Probability

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

If $A$ and $B$ are mutually exclusive events, then they will be independent also.

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 True or False Type Questions as classified in NCERT Exemplar

Sol: $F a l s e$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

If $A$ and $B$ are independent events, then $A^{'}$ and $B^{'}$ are also independent.

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 True or False Type Questions as classified in NCERT Exemplar

Sol: $T r u e$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Let $P (A) > 0$ and $P (B) > 0$ . Then A and B can be both mutually exclusive and independent.

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 True or False Type Questions as classified in NCERT Exemplar

Sol: $F a l s e$

Upvote

New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement, at most one is defective?

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

A and B are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4}$ , respectively. If the probability of their making a common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer being correct is:

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:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t t h a t b o t h o f t h e m s o l v e t h e p r o b l e m \\ ∴ P (E_{1}) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \\ a n d E_{2} b e t h e e v e n t t h a t b o t h o f t h e m s a m e i n c o r r e c t l y t h e p r o b l e m . \\ P (E_{2}) = (1 - \frac{1}{3}) \times (1 - \frac{1}{4}) = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2} \\ L e t H b e t h e e v e n t t h a t b o t h o f t h e m g e t t h e s a m e a n s w e r . \\ H e r e, P (H / E_{1}) = 1, P (H / E_{2}) = \frac{1}{20} \\ ∴ P (E_{1} / H) = \frac{P (E_{1}) . P (H / E_{1})}{P (E_{1}) . P (H / E_{1}) + P (E_{2}) . P (H / E_{2})} \\ = \frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1 + \frac{1}{2} \times \frac{1}{20}} = \frac{\frac{1}{12}}{\frac{1}{12} + \frac{1}{40}} = \frac{\frac{1}{12}}{\frac{10 + 3}{120}} = \frac{1 / 12}{13 / 120} = \frac{10}{13} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

In a college, 30% of students fail in physics, 25% fail in mathematics, and 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is:

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:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t t h a t t h e s t u d e n t s f a i l s i n P h y s i c s a n d E_{2} b e t h e e v e n t t h a t t h e s t u d e n t s \\ f a i l s i n M a t h e m a t i c s . \\ ∴ P (E_{1}) = \frac{30}{100}, P (E_{2}) = \frac{25}{100} a n d P (E_{1} \cap E_{2}) = \frac{10}{100} \\ P (E_{1} / E_{2}) = \frac{P (E_{1} \cap E_{2})}{P (E_{2})} = \frac{10 / 100}{25 / 100} = \frac{2}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Suppose a random variable $X$ follows the binomial distribution with parameters $n$ and $p$ , where $0 < p < 1$ . If $\frac{P (x = r)}{P (x = n - r)}$ is independent of $n$ and $r$ , then $p$ 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

Upvote

New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

For the following probability distribution:

$X$ : 1 2 3 4
$P (X)$ : $\frac{1}{10} \frac{1}{5} \frac{3}{10} \frac{2}{5}$
$E (X^{2})$ is equal to:

(a) 3

(b) 5

(c) 7

(d) 10

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:

$\begin{array}{l} W e k n o w t h a t \\ E (X^{2}) = \sum_{i = 1}^{n} P_{i} X_{i}^{2} \\ = 1 \times \frac{1}{10} + 4 \times \frac{1}{5} + 9 \times \frac{3}{10} + 16 \times \frac{2}{5} \\ = \frac{1}{10} + \frac{4}{5} + \frac{27}{10} + \frac{32}{5} = \frac{28}{10} + \frac{36}{5} = \frac{100}{10} = 10 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

The probability distribution of a discrete random variable $X$ is given below:

$X$ : 2 3 4 5
$P (X)$ : $5 / k, 7 / k, 9 / k, 11 / k$
The value of $k$ is:

(a) 8

(b) 16

(c) 32

(d) 48

14 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:

$\begin{array}{l} W e k n o w t h a t \sum_{i = 1}^{n} P (X_{i}) = 1 \\ ∴ \frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1 \\ \frac{32}{k} = 1 \Rightarrow k = 32 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$