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Question: Let \[{H_1},{H_2},....,{H_n}\]be mutually exclusive and exhaustive events with\[P({H_i}) > 0,\]\[i =...

Let H1,H2,....,Hn{H_1},{H_2},....,{H_n}be mutually exclusive and exhaustive events withP({H_i}) > 0,$$$$i = 1,2,.....n . Let EE be any other event with0<P(E)<10 < P(E) < 1.
Statement 1: P(HiE)>P(EHi).P(Hi)P({H_i}|E) > P(E|{H_i}).P({H_i}), for i=1,2,.....ni = 1,2,.....n
Statement 2: i=1nP(Hi)=1\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.

Explanation

Solution

If H1,H2,....,Hn{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(AB)=P(AB)P(B)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, H1,H2,....,Hn{H_1},{H_2},....,{H_n}Is mutually exclusive and exhaustive events, sum of probability is equal to one, that is:
H1H2H3...........Hn=S\Rightarrow {H_1} \cup {H_2} \cup {H_3} \cup ........... \cup {H_n} = S(Sample)
P(H1)+P(H2)+P(H3)+.........+P(Hn)=1\Rightarrow P({H_1}) + P({H_2}) + P({H_3}) + ......... + P({H_n}) = 1
If we express in the summation form we have,
i=1nP(Hi)=1\Rightarrow \sum\limits_{i = 1}^n {P({H_i}) = 1}.
Hence statement 2 is true.
Let’s take the statement 1, that is
P(HiE)>P(EHi)×P(Hi)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(AB)=P(AB)P(B)P\left( {A|B} \right) = \dfrac{{P(A \cap B)}}{{P(B)}}, applying in above we get:
P(HiE)P(E)>P(EHi)P(Hi)×P(Hi)\Rightarrow \dfrac{{P({H_i} \cap E)}}{{P(E)}} > \dfrac{{P(E \cap {H_i})}}{{P({H_i})}} \times P({H_i})
Cancelling P(Hi)P({H_i}) we get,
P(HiE)P(E)>P(EHi)\Rightarrow \dfrac{{P({H_i} \cap E)}}{{P(E)}} > P(E \cap {H_i})
Given, 0<P(E)<10 < P(E) < 1
It is obvious that 1P(E)\dfrac{1}{{P(E)}}is larger, that is greater than 1
Hence, P(HiE)P(E)>P(EHi)\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.