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Home » State true or false: (give reason for your answer) (v) and  are mutually exclusive. (vi)  ,  , are mutually exclusive and exhaustive.
State true or false: (give reason for your answer) (v) A and {B^I} are mutually exclusive. (vi) {A^I} , {B^I} , C are mutually exclusive and exhaustive.
CBSE | Class 11 | Exercise 16.2 | Maths | NCERT | Probability
State true or false: (give reason for your answer) (v) A and {B^I} are mutually exclusive. (vi) {A^I} , {B^I} , C are mutually exclusive and exhaustive.

Assume that 1, 2, 3, 4, 5 and 6 are the outcomes possible when a die is thrown.

Given that a pair of die is thrown, then the sample space will be,

S = \left\{ \begin{gathered} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \hfill \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \hfill \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \hfill \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \hfill \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \hfill \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \hfill \\ \end{gathered} \right\}

A: getting an even number on the first die.

The sample space of event A is,

A = \left\{ \begin{gathered} (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \hfill \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \hfill \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \hfill \\ \end{gathered} \right\}

B: getting an odd number on the first die.

The sample space of event B is,

B = \left\{ \begin{gathered} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \hfill \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \hfill \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \hfill \\ \end{gathered} \right\}

C: getting the sum of the numbers on the dice \leqslant 5.

The sample space of event C is,

C = \{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(3,1),(3,2),(2,3),(4,1)\}

 

(v) A and {B^I} are mutually exclusive.

As, A\; \cap \;{B^I}\; = {\text{ }}A\; \cap \;A

A\; \cap \;{B^I}\; = {\text{ }}A\;

Thus, A\; \cap \;{B^I}\; \ne \;\varphi.

Hence, A and {B^I} are not mutually exclusive.

Therefore, the given statement is false.

(vi) {A^I} , {B^I} , C are mutually exclusive and exhaustive.

We know, the sample space of

{A^I} = \left\{ \begin{gathered} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \hfill \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \hfill \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \hfill \\ \end{gathered} \right\},

B = \left\{ \begin{gathered} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \hfill \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \hfill \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \hfill \\ \end{gathered} \right\} and

C = \left\{ \begin{gathered} (1,1),(1,2),(1,3),(1,4),(2,1), \hfill \\ (2,2),(2,3),(3,1),(3,2),(4,1), \hfill \\ \end{gathered} \right\}

{A^I} \cap {C^I} = \varphi

So, there are no common elements in {A^I} and {B^I}.

Thus, {A^I} , {B^I} are mutually exclusive.

{B^I} \cap C = \left\{ {(2,1),(2,2),(2,3),(4,1)} \right\}

{B^I} \cap C \ne \varphi

So,{B^I} , C are not mutually exclusive.

Thus, {A^I} , {B^I} , C are not mutually exclusive and exhaustive.

Therefore, the given statement is false.

i

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