Consider the following statements: 1. If A and B are mutually exclusive events, then it is possible that P(A)=P(B)=0.6 . 2. If A and B are any two events such that P(A B)=1 , then P( B A )=1 . Which of the above

Correct answer: B. 2 only (i) Since A \& B are mutually exclusive. aligned & P(A B)=P(A)+P(B)-P(A B) & =.6+.6-0 & P(A B)=1.2 aligned statement 1 is wrong. aligned & (ii) P((A)/(B))=(P(A B))/(P(B)) & 1=(P(A B))/(P(B)) & P(B)=P(A B) & B A aligned now, aligned & P( B A )= P( B A ) P( A ) & =(1-P(A B))/(1-P(A)) & ( A B=A) & =(1-P(A))/(1-P(A))=1 aligned statement 2 is correct.

NDA 2019

Exam: NDA
Year: 2019
Subject: Mathematics
Chapter: Probability

NDA 2019 · Mathematics · Probability

Consider the following statements: 1. If A and B are mutually...

Consider the following statements: 1. If A and B are mutually exclusive events, then it is possible that

\mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})=0.6

. 2. If

A

and

B

are any two events such that

P(A \mid B)=1

, then

\mathrm{P}(\overline{\mathrm{B}} \mid \overline{\mathrm{A}})=1

. Which of the above statements is/are correct?

Options

A1 only
B2 only
CBoth 1 and 2
DNeither 1 nor 2

Correct answer

B. 2 only

Step-by-step solution

(i) Since A \& B

are mutually exclusive.

\begin{aligned} & \therefore P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & =.6+.6-0 \\ & \Rightarrow P(A \cup B)=1.2 \end{aligned}

\therefore

statement 1 is wrong.

\begin{aligned} & \text{ (ii) } \because P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)} \\ & \Rightarrow 1=\frac{P(A \cap B)}{P(B)} \\ & \Rightarrow P(B)=P(A \cap B) \\ & \Rightarrow B \subseteq A \end{aligned}

now,

\begin{aligned} & P\left(\frac{\bar{B}}{\bar{A}}\right)=\frac{P(\bar{B} \cap \bar{A})}{P(\bar{A})} \\ & =\frac{1-P(A \cup B)}{1-P(A)} \\ & (\because A \cup B=A) \\ & =\frac{1-P(A)}{1-P(A)}=1 \end{aligned}

\therefore$ statement 2 is correct.

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