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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

If $P (A) = \frac{2}{5}$ , $P (B) = \frac{3}{10}$ , and $P (A \cap B) = \frac{1}{5}$ , then $P (A ∣ B) \cdot P (B^{'} ∣ A^{'})$ is equal to:

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = \frac{2}{5}, P (B) = \frac{3}{10} a n d P (A \cap B) = \frac{1}{5} \\ P (A') = 1 - \frac{2}{5} = \frac{3}{5}, P (B') = 1 - \frac{3}{10} = \frac{7}{10} \\ a n d P (A' \cap B') = 1 - P (A \cup B) = 1 - [P (A) + P (B) - P (A \cap B)] \\ = 1 - [\frac{2}{5} + \frac{3}{10} - \frac{1}{5}] = 1 - [\frac{1}{5} + \frac{3}{10}] = 1 - \frac{5}{10} = \frac{1}{2} \\ ∴ P (A' / B') = \frac{P (A' \cap B')}{P (B')} = \frac{\frac{1}{2}}{\frac{7}{10}} = \frac{5}{7} \\ a n d P (B' / A') = \frac{P (A' \cap B')}{P (A')} = \frac{\frac{1}{2}}{\frac{3}{5}} = \frac{5}{6} \\ ∴ P (A' / B') . P (B' / A') = \frac{5}{7} \times \frac{5}{6} = \frac{25}{42} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

If $P (A) = \frac{3}{10}$ , $P (B) = \frac{2}{5}$ , and $P (A \cup B) = \frac{3}{5}$ , then $P (B ? A) + P (A ? B)$ equals:

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (B) = \frac{2}{5}, P (A) = \frac{3}{10} a n d P (A \cup B) = \frac{3}{5} \\ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \frac{3}{5} = \frac{3}{10} + \frac{2}{5} - P (A \cap B) \\ P (A \cap B) = \frac{3}{10} + \frac{2}{5} - \frac{3}{5} = \frac{3 + 4 - 6}{10} = \frac{1}{10} \\ N o w P (A / B) + P (B / A) = \frac{P (A \cap B)}{P (B)} + \frac{P (A \cap B)}{P (A)} = \frac{\frac{1}{10}}{\frac{2}{5}} + \frac{\frac{1}{10}}{\frac{3}{10}} = \frac{1}{4} + \frac{1}{3} = \frac{7}{12} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

If $P (A \cap B) = \frac{7}{10}$ and $P (B) = \frac{17}{20}$ , then $P (A ? B)$ equals:

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

Sol:

$\begin{array}{l} G i v e n t h a t : P (B) = \frac{17}{20}, a n d P (A \cap B) = \frac{7}{10} \\ ∴ P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{\frac{7}{10}}{\frac{17}{20}} = \frac{14}{17} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Choose the correct answer from the given four options in each of the Exercises 1 to 38:

Q1. If $P (A) = \frac{4}{5}$ , and $P (A \cap B) = \frac{7}{10}$ , then $P (B ? A)$ is equal to:

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = \frac{4}{5}, a n d P (A \cap B) = \frac{7}{10} \\ ∴ P (B / A) = \frac{P (A \cap B)}{P (A)} = \frac{\frac{7}{10}}{\frac{4}{5}} = \frac{7}{8} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on $k$ .

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t A b e t h e e v e n t h a v i n g m w h i t e a n d n b l a c k b a l l s . \\ E_{1} = {f i r s t b a l l d r a w n o f w h i t e c o l o u r} \\ E_{2} = {f i r s t b a l l d r a w n o f b l a c k c o l o u r} \\ E_{3} = {second b a l l d r a w n o f w h i t e c o l o u r} \\ P (E_{1}) = \frac{m}{m + n}, P (E_{2}) = \frac{n}{m + n}, P (E_{3} / E_{1}) = \frac{m + k}{m + n + k}, P (E_{3} / E_{2}) = \frac{m}{m + n + k} \\ N o w, P (E_{3}) = P (E_{1}) . P (E_{3} / E_{1}) + P (E_{2}) . P (E_{3} / E_{2}) \\ = \frac{m}{m + n} \times \frac{m + k}{m + n + k} + \frac{n}{m + n} \times \frac{m}{m + n + k} \\ = \frac{m}{m + n + k} [\frac{m + k}{m + n} + \frac{n}{m + n}] \\ = \frac{m}{m + n + k} [\frac{m + n + k}{m + n}] = \frac{m}{m + n} \\ H e n c e, t h e p r o b a b i l i t y o f d r a w n i n g a w h i t e b a l l d o e s n o t d e p e n d u p o n k . \end{array}$

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