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Class 12th Probability

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, the probability of getting exactly one red ball is

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : B a g c o n t a i n s 5 r e d a n d 3 b l u e b a l l s . \\ o f g e t t i n g e x a c t l y o n e r e d b a l l i f 3 b a l l s a r e r a n d o m l y d r a w n w i t h o u t r e p l a c e m e n t . \\ P (R) . P (B) . P (B) + P (B) . P (R) . P (B) + P (B) . P (B) . P (R) \\ = \frac{5}{8} . \frac{3}{7} . \frac{2}{6} + \frac{3}{8} . \frac{5}{7} . \frac{2}{6} + \frac{3}{8} . \frac{2}{7} . \frac{5}{6} = \frac{30}{336} + \frac{30}{336} + \frac{30}{336} = \frac{90}{336} = \frac{15}{56} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Class 12th Probability

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:

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alok kumar singh

Contributor-Level 10

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

Class 12th Probability

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

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

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

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

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

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alok kumar singh

Contributor-Level 10

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

Class 12th Probability

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:

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alok kumar singh

Contributor-Level 10

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

Class 12th Probability

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

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

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New answer posted

a year ago

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:

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Class 12th Probability

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

$(a) 1 - (∣)$

$(b) 1 - (' ∣)$

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

$P(A') ∣P (B')$

0 Follower 17 Views

alok kumar singh

Contributor-Level 10

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

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:

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

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

Class 12th NEET

How does the Shiksha's NCERT Exemplar help in mastering Chapter 12 – Atoms?

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Pallavi Pathak

Contributor-Level 10

The NCERT Exemplar given on Shiksha's website provides a variety of numerical, conceptual and application-based questions around energy transitions, atomic models, and spectral series. It is a powerful study material for CBSE Board examination preparation and competitive exams like NEET and JEE.

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New answer posted

a year ago

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:

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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