Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Suppose 10,000 tickets are sold in a lottery, each for Re 1. First prize is of Rs 3000 and the second prize is of Rs 2000. There are three third prizes of Rs 500 each. If you buy one ticket, what is your expectation?

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:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Three dice are thrown at the same time. Find the probability of getting three twos, if it is known that the sum of the numbers on the dice was six.

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e d i c e i s t h r o w n t h r e e t i m e s \\ ∴ S a m p l e s p a c e n (S) = {(6)}^{3} = 216 \\ L e t E_{1} b e t h e e v e n t w h e n t h e s u m o f n u m b e r s o n t h e d i c e w a s s i x a n d E_{2} b e t h e e v e n t w h e n \\ t h r e e t w o' s o c c u r . \\ \Rightarrow E_{1} = {(1, 1, 4), (1, 2, 3), (1, 3, 2), (1, 4, 1), (2, 1, 3), (2, 2, 2), (2, 3, 1), (3, 1, 2), (3, 2, 1), (4, 1, 1)} \\ \Rightarrow n (E_{1}) = 10 a n d n (E_{2}) = 1 [? E_{2} = {2, 2, 2}] \\ ∴ P (E_{2} / E_{1}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{1 / 216}{10 / 216} = \frac{1}{10} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

In a dice game, a player pays a stake of Re1 for each throw of a die. She receives Rs 5 if the die shows a 3, Rs 2 if the die shows a 1 or 6, and nothing otherwise. What is the player’s expected profit per throw over a long series of throws?

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:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If X is the number of tails in three tosses of a coin, determine the standard deviation of X.

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

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Prove that:

$i. P (A) = P (A \cap B) + P (A \cap B^{'})$

$ii. P (A \cup B) = P (A \cap B) + P (A \cap B^{'}) + P (A^{'} \cap B)$

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:

$\begin{array}{l} (i) T o p r o v e : P (A) = P (A \cap B) + P (A \cap \bar{B}) \\ R . H . S . = P (A \cap B) + P (A \cap \bar{B}) \\ = P (A) . P (B) + P (A) . P (\bar{B}) = P (A) [P (B) + P (\bar{B})] \\ = P (A) . 1 = P (A) = L . H . S \\ H e n c e p r o v e d . \\ (i i) T o p r o v e : P (A \cup B) = P (A \cap B) + P (A \cap \bar{B}) + P (\bar{A} \cap B) \\ R . H . S . = P (A \cap B) + P (A \cap \bar{B}) + P (\bar{A} \cap B) \\ = P (A) . P (B) + P (A) . P (\bar{B}) + P (\bar{A}) . P (B) \\ = P (A) . P (B) + P (A) [1 - P (B)] + [1 - P (A)] . P (B) \\ = P (A) . P (B) + P (A) - P (A) . P (B) + P (B) - P (A) . P (B) \\ = P (A) + P (B) - P (A \cap B) = P (A \cup B) = L . H . S . \\ H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A discrete random variable X has the probability distribution given as below:

X

0.5

1

1.5

2

P(X)

k

k²

2k²

k

i. Find the value of k

ii. Determine the mean of the distribution.

18 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:

$\begin{array}{l} F o r a p r o b a b i l i t y d i s t r i b u t i o n, w e k n o w t h a t i f P_{i} \geq 0 \\ (i) \sum_{i = 1}^{n} P_{i} = 1 \Rightarrow k + k^{2} + 2 k^{2} + k = 1 \\ \Rightarrow 3 k^{2} + 2 k - 1 = 0 \Rightarrow 3 k^{2} + 3 k - k - 1 = 0 \\ \Rightarrow 3 k (k + 1) - 1 (k + 1) = 0 \Rightarrow (3 k - 1) (k + 1) = 0 \\ ∴ k = \frac{1}{3} a n d k = - 1 \\ B u t k \geq 0 ∴ k = \frac{1}{3} \\ (i i) M e a n o f t h e d i s t r i b u t i o n \\ E (X) = \sum_{i = 1}^{n} X_{i} P_{i} = 0.5 k + 1.5 k^{2} + 1.5 (2 k^{2}) + 2 k \\ = \frac{k}{2} + k^{2} + 3 k^{2} + 2 k = 4 k^{2} + \frac{5}{2} k \\ = 4 {(\frac{1}{3})}^{2} + \frac{5}{2} (\frac{1}{3}) = \frac{4}{9} + \frac{5}{6} = \frac{23}{18} \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Let $E_{1}$ and $E_{2}$ be two independent events such that $P (E_{1}) = p_{1}$ and $P (E_{2}) = p_{2}$ .

Describe in words the events whose probabilities are:

${i. p}_{1} p_{2}$

$ii. (1 - p_{1}) p_{2}$

$iii. 1 - (1 - p_{1}) (1 - p_{2})$

${iv. p}_{1} + p_{2} - 2 p_{1} p_{2}$

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

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Three events A, B, and C have probabilities $P (A) = \frac{2}{5}$ , $\frac{1}{3}$ and $\frac{1}{2}$ , respectively. Given that $P (A \cap C) = \frac{1}{5}$ and $P (B \cap C) = \frac{1}{4}$ , find the values of $P (C ? B)$ and $P (A^{'} \cap C^{'})$ .

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e P (A) = \frac{2}{5}, P (B) = \frac{1}{3} a n d P (C) = \frac{1}{2} \\ P (A \cap C) = \frac{1}{5} a n d P (B \cap C) = \frac{1}{4} \\ ∴ P (C / B) = \frac{P (B \cap C)}{P (B)} = \frac{1 / 4}{1 / 3} = \frac{3}{4} \\ P (A' \cap C') = 1 - P (A \cup C) = 1 - [P (A) + P (C) - P (A \cap C)] \\ = 1 - [\frac{2}{5} + \frac{1}{2} - \frac{1}{5}] = 1 - \frac{7}{10} = \frac{3}{10} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t i e s a r e \frac{3}{4} a n d \frac{3}{10} . \end{array}$

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New question posted•a year ago

Ncert Solutions Maths class 12th Probability

<p><strong>Three events A, B, and C have probabilities <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><mi>A</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mn>2</mn></mrow><mrow><mn>5</mn></mrow></mfrac></mrow></math> , <math><mrow><mo> </mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></math> and <math><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></math> , respectively. Given that <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><mi>A</mi><mo>∩</mo><mi>C</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>5</mn></mrow></mfrac></mrow></math> and <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><mi>B</mi><mo>∩</mo><mi>C</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>4</mn></mrow></mfrac></mrow></math> , find the values of <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><mi>C</mi><mo>?</mo><mi>B</mi></mrow><mo>)</mo></mrow></mrow></math> and <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><msup><mi>A</mi><mo>′</mo></msup><mo>∩</mo><msup><mi>C</mi><mo>′</mo></msup></mrow><mo>)</mo></mrow></mrow></math> .</strong></p>

4 Views|Posted a year ago

Asked by Alok Kumar Singh

New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A and B are two events such that $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ , and $P (A \cap B) = \frac{1}{4}$ . Find:

$i. P(A ∣ B$

$ii. (B∣ A)$

$iii. (A' ∣ B)$

$iv. (A' ∣ B')$

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:

$\begin{array}{l} W e h a v e P (A) = \frac{1}{2}, P (B) = \frac{1}{3} a n d P (A \cap B) = \frac{1}{4} \\ P (A') = 1 - \frac{1}{2} = \frac{1}{2}, P (B') = 1 - \frac{1}{3} = \frac{2}{3} \\ P (A' \cap B') = 1 - P (A \cup B) = 1 - [P (A) + P (B) - P (A \cap B)] \\ = 1 - [\frac{1}{2} + \frac{1}{3} - \frac{1}{4}] = 1 - [\frac{6 + 4 - 3}{12}] = 1 - \frac{7}{12} = \frac{5}{12} \\ (i) P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{1 / 4}{1 / 3} = \frac{3}{4} \\ (i i) P (B / A) = \frac{P (A \cap B)}{P (A)} = \frac{1 / 4}{1 / 2} = \frac{1}{2} \\ (i i i) P (A' / B) = \frac{P (A' \cap B)}{P (B)} = \frac{P (B) - P (A \cap B)}{P (B)} = 1 - \frac{P (A \cap B)}{P (B)} \\ = 1 - \frac{1 / 4}{1 / 3} = 1 - \frac{3}{4} = \frac{1}{4} \\ (i v) P (A' / B') = \frac{P (A' \cap B')}{P (B')} = \frac{5 / 12}{2 / 3} = \frac{5}{12} \times \frac{3}{2} = \frac{5}{8} \end{array}$

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X	0.5	1	1.5	2
P(X)	k	k²	2k²	k

Ncert Solutions Maths class 12th

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