Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

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

X

0

1

2

3

P(X)

k

i. Determine the value of k.

ii. Determine P $(x \leq 2)$ and P $(x > 2)$ .

iii. Find $P(x \leq 2) +P (x > 2)$ .

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) W e k n o w t h a t P (0) + P (1) + P (2) + P (3) = 1 \\ \Rightarrow k + \frac{k}{2} + \frac{k}{4} + \frac{k}{8} = 1 \\ \Rightarrow \frac{8 k + 4 k + 2 k + k}{8} = 1 \Rightarrow 15 k = 8 \\ ∴ k = \frac{8}{15} \\ H e r e, n = 8, p = \frac{1}{10}, q = 1 - \frac{1}{10} = \frac{9}{10} \\ (i i) P (X \leq 2) = P (X = 0) + P (X = 1) + P (X = 2) \\ = k + \frac{k}{2} + \frac{k}{4} = \frac{7 k}{4} = \frac{7}{4} \times \frac{8}{15} = \frac{14}{15} \\ a n d P (X > 2) = P (X = 3) = \frac{k}{8} = \frac{1}{8} \times \frac{8}{15} = \frac{1}{15} \\ (i i i) P (X \leq 2) + P (X > 2) = \frac{14}{15} + \frac{1}{15} = \frac{15}{15} = 1 . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Consider the probability distribution of a random variable X:

X

0

1

2

3

4

P(X)

0.1

0.25

0.3

0.2

0.15

Calculate:(i) $V (\frac{X}{2})$ (ii) Variance of X.

11 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:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch?

13 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:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, n = 7, p = 0.25 = \frac{25}{100} = \frac{1}{4}, q = 1 - \frac{1}{4} = \frac{3}{4} \\ ∴ P (X \geq 2) = 1 - [P (x = 0) + P (x = 1)] \\ = 1 - [{}_{7}{C_{0}} {(\frac{1}{4})}^{0} {(\frac{3}{4})}^{7} + {}_{7}{C_{1}} {(\frac{1}{4})}^{1} {(\frac{3}{4})}^{6}] \\ = 1 - [{(\frac{3}{4})}^{7} + \frac{7}{4} {(\frac{3}{4})}^{6}] = 1 - {(\frac{3}{4})}^{6} [\frac{3}{4} + \frac{7}{4}] \\ = 1 - {(\frac{3}{4})}^{6} (\frac{10}{4}) = 1 - \frac{729}{4096} \times \frac{10}{4} = 1 - \frac{7290}{16384} \\ = \frac{16384 - 7290}{16384} = \frac{9094}{16384} = \frac{4547}{8192} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{4547}{8192} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Ten coins are tossed. What is the probability of getting at least 8 heads?

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.

12 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:

$\begin{array}{l} H e r e, p = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2} \\ \Rightarrow q = 1 - \frac{1}{2} = \frac{1}{2} a n d n = 5 \\ ∴ P (x = r) = {}_{n}{C_{r}} p^{r} q^{n - r} \\ = {}_{5}{C_{3}} {(\frac{1}{2})}^{3} {(\frac{1}{2})}^{5 - 3} = \frac{5!}{3! 2!} . {(\frac{1}{2})}^{3} . {(\frac{1}{2})}^{2} = 10 . \frac{1}{8} . \frac{1}{4} = \frac{5}{16} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{5}{16} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are kings?

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1}, E_{2}, E_{3} a n d E_{4} b e t h e e v e n t s t h a t I, I I, I I I a n d I V c a r d i s K i n g r e s p e c t i v e l y . \\ ∴ P (E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) \\ = P (E_{1}) . P (E_{2} / E_{1}) . P [\frac{E_{3}}{E_{1} \cap E_{2}}] . P [\frac{E_{4}}{(E_{1} \cap E_{2} \cap E_{3} \cap E_{4})}] \\ = \frac{4}{52} \times \frac{3}{51} + \frac{2}{50} \times \frac{1}{49} = \frac{24}{52.51.50.49} = \frac{1}{13.17.25.49} = \frac{1}{27075} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{1}{27075} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of the second ball being blue?

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:

$\begin{array}{l} G i v e n t h a t t h e b o x h a s 5 b l u e a n d 4 r e d b a l l s . \\ L e t E_{1} = T h e e v e n t t h a t I b a l l d r a w n i s b l u e \\ E_{2} = T h e e v e n t t h a t I b a l l d r a w n i s r e d \\ a n d E = T h e e v e n t t h a t a I I b a l l d r a w n i s b l u e \\ ∴ P (E_{1}) = \frac{5}{9}, P (E_{2}) = \frac{4}{9}, P (E / E_{1}) = \frac{4}{8} a n d P (E / E_{2}) = \frac{5}{8} \\ ∴ P (E) = P (E_{1}) . P (E / E_{1}) + P (E_{2}) . P (E / E_{2}) \\ = \frac{5}{9} \times \frac{4}{8} + \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} + \frac{20}{72} = \frac{40}{72} = \frac{5}{9} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{5}{9} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Bag I contains 3 black and 2 white balls; Bag II contains 2 black and 4 white balls. A bag and a ball are selected at random. Determine the probability of selecting a black ball.

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t b a g I = {3 B, 2 W} \\ a n d b a g I I = {2 B, 4 W} \\ L e t E_{1} = T h e e v e n t t h a t b a g I i s s e l e c t e d \\ E_{2} = T h e e v e n t t h a t b a g I I i s s e l e c t e d \\ a n d E = T h e e v e n t t h a t a b l a c k b a l l i s s e l e c t e d \\ ∴ P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{2}, P (E / E_{1}) = \frac{3}{5} a n d P (E / E_{2}) = \frac{1}{3} \\ ∴ P (E) = P (E_{1}) . P (E / E_{1}) + P (E_{2}) . P (E / E_{2}) \\ = \frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{1}{3} = \frac{3}{10} + \frac{1}{6} = \frac{9 + 5}{30} = \frac{14}{30} = \frac{7}{15} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{7}{15} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t W_{1} a n d W_{2} b e t w o b a g s c o n t a i n i n g (4 W, 5 B) a n d (9 W, 7 B) b a l l s r e s p e c t i v e l y . \\ L e t E_{1} b e t h e e v e n t t h a t t h e t r a n s f e r r e d b a l l f r o m t h e b a g W_{1} t o W_{2} i s w h i t e a n d \\ E_{2} b e t h e e v e n t t h a t t h e t r a n s f e r r e d b a l l i s b l a c k . \\ a n d E b e t h e e v e n t t h a t t h e b a l l d r a w n f r o m t h e second b a g i s w h i t e . \\ ∴ P (E_{1}) = \frac{4}{9}, P (E_{2}) = \frac{5}{9}, P (E / E_{1}) = \frac{10}{17} a n d P (E / E_{2}) = \frac{9}{17} \\ ∴ P (E) = P (E_{1}) . P (E / E_{1}) + P (E_{2}) . P (E / E_{2}) \\ = \frac{4}{9} \times \frac{10}{17} + \frac{5}{9} \times \frac{9}{17} = \frac{40}{153} + \frac{45}{153} = \frac{85}{153} = \frac{5}{9} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{5}{9} . \end{array}$

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X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

Ncert Solutions Maths class 12th

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