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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A die is tossed twice. A ‘success’ is getting an even number on a toss. Find the variance of the number of successes.

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

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two cards are drawn successively without replacement from a well-shuffled deck of cards. Find the mean and standard deviation of the random variable $X$ , where $X$ is the number of aces

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 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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A bag contains $(2 n + 1)$ coins. It is known that $n$ of these coins have a head on both sides, while the rest of the coins are fair. A coin is picked at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$ , determine the value of $n$ .

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 Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t n c o i n s a r e t w o h e a d e d c o i n s a n d t h e r e m a i n i n g (n + 1) c o i n s a r e f a i r . \\ L e t E_{1} : t h e e v e n t t h a t u n f a i r c o i n i s s e l e c t e d . \\ E_{2} : t h e e v e n t t h a t f a i r c o i n i s s e l e c t e d . \\ E_{3} : t h e e v e n t t h a t t h e t o s s r e s u l t i s a h e a d . \\ ∴ P (E_{1}) = \frac{n}{2 n + 1} a n d P (E_{2}) = \frac{n + 1}{2 n + 1} \\ P (E / E_{1}) = 1 (s u r e e v e n t) a n d P (E / E_{2}) = \frac{1}{2} \\ ∴ P (E) = P (E_{1}) . P (E / E_{1}) + P (E_{2}) . P (E / E_{2}) \\ = \frac{n}{2 n + 1} . 1 + \frac{n + 1}{2 n + 1} . \frac{1}{2} = \frac{1}{2 n + 1} (n + \frac{n + 1}{2}) \\ = \frac{1}{2 n + 1} (\frac{2 n + n + 1}{2}) = \frac{3 n + 1}{2 (2 n + 1)} \\ B u t P (E) = \frac{31}{42} (g i v e n) \\ ∴ \frac{3 n + 1}{2 (2 n + 1)} = \frac{31}{42} \Rightarrow \frac{3 n + 1}{2 n + 1} = \frac{31}{21} \\ \Rightarrow 63 n + 21 = 62 n + 31 \\ \Rightarrow n = 10 \\ H e n c e, t h e r e q u i r e d v a l u e o f n i s 10 . \end{array}$

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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 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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Upvote

New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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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