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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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?

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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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

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