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 discrete random variable $x$ is given as under:

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Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

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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 as under:

X

1

2

4

2A

3A

5A

P(X)
1/2 1/5 3/25 1/10 1/25 1/25

Calculate:

The value of if $E(X) = 2.94$ (ii)Variance of X

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 Long Answer 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

Let $X$ be a discrete random variable whose probability distribution is defined as follows:

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Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long 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

An item is manufactured by three machines A, B, and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on A, 30% on B, and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective, and 3% of these produced on C are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?

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 Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} : t h e e v e n t t h a t t h e i t e m i s m a n u f a c t u r e d o n m a c h i n e A \\ E_{2} : t h e e v e n t t h a t t h e i t e m i s m a n u f a c t u r e d o n m a c h i n e B \\ E_{3} : t h e e v e n t t h a t t h e i t e m i s m a n u f a c t u r e d o n m a c h i n e C \\ L e t H b e t h e e v e n t t h a t t h e s e l e c t e d i t e m i s d e f e c t i v e . \\ Using B a y e s' T h e o r m \\ ∴ P (E_{1}) = \frac{50}{100}, P (E_{2}) = \frac{30}{100} a n d P (E_{3}) = \frac{20}{100} \\ P (H / E_{1}) = \frac{2}{100}, P (H / E_{2}) = \frac{2}{100} a n d P (H / E_{3}) = \frac{3}{100} \\ ∴ P (E_{1} / H) = \frac{P (E_{1}) . P (H / E_{1})}{P (E_{1}) . P (H / E_{1}) + P (E_{2}) . P (H / E_{2}) + P (E_{3}) . P (H / E_{3})} \\ = \frac{\frac{50}{100} \times \frac{2}{100}}{\frac{50}{100} \times \frac{2}{100} + \frac{30}{100} \times \frac{2}{100} + \frac{20}{100} \times \frac{3}{100}} = \frac{100}{100 + 60 + 60} = \frac{100}{220} = \frac{10}{22} = \frac{5}{11} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{5}{11} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

By examining the chest X-ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of a healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?

23 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} : t h e e v e n t t h a t a p e r s o n h a s T B \\ E_{2} : t h e e v e n t t h a t a p e r s o n d o e s n o t h a v e T B \\ L e t H b e t h e e v e n t t h a t t h e p e r s o n i s d i a g n o s e d t o h a v e T B . \\ B a y e s' T h e o r m \\ ∴ P (E_{1}) = \frac{1}{1000} = 0.001, P (E_{2}) = 1 - \frac{1}{1000} = \frac{999}{1000} = 0.999 \\ P (H / E_{1}) = 0.99, P (H / E_{2}) = 0.001 \\ ∴ P (E_{1} / H) = \frac{P (E_{1}) . P (H / E_{1})}{P (E_{1}) . P (H / E_{1}) + P (E_{2}) . P (H / E_{2})} \\ = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.001} = \frac{0.99}{0.99 + 0.999} \\ = \frac{0.990}{0.990 + 0.999} = \frac{990}{1989} = \frac{110}{221} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{110}{221} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

There are three urns containing: 2 white and 3 black balls 3 white and 2 black balls 4 white and 1 black ball respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn, and it is found to be white. Find the probability that the ball drawn was from the second urn.

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 Long Answer Type Questions as classified in NCERT Exemplar

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3, a ball is taken from the first bag; if it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

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 Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t o f s e l e c t i n g B a g I \\ E_{2} b e t h e e v e n t o f s e l e c t i n g B a g I I \\ a n d E_{3} b e t h e e v e n t t h a t b l a c k b a l l i s s e l e c t e d \\ ∴ P (E_{1}) = \frac{2}{6} = \frac{1}{3} a n d P (E_{2}) = 1 - \frac{1}{3} = \frac{2}{3} \\ P (E_{3} / E_{1}) = \frac{3}{7} a n d P (E_{3} / E_{2}) = \frac{4}{7} \\ ∴ P (E_{3}) = P (E_{1}) . P (E_{3} / E_{1}) + P (E_{2}) . P (E_{3} / E_{2}) \\ = \frac{1}{3} . \frac{3}{7} + \frac{2}{3} . \frac{4}{7} = \frac{3 + 8}{21} = \frac{11}{21} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{11}{21} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letters "TA" are visible. What is the probability that the letter came from TATA NAGAR?

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 Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} : T h e e v e n t t h a t t h e l e t t e r c o m e s f r o m T A T A N A G A R \\ a n d E_{2} : T h e e v e n t t h a t t h e l e t t e r c o m e s f r o m C A L C U T T A \\ A l s o, E_{3} : T h e e v e n t t h a t o n t h e l e t t e r, t w o l e t t e r s T A a r e v i s i b l e . \\ ∴ P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{2} a n d P (\frac{E_{3}}{E_{1}}) = \frac{2}{8} a n d P (\frac{E_{3}}{E_{2}}) = \frac{1}{7} \\ [? F o r T A T A N A G A R, t h e t w o consecutive l e t t e r s v i s i b l e a r e T A, A T, T A, A N, N A, A G, G A, A R] \\ ∴ P (E_{3} / E_{1}) = \frac{2}{8} \\ a n d [F o r C A L C U T T A, t h e t w o consecutive l e t t e r s v i s i b l e a r e C A, A L, L C, C U, U T, T T a n d T A] \\ S o, P (E_{3} / E_{2}) = \frac{1}{7} \\ N o w using B a y e s' T h e r o r m, w e h a v e \\ ∴ P (E_{1} / E_{3}) = \frac{P (E_{1}) . P (E_{3} / E_{1})}{P (E_{1}) . P (E_{3} / E_{1}) + P (E_{2}) . P (E_{3} / E_{2})} \\ = \frac{\frac{1}{2} . \frac{2}{8}}{\frac{1}{2} . \frac{2}{8} + \frac{1}{2} . \frac{1}{7}} = \frac{\frac{1}{8}}{\frac{1}{8} + \frac{1}{14}} = \frac{\frac{1}{8}}{\frac{7 + 4}{56}} = \frac{7}{11} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{7}{11} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A shopkeeper sells three types of flower seeds: $A_{1}$ , $A_{2}$ , and $A_{3}$ . They are sold as a mixture where the proportions are 4:4:2, respectively. The germination rates of the three types of seeds are 45%, 60%, and 35%. Calculate the probability:

(i) Of a randomly chosen seed to germinate

(ii) That it will not germinate given that the seed is of type $A_{3}$

(iii) That it is of type $A_{2}$ given that a randomly chosen seed does not germinate.

30 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t A_{1} : A_{2} : A_{3} = 4 : 4 : 2 \\ ∴ P (A_{1}) = \frac{4}{10}, P (A_{2}) = \frac{4}{10} a n d P (A_{3}) = \frac{2}{10} w h e r e A_{1}, A_{2} a n d A_{3} a r e t h e t h r e e t y p e s o f s e e d s . \\ L e t E b e t h e e v e n t s t h a t a s e e d a n d \bar{E} b e t h e e v e n t s t h a t a s e e d d o e s n o t \\ ∴ P (\frac{E}{A_{1}}) = \frac{45}{100}, P (\frac{E}{A_{2}}) = \frac{60}{100} a n d P (\frac{E}{A_{3}}) = \frac{35}{100} \\ a n d P (\frac{\bar{E}}{A_{1}}) = \frac{55}{100}, P (\frac{\bar{E}}{A_{2}}) = \frac{40}{100} a n d P (\frac{\bar{E}}{A_{3}}) = \frac{65}{100} \\ (i) P (E) = P (A_{1}) . P (\frac{E}{A_{1}}) + P (A_{2}) . P (\frac{E}{A_{2}}) + P (A_{3}) . P (\frac{E}{A_{3}}) \\ = \frac{4}{10} . \frac{45}{100} + \frac{4}{10} . \frac{60}{100} + \frac{2}{10} . \frac{35}{100} \\ = \frac{180}{1000} + \frac{240}{1000} + \frac{70}{1000} = \frac{490}{1000} = 0.49 \\ (i i) P (\bar{E} / A_{3}) = 1 - P (E / A_{3}) = 1 - \frac{35}{100} = \frac{65}{100} = 0.65 \\ (i i i), B a y e s' T h e o r e m, w e g e t \\ P (A_{2} / \bar{E}) = \frac{P (A_{2}) . P (\bar{E} / A_{2})}{P (A_{1}) . P (\bar{E} / A_{1}) + P (A_{2}) . P (\bar{E} / A_{2}) + P (A_{3}) . P (\bar{E} / A_{3})} \\ = \frac{\frac{4}{10} . \frac{40}{100}}{\frac{4}{10} . \frac{55}{100} + \frac{4}{10} . \frac{40}{100} + \frac{2}{10} . \frac{65}{100}} = \frac{\frac{160}{1000}}{\frac{220}{1000} + \frac{160}{1000} + \frac{130}{1000}} \\ = \frac{160}{220 + 160 + 130} = \frac{160}{510} = \frac{16}{51} = 0.314 \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{16}{51} o r 0.314 \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Refer to Question 41 above. If a white ball is selected, what is the probability that it came from

$i.$ $B a g I I$

$ii. B a g I I I$

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 Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} Referring t o E x e r c i s e Q . 41, w e w i l l u s e h e r e, B a y e s' T h e o r e m \\ (i) P (E_{2} / F) = \frac{P (E_{2}) . P (F / E_{2})}{P (E_{1}) . P (F / E_{1}) + P (E_{2}) . P (F / E_{2}) + P (E_{3}) . P (F / E_{3})} \\ = \frac{\frac{2}{6} . \frac{1}{3}}{\frac{1}{6} . 0 + \frac{2}{6} . \frac{1}{3} + \frac{3}{6} . 1} = \frac{\frac{2}{18}}{\frac{2}{18} + \frac{3}{6}} = \frac{2}{18} \times \frac{18}{11} = \frac{2}{11} \\ (i i) P (E_{3} / F) = \frac{P (E_{3}) . P (F / E_{3})}{P (E_{1}) . P (F / E_{1}) + P (E_{2}) . P (F / E_{2}) + P (E_{3}) . P (F / E_{3})} \\ = \frac{\frac{3}{6} . 1}{\frac{1}{6} . 0 + \frac{2}{6} . \frac{1}{3} + \frac{3}{6} . 1} = \frac{\frac{3}{6}}{\frac{2}{18} + \frac{3}{6}} = \frac{3}{6} \times \frac{18}{11} = \frac{9}{11} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t i e s a r e \frac{2}{11} a n d \frac{9}{11} \end{array}$

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Ncert Solutions Maths class 12th

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