Ncert Solutions Maths class 12th

Get insights from 2.5k questions on Ncert Solutions Maths class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Maths class 12th

2.5k
Questions
0
Discussions
0
Active Users
65
Followers

Have a question about Ncert Solutions Maths class 12th? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If $A$ and $B$ are two independent events with $P (A) = \frac{3}{5}$ and $P (B) = \frac{4}{9}$ , then $P (A^{'} \cap B^{'})$ equals:

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : A a n d B a r e i n d e p e n d e n t e v e n t s \\ s u c h t h a t P (A) = \frac{3}{5} ∴ P (A') = 1 - \frac{3}{5} = \frac{2}{5} \\ P (B) = \frac{4}{9} ∴ P (B') = 1 - \frac{4}{9} = \frac{5}{9} \\ ∴ P (A' \cap B') = P (A') . P (B') = \frac{2}{5} . \frac{5}{9} = \frac{2}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If $A$ and $B$ are such events that $P (A) > 0$ and $P (B) \neq 1$ , then $P (A^{'} ∣ B^{'})$ equals:

$(a) 1 - (∣)$

$(b) 1 - (' ∣)$

$(c) 1 - \frac{(\cap)}{(')}$

$P(A') ∣P (B')$

18 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) > 0 a n d P (B) \neq 1 \\ ∴ P (A' / B') = \frac{P (A' \cap B')}{P (B')} = \frac{1 - P (A \cup B)}{P (B')} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Let $P (A) = \frac{7}{13}$ , $P (B) = \frac{9}{13}$ , and $P (A \cap B) = \frac{4}{13}$ . Then $P (A^{'} ∣ B)$ is equal to:

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ P (A) = \frac{7}{13}, P (B) = \frac{9}{13}, P (A \cap B) = \frac{4}{13} \\ P (A' / B) = \frac{P (A' \cap B)}{P (B)} = \frac{P (B) - P (A \cap B)}{P (B)} \\ = \frac{\frac{9}{13} - \frac{4}{13}}{\frac{9}{13}} = \frac{\frac{5}{13}}{\frac{9}{13}} = \frac{5}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If $P (B) = \frac{3}{5}$ , $P (A / B) = \frac{1}{2}$ , and $P (A \cup B) = \frac{4}{5}$ , then $P (A \cup B^{'}) + P (A^{'} \cup B)$ equals:

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ P (B) = \frac{3}{5}, P (A / B) = \frac{1}{2}, P (A \cup B) = \frac{4}{5} \\ P (A / B) = \frac{P (A \cap B)}{P (B)} \\ \Rightarrow \frac{1}{2} = \frac{P (A \cap B)}{\frac{3}{5}} \Rightarrow P (A \cap B) = \frac{3}{10} \\ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \frac{4}{5} = P (A) + \frac{3}{5} - \frac{3}{10} \\ P (A) = \frac{4}{5} - \frac{3}{5} + \frac{3}{10} = \frac{1}{5} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2} \\ N o w P (A \cup B)' + P (A' \cup B) \\ = 1 - P (A \cup B) + 1 - P (A \cap B') \\ = 2 - \frac{4}{5} - P (A) . P (B') \\ = \frac{6}{5} - \frac{1}{2} . (1 - \frac{3}{5}) = \frac{6}{5} - \frac{1}{2} \times \frac{2}{5} = \frac{6}{5} - \frac{1}{5} = \frac{5}{5} = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

In Exercise 64 above, P(B | $A^{'}$ ) is equal to

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A c c o r d i n g t o E x e r c i s e 64, w e h a v e \\ P (B) = \frac{3}{5}, P (A / B) = \frac{1}{2}, P (A \cup B) = \frac{4}{5} \\ P (B / A') = \frac{P (B \cap A')}{P (A')} = \frac{P (B) - P (A \cap B)}{1 - P (A)} = \frac{\frac{3}{5} - \frac{3}{10}}{1 - \frac{1}{2}} = \frac{\frac{3}{10}}{\frac{1}{2}} = \frac{3}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

You are given that $A$ and $B$ are two events such that $P (B) = \frac{3}{5}$ , $P (A / B) = \frac{1}{2}$ , and $P (A \cup B) = \frac{4}{5}$ , then $P (A)$ equals

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (B) = \frac{3}{5}, P (A / B) = \frac{1}{2}, P (A \cup B) = \frac{4}{5} \\ W e k n o w t h a t P (A / B) = \frac{P (A \cap B)}{P (B)} \Rightarrow \frac{1}{2} = \frac{P (A \cap B)}{3 / 5} \\ ∴ P (A \cap B) = \frac{3}{10} \\ N o w P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \frac{4}{5} = P (A) + \frac{3}{5} - \frac{3}{10} \\ \Rightarrow P (A) = \frac{4}{5} - \frac{3}{5} + \frac{3}{10} = \frac{1}{5} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A and B are events such that $P (A) = 0.4$ , $P (B) = 0.3$ , and $P (A \cup B) = 0.5$ . Then $P (B' \cap A)$ equals

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (B) = 0.4, P (B) = 0.3, P (A \cup B) = 0.5 \\ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ 0.5 = 0.4 + 0.3 - P (A \cap B) \\ \Rightarrow P (A \cap B) = 0.4 + 0.3 - 0.5 = 0.2 \\ ∴ P (B' \cap A) = P (A) - P (A \cap B) \\ = 0.4 - 0.2 = 0.2 = \frac{1}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If $A$ and $B$ are two events and $A \neq ϕ$ , $B \neq ϕ$ , then

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : A = ? a n d B = ?, \\ t h e n P (A / B) = \frac{P (A \cap B)}{P (B)} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If $P (A) = 0.4$ , $P (B) = 0.8$ , and $P (B / A) = 0.6$ , then $P (A \cup B)$ is equal to:

(a) 0.24

(b) 0.3

(c) 0.48

(d) 0.96

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If A and B are two events such that $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ , and $P (A ? B) = \frac{1}{4}$ , then $P (A^{'} \cap B^{'})$ equals:

16 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = \frac{1}{2}, P (B) = \frac{1}{3} a n d P (A / B) = \frac{1}{4} \\ P (A / B) = \frac{P (A \cap B)}{P (B)} \Rightarrow \frac{1}{4} = \frac{P (A \cap B)}{\frac{1}{3}} \\ ∴ P (A \cap B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} \\ N o w P (A' \cap B') = 1 - P (A \cup B) \\ = 1 - [P (A) + P (B) - P (A \cap B)] \\ = 1 - [\frac{1}{2} + \frac{1}{3} - \frac{1}{12}] = 1 - [\frac{5}{6} - \frac{1}{12}] \\ = 1 - \frac{9}{12} = \frac{3}{12} = \frac{1}{4} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Upvote

Ncert Solutions Maths class 12th

Questions

Discussions

Active Users

Followers