Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

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

Ncert Solutions Maths class 12th Probability

For the following probability distribution:

$X$ : -4 -3 -2 -1 0
$P (X)$ : 0.1 0.2 0.3 0.2 0.2
$E (X)$ is equal to:

(a) 0

(b) -1

(c) -2

(d) -1.8

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:

$\begin{array}{l} W e k n o w t h a t E (X) = \sum_{i = 1}^{n} X_{i} P_{i} \\ = (- 4) (0.1) + (- 3) (0.2) + (- 2) (0.3) + (- 1) (0.2) + 0 (0.2) \\ = - 0.4 - 0.6 - 0.6 - 0.2 = - 1.8 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

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:

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

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons, 4 are swimmers is:

^{(a) 5}C₄ (0.7)⁴ (0.3)

^{(b) 5}C₁ (0.7) (0.3)⁴

^{(c) 5}C₁ (0.7) (0.3)⁴

(d) (0.7)⁴ (0.3)

2 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: $\begin{array}{l} \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is

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:

$\begin{array}{l} H e r e, n = 10, p = \frac{1}{2}, q = 1 - \frac{1}{2} = \frac{1}{2} (f o r t r u e / f a l s e q u e s t i o n s) \\ a n d r \geq 8 i . e ., 8, 9, 10 \\ ∴ P (X \geq 8) = P (x = 8) + P (x = 9) + P (x = 10) \\ = {}_{10}{C_{8}} {(\frac{1}{2})}^{8} {(\frac{1}{2})}^{2} + {}_{10}{C_{9}} {(\frac{1}{2})}^{9} (\frac{1}{2}) + {}_{10}{C_{10}} {(\frac{1}{2})}^{10} {(\frac{1}{2})}^{0} \\ = 45 . {(\frac{1}{2})}^{10} + 10 . {(\frac{1}{2})}^{10} + {(\frac{1}{2})}^{10} = {(\frac{1}{2})}^{10} (45 + 10 + 1) \\ = 56 \times \frac{1}{1024} = \frac{7}{128} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Two cards are drawn from a well-shuffled deck of 52 playing cards with replacement. The probability that both cards are queens is:

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:

$\begin{array}{l} P r o b a b i l i t y o f g e t t i n g Q u e e n = \frac{4}{52} \\ S o, t h e r e q u i r e d p r o b a b i l i t y = P (Q u e e n) . P (Q u e e n) \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{4}{52} . \frac{4}{52} = \frac{1}{13} . \frac{1}{13} (w i t h r e p l a c e m e n t) \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Which one is not a requirement of a binomial distribution?

(a) There are 2 outcomes for each trial

(b) There is a fixed number of trials

(c) The outcomes must be dependent on each other

(d) The probability of success must be the same for all the trials

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:

$\begin{array}{l} W e k n o w t h a t f o r a B i n o m i a l d i s t r i b u t i o n, t h e o u t c o m e s m u s t n o t b e d e p e n d e n t o n e a c h o t h e r . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Two dice are thrown. If it is known that the sum of the numbers on the dice was less than 6, the probability of getting a sum of 3, 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:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t s h o w i n g t h e s u m o f t h e n u m b e r s o n t h e t w o d i c e w a s l e s s t h a n 6 \\ ∴ E_{1} = {(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (2, 2), (2, 3), (3, 2)} \\ n (E_{1}) = 10 \\ . E_{2} b e t h e e v e n t t h a t t h e s u m o f t h e n u m b e r s i s 3 . \\ ∴ E_{2} = {(1, 2), (2, 1)} \\ n (E_{2}) = 2 a n d n (E_{1} \cap E_{2}) = 2 \\ ∴ P (E_{2} / E_{1}) = \frac{n (E_{1} \cap E_{2})}{n (E_{1})} = \frac{2}{10} = \frac{1}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Eight coins are tossed together. The probability of getting exactly 3 heads is

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: $\begin{array}{l} \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is

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:

$\begin{array}{l} p r o b a b i l i t y = P (d e a d) . P (d e a d) \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{3}{8} . \frac{2}{7} = \frac{3}{28} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Ncert Solutions Maths class 12th

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