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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Refer to Question 74 above. The probability that exactly two of the three balls were red, with the first ball being red, is

17 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:

$\begin{array}{l} A c c o r d i n g t o Q u e s t i o n 74, \\ L e t E_{1} b e e v e n t t h a t f i r s t b a l l i s r e d . \\ E_{2} b e e v e n t t h a t e x a c t l y t w o o f t h e t h r e e b a l l s a r e r e d . \\ ∴ P (E_{1}) = P (R) . P (R) . P (B) + P (R) . P (R) . P (R) + P (R) . P (B) . P (R) + P (R) . P (B) . P (B) \\ = \frac{5}{8} . \frac{4}{7} . \frac{3}{6} + \frac{5}{8} . \frac{4}{7} . \frac{3}{6} + \frac{5}{8} . \frac{3}{7} . \frac{4}{6} + \frac{5}{8} . \frac{3}{7} . \frac{2}{6} \\ = \frac{60}{336} + \frac{60}{336} + \frac{60}{336} + \frac{30}{336} = \frac{210}{336} \\ P (E_{1} \cap E_{2}) = P (R) . P (B) . P (R) + P (R) . P (R) . P (B) \\ = \frac{5}{8} . \frac{3}{7} . \frac{4}{6} + \frac{5}{8} . \frac{4}{7} . \frac{3}{6} = \frac{60}{336} + \frac{60}{336} = \frac{120}{336} \\ ∴ P (E_{2} / E_{1}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{120 / 336}{210 / 336} = \frac{4}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two events $E$ and $F$ are independent. If $P (E) = 0.3$ , $P (E \cup F) = 0.5$ , then $P (E ∣ F) - P (F ∣ E)$ 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:

$\begin{array}{l} G i v e n t h a t : E a n d F a r e t w o i n d e p e n d e n t e v e n t s s u c h t h a t \\ P (E) = 0.3, a n d P (E \cup F) = 0.5 \\ ∴ P (E \cup F) = P (E) + P (F) - P (E \cap F) \\ 0.5 = 0.3 + P (F) - P (E) . P (F) \\ \Rightarrow 0.5 - 0.3 = P (F) [1 - P (E)] \Rightarrow 0.2 = P (F) (1 - 0.3) \\ \Rightarrow 0.2 = P (F) . (0.7) \\ ∴ P (F) = \frac{0.2}{0.7} = \frac{2}{7} \\ N o w P (E / F) - P (F / E) = \frac{P (E \cap F)}{P (F)} . \frac{P (E \cap F)}{P (E)} \\ = \frac{P (E) . P (F)}{P (F)} . \frac{P (E) . P (F)}{P (E)} = P (E) - P (F) = \frac{3}{10} - \frac{2}{7} = \frac{1}{70} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

If the events $A$ and $B$ are independent, then $P (A \cap B)$ is equal to:

$(a) P(A) +P (B)$

$(b) P(A) -P (B)$

$(c) P(A) \cdot P(B)$

$(d) P(A) ? P(B)$

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:

$\begin{array}{l} A a n d B a r e t w o i n d e p e n d e n t e v e n t s \\ ∴ P (A \cap B) = P (A) . P (B) \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Let $A$ and $B$ be two events such that $P (A) = \frac{3}{8}$ , $P (B) = \frac{5}{8}$ , and $P (A \cup B) = \frac{3}{4}$ . Then $P (A ∣ B) \cdot P (A^{'} ∣ B)$ 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:

$\begin{array}{l} G i v e n t h a t : P (A) = \frac{3}{8}, P (B) = \frac{5}{8} a n d P (A \cup B) = \frac{3}{4} \\ ∴ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \frac{3}{4} = \frac{3}{8} + \frac{5}{8} - P (A \cap B) \\ \Rightarrow P (A \cap B) = \frac{3}{8} + \frac{5}{8} - \frac{3}{4} = \frac{1}{4} \\ N o w P (A / B) . P (A' / B) = \frac{P (A \cap B)}{P (B)} . \frac{P (A' \cap B)}{P (B)} \\ = \frac{P (A \cap B)}{P (B)} . \frac{P (B) - P (A \cap B)}{P (B)} \\ = \frac{\frac{1}{4}}{5} . \frac{(\frac{5}{8} - \frac{1}{4})}{5} = \frac{2}{5} . \frac{3}{5} = \frac{6}{25} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

If two events are independent, then

(a) They must be mutually exclusive

(b) The sum of their probabilities must be equal to 1

(c) (a)and (b) both are correct

(d) None of the above is correct

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

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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 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:

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

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

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