of them hit the plane?
- If it is known that the shell fired from exactly one of them hit the plane, then what is the probability that it was fired from B?
