Let \[{H_1},{H_2},....,{H_n}\]be mutually exclusive and exhaustive events with\[P({H_i}) > 0,\]\[i = 1,2,.....n\] . Let \[E\] be any other event with\[0 < P(E) < 1\].
Statement 1: \[P({H_i}|E) > P(E|{H_i}).P({H_i})\], for \[i = 1,2,.....n\]
Statement 2: \[\sum\limits_{i = 1}^n {P({H_i}) = 1} \]
A. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
B. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
C. Statement 1 is true, statement 2 is false.
D. Statement 1 is false, statement 2 is true.
Answer
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Hint: If \[{H_1},{H_2},....,{H_n}\]be mutually exclusive and exhaustive events, then the sum of the probability is equal to one. Using this we will have statement I. we know the formula for conditional probability is \[P\left( {\dfrac{A}{B}} \right) = \dfrac{{P(A \cap B)}}{{P(B)}}\]. Using these we will say whether statement 1 and statement 2 is correct or not.
Complete step by step answer:
Given, \[{H_1},{H_2},....,{H_n}\]Is mutually exclusive and exhaustive events, sum of probability is equal to one, that is:
\[ \Rightarrow {H_1} \cup {H_2} \cup {H_3} \cup ........... \cup {H_n} = S\](Sample)
\[ \Rightarrow P({H_1}) + P({H_2}) + P({H_3}) + ......... + P({H_n}) = 1\]
If we express in the summation form we have,
\[ \Rightarrow \sum\limits_{i = 1}^n {P({H_i}) = 1} \].
Hence statement 2 is true.
Let’s take the statement 1, that is
\[P\left( {{H_i}|E} \right) > P\left( {E|{H_i}} \right) \times P({H_i})\]
By the definition of conditional probability, we have \[P\left( {A|B} \right) = \dfrac{{P(A \cap B)}}{{P(B)}}\], applying in above we get:
\[ \Rightarrow \dfrac{{P({H_i} \cap E)}}{{P(E)}} > \dfrac{{P(E \cap {H_i})}}{{P({H_i})}} \times P({H_i})\]
Cancelling \[P({H_i})\] we get,
\[ \Rightarrow \dfrac{{P({H_i} \cap E)}}{{P(E)}} > P(E \cap {H_i})\]
Given, \[0 < P(E) < 1\]
It is obvious that \[\dfrac{1}{{P(E)}}\]is larger, that is greater than 1
Hence, \[\dfrac{{P({H_i} \cap E)}}{{P(E)}} > P(E \cap {H_i})\] is true.
Hence statement 1 is correct.
Since statement 1 and statement 2 are not related. No statement is an explanation of the other.
Hence, the most convenient answer is option (B).
Note: Conditional probability refers to the chances that some outcome occurs given that another event has also occurred. Two events A and B are said to be mutually exclusive if the occurrence of A prohibits the occurrence of B (Vice versa). Two events A and B are said to be exhaustive if at least one of them will definitely occur.
Complete step by step answer:
Given, \[{H_1},{H_2},....,{H_n}\]Is mutually exclusive and exhaustive events, sum of probability is equal to one, that is:
\[ \Rightarrow {H_1} \cup {H_2} \cup {H_3} \cup ........... \cup {H_n} = S\](Sample)
\[ \Rightarrow P({H_1}) + P({H_2}) + P({H_3}) + ......... + P({H_n}) = 1\]
If we express in the summation form we have,
\[ \Rightarrow \sum\limits_{i = 1}^n {P({H_i}) = 1} \].
Hence statement 2 is true.
Let’s take the statement 1, that is
\[P\left( {{H_i}|E} \right) > P\left( {E|{H_i}} \right) \times P({H_i})\]
By the definition of conditional probability, we have \[P\left( {A|B} \right) = \dfrac{{P(A \cap B)}}{{P(B)}}\], applying in above we get:
\[ \Rightarrow \dfrac{{P({H_i} \cap E)}}{{P(E)}} > \dfrac{{P(E \cap {H_i})}}{{P({H_i})}} \times P({H_i})\]
Cancelling \[P({H_i})\] we get,
\[ \Rightarrow \dfrac{{P({H_i} \cap E)}}{{P(E)}} > P(E \cap {H_i})\]
Given, \[0 < P(E) < 1\]
It is obvious that \[\dfrac{1}{{P(E)}}\]is larger, that is greater than 1
Hence, \[\dfrac{{P({H_i} \cap E)}}{{P(E)}} > P(E \cap {H_i})\] is true.
Hence statement 1 is correct.
Since statement 1 and statement 2 are not related. No statement is an explanation of the other.
Hence, the most convenient answer is option (B).
Note: Conditional probability refers to the chances that some outcome occurs given that another event has also occurred. Two events A and B are said to be mutually exclusive if the occurrence of A prohibits the occurrence of B (Vice versa). Two events A and B are said to be exhaustive if at least one of them will definitely occur.
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