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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Mathematics
Chapter
Probability

JEE Advanced 2007 · Paper 1 · Mathematics · Probability

Let H_1, H_2, , H_n be mutually exclusive events with P(H_i)>0,...

Let H1,H2,…,HnH_1, H_2, \ldots, H_n be mutually exclusive events with P(Hi)>0,i=1,2,…,nP\left(H_i\right)>0, i=1,2, \ldots, n. Let EE be any other event with 0<P(E)<10 < P(E) < 1. Statement I P(Hi/E)>P(E/Hi)P(Hi)P\left(H_i / E\right)>P\left(E / H_i\right) P\left(H_i\right) for i=1,2,…,ni=1,2, \ldots, n. Statement II $\sum_{i=1}^n P\left(H_i\right)=1

Options

  1. AStatement I is true, Statement II is true; Statement II is a correct explanation for Statement I
  2. BStatement I is true, Statement II is true; Statement II is not a correct explanation for Statement II
  3. CStatement I is true, Statement II is false
  4. DStatement I is false, Statement II is true

Correct answer

D. Statement I is false, Statement II is true

Step-by-step solution

Statement : I If PP\left(H_i \cap E\right)=0forsomefor somei,thenP, then P\left(\frac{H_i}{E}\right)=P(EHi)=P\left(\frac{E}{H_i}\right)=0 If P(Hi∩E)≠0P\left(H_i \cap E\right) \neq 0 for ∀i=1,2,….n\forall i=1,2, \ldots . n, then P(HiE)=P(Hi∩E)P(Hi)×P(Hi)P(E)=P(EHi)×P(Hi)P(E)>P(EHi)\cdotP(Hi) [as 0<P(E)<1]\begin{aligned}P\left(\frac{H_i}{E}\right)& =\frac{P\left(H_i \cap E\right)}{P\left(H_i\right)} \times \frac{P\left(H_i\right)}{P(E)}\\ & \left.=\frac{P\left(\frac{E}{H_i}\right) \times P\left(H_i\right)}{P(E)}>P\left(\frac{E}{H_i}\right) \cdotP\left(H_i\right) \quad \text{ [as }0< P(E) < 1\right] \end{aligned}Hence, Statement I may not always be true. Statement II Clearly, H1∪H2∪….∪Hn=SH_1 \cup H_2 \cup \ldots . \cup H_n=S (sample space) \Rightarrow \quad P(H1)P\left(H_1\right)+P\left(H_2\right)$+\ldots+P\left(H_n\right)=1 . Hence, Statement II is true.
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