Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

If $A = {1, 2, 3, 4}$ , define relations on $A$ which have properties of being:

(a) reflexive, transitive but not symmetric

(b) symmetric but neither reflexive nor transitive

(c) reflexive, symmetric and transitive.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$G i v e n t h a t :$
$\begin{array}{l} A = {1, 2, 3} \\ a n d R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} \\ H e r e, 1 R 1, 2 R 2 a n d 3 R 3, S o, R i s r e f l e x i v e . \\ 1 R 2 b u t 2 R 1 o r 2 R 3 b u t 3 R 2, \\ S o, R i s n o t s y m m e t r i c . \\ 1 R 1 a n d 1 R 2 \Rightarrow 1 R 3, S o, R i s t r a n s i t i v e . \\ H e n c e, t h e c o r r e c t a n s w e r i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Let $A$ and $B$ be two events. If $P (A ? B) = P (A)$ , then $A$ is _____ of $B$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} ? P (A / B) = \frac{P (A \cap B)}{P (B)} \\ \Rightarrow P (A) = \frac{P (A \cap B)}{P (B)} \\ \Rightarrow P (A \cap B) = P (A) . P (B) \\ S o, A i s i n d e p e n d e n t o f B . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Let $X$ be a random variable taking values $x_{1}, x_{2}, \dots, x_{n}$ with probabilities $p_{1}, p_{2}, \dots, p_{n}$ , respectively. Then $(X) =$ _____

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} V a r (X) = E (X^{2}) - {[E (X)]}^{2} \\ = {\sum_{}^{} X^{2} P (X) - [\sum_{}^{} X . P (X)]}^{2} = \sum_{}^{} p_{i} x_{i}^{2} - {(\sum_{}^{} p_{i} x_{i})}^{2} \\ H e n c e, V a r (X) i s e q u a l t o \sum_{}^{} p_{i} x_{i}^{2} - {(\sum_{}^{} p_{i} x_{i})}^{2} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If $X$ follows binomial distribution with parameters $n = 5$ , $p$ , and $P (X = 2) = 9$ , $P (X = 3)$ , then $p =$ _____

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If $A$ and $B$ are such that:

$P (A^{'} \cup B^{'}) = \frac{2}{3}$ and $P (A \cup B) = \frac{5}{9}$ , then $P (A^{'}) + P (B^{'}) =$ _____

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, P (A' \cup B') = \frac{2}{3} a n d P (A \cup B) = \frac{5}{9} \\ ∴ 1 - P (A \cap B) = \frac{2}{3} \\ \Rightarrow P (A \cap B) = 1 - \frac{2}{3} = \frac{1}{3} \\ N o w P (A') + P (B') = 1 - P (A) + 1 - P (B) = 2 - [P (A) + P (B)] \\ = 2 - [P (A \cup B) + P (A \cap B)] \\ = 2 - [\frac{5}{9} + \frac{1}{3}] = 2 - \frac{8}{9} = \frac{10}{9} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{10}{9} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If $A$ and $B$ are two events such that:

$P (A ? B) = p$ , $P (A) = p$ , $P (B) = \frac{1}{3}$ , and $P (A \cup B) = \frac{5}{9}$ , then $p =$ _____

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = p, P (B) = \frac{1}{3} a n d P (A \cup B) = \frac{5}{9} \\ P (A / B) = \frac{P (A \cap B)}{P (B)} = p \Rightarrow P (A \cap B) = p . P (B) = p . \frac{1}{3} \\ a n d P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \frac{5}{9} = p + \frac{1}{3} - \frac{p}{3} \Rightarrow \frac{5}{9} - \frac{1}{3} = \frac{2 p}{3} \Rightarrow \frac{2}{9} = \frac{2 p}{3} \Rightarrow p = \frac{1}{3} \\ H e n c e, p i s e q u a l t o \frac{1}{3} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If $A$ , $B$ , and $C$ are three independent events such that $P (A) = P (B) = P (C) = p$ , then:

$P (Aleasttwoof A, B, Coccur) = 3 p^{2} - 2 p^{3}$ .

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

$\begin{array}{l} T r u e \\ Since P (a t l e a s t t w o o f A, B a n d C o c c u r) \\ = p \times p \times (1 - p) + (1 - p) . p . p + p (1 - p) . p + p . p . p \\ = 3 p^{2} (1 - p) + p^{3} = 3 p^{2} - 3 p^{3} + p^{3} = 3 p^{2} - 2 p^{3} \end{array}$