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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two dice are tossed. Find whether the following two events $A$ and $B$ are independent: $A = {(x, y) : x + y = 11}$ , $B = {(x, y) : x \neq 5}$ where $(x, y)$ denotes a typical sample point.

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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:

$\begin{array}{l} G i v e n t h a t : \\ A = {(x, y) : x + y = 11} a n d B = {(x, y) : x \neq 5} \\ ∴ A = {(5, 6), (6, 5)} \\ B = {\begin{cases} (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), \\ (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \end{cases}} \\ \Rightarrow n (A) = 2, n (B) = 30 a n d n (A \cap B) = 1 \\ ∴ P (A) = \frac{2}{36} = \frac{1}{18} a n d P (B) = \frac{30}{36} = \frac{5}{6} \\ ∴ P (A) . P (B) = \frac{1}{18} . \frac{5}{6} = \frac{5}{108} a n d P (A \cap B) = \frac{1}{36} \\ , P (A) . P (B) \neq P (A \cap B) \\ H e n c e, A a n d B a r e n o t i n d e p e n d e n t . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A and B throw a pair of dice alternately. A wins the game if he gets a total of 6 and B wins if she gets a total of 7. If A starts the game, find the probability of winning the game by A in the third throw of the pair of dice.

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:

$\begin{array}{l} L e t A_{1} b e t h e e v e n t o f g e t t i n g a t o t a l o f 6 . \\ = {(2, 4), (4, 2), (1, 5), (5, 1), (3, 3)} \\ L e t B_{1} b e t h e e v e n t o f g e t t i n g a t o t a l o f 7 . \\ = {(2, 5), (5, 2), (1, 6), (6, 1), (3, 4), (4, 3)} \\ L e t P (A_{1}) i s t h e p r o b a b i l i t y, i f A w i n s i n a t h r o w = \frac{5}{36} \\ a n d P (B_{1}) i s t h e p r o b a b i l i t y, i f B w i n s i n a t h r o w = \frac{1}{36} \\ ∴ T h e r e q u i r e d p r o b a b i l i t y o f w i n n i n g A i n h i s t h i r d t h r o w \\ = P (\bar{A_{1}}) . P (\bar{B_{1}}) . P (A_{1}) = \frac{31}{36} . \frac{5}{6} . \frac{5}{36} = \frac{775}{7776} . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Find the variance of the distribution:

x

0

1

2

3

4

5

P(x)
1/6 5/18 2/9 1/6 1/9 1/18

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:

$\begin{array}{l} W e k n o w t h a t : V a r (X) = E (X^{2}) - {[E (X)]}^{2} \\ E (X) = \sum_{i = 1}^{n} p_{i} x_{i} \\ = 0 \times \frac{1}{6} + 1 \times \frac{5}{18} + 2 \times \frac{2}{9} + 3 \times \frac{1}{6} + 4 \times \frac{1}{9} + 5 \times \frac{1}{18} \\ = 0 + \frac{5}{18} + \frac{4}{9} + \frac{3}{6} + \frac{4}{9} + \frac{5}{18} = \frac{5 + 8 + 9 + 8 + 5}{18} = \frac{35}{18} \\ a n d E (X^{2}) = 0 \times \frac{1}{6} + 1 \times \frac{5}{18} + 4 \times \frac{2}{9} + 9 \times \frac{1}{6} + 16 \times \frac{1}{9} + 25 \times \frac{1}{18} \\ = \frac{5}{18} + \frac{8}{9} + \frac{9}{6} + \frac{16}{9} + \frac{25}{18} = \frac{5 + 16 + 27 + 32 + 25}{18} = \frac{105}{18} \\ ∴ V a r (X) = \frac{105}{18} - \frac{35}{18} \times \frac{35}{18} = \frac{1890 - 1225}{324} = \frac{665}{324} \\ H e n c e, t h e r e q u i r e d i s \frac{665}{324} . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.

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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

Two natural numbers $r$ , $s$ are drawn one at a time, without replacement from the set $S = {1, 2, 3, . . ., n}$ . Find $P [r \leq p \leq s ∣ p \in S]$

10 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:

$\begin{array}{l} G i v e n t h a t : S = {1, 2, 3, \dots, n} \\ ∴ P (r \leq p / s \leq p) = \frac{P (P \cap S)}{P (S)} = \frac{p - 1}{n} \times \frac{n}{n - 1} = \frac{p - 1}{n - 1} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{p - 1}{n - 1} . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Suppose that 6% of the people with blood group O are left-handed and 10% of those with other blood groups are left-handed. 30% of the people have blood group O. If a left-handed person is selected at random, what is the probability that he/she will have blood group O?

10 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:

$\begin{array}{l} L e t E_{1} = T h e e v e n t t h a t a p e r s o n s e l e c t e d i s o f b l o o d g r o u p O \\ E_{2} = T h e e v e n t t h a t a p e r s o n s e l e c t e d i s o f o t h e r g r o u p \\ a n d H = T h e e v e n t t h a t s e l e c t e d p e r s o n i s l e f t h a n d e d . \\ P (E_{1}) = 0.30, P (E_{2}) = 0.70, P (H / E_{1}) = 0.06, P (H / E_{2}) = 0.10 \\ ∴ Using B a y e' s T h e o r m, w e g e t \\ P (E_{1} / H) = \frac{P (E_{1}) . P (H / E_{1})}{P (E_{1}) . P (H / E_{1}) + P (E_{2}) . P (H / E_{2})} \\ = \frac{0.30 \times 0.06}{0.30 \times 0.06 + 0.70 \times 0.10} = \frac{0.018}{0.018 + 0.070} = \frac{0.018}{0.088} = \frac{9}{44} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{9}{44} . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

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:

$\begin{array}{l} L e t E_{1} = E v e n t t h a t t h e c o i n i s f a i r \\ E_{2} = E v e n t t h a t t h e c o i n i s 2 - h e a d e d \\ a n d H = E v e n t t h a t t h e t o s s e d c o i n g e t s h e a d . \\ P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{2}, P (H / E_{1}) = \frac{1}{2}, P (H / E_{2}) = 1 \\ ∴ Using B a y e' s T h e o r m, w e g e t \\ P (E_{1} / H) = \frac{P (E_{1}) . P (H / E_{1})}{P (E_{1}) . P (H / E_{1}) + P (E_{2}) . P (H / E_{2})} \\ = \frac{\frac{1}{2} . \frac{1}{2}}{\frac{1}{2} . \frac{1}{2} + \frac{1}{2} . 1} = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{1}{3} . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A factory produces bulbs. The probability that any one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10. From a single box, find the probability that:

i. one of the bulbs is defective

ii. exactly two bulbs are defective

iii. more than 8 bulbs work properly

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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t X b e t h e r a n d o m d e n o t i n g a b u l b t o b e d e f e c t i v e . \\ H e r e, n = 10, p = \frac{1}{50}, q = 1 - \frac{1}{50} = \frac{49}{50} \\ W e k n o w t h a t P (X = r) = {}_{n}{C_{r}} p^{r} q^{n - r} \\ (i) N o n e o f t h e b u l b s i s d e f e c t i v e, i . e ., r = 0 \\ P (x = 0) = {}_{10}{C_{0}} {(\frac{1}{50})}^{0} {(\frac{49}{50})}^{10 - 0} = {(\frac{49}{50})}^{10} \\ (i i) E x a c t l y t w o b u l b s a r e d e f e c t i v e, i . e ., r = 2 \\ ∴ P (x = 2) = {}_{10}{C_{2}} {(\frac{1}{50})}^{2} {(\frac{49}{50})}^{10 - 2} = 45 . \frac{{(49)}^{8}}{{(50)}^{10}} = 45 \times {(\frac{1}{50})}^{10} \times {(49)}^{8} \\ (i i i) M o r e t h a n 8 b u l b s w o r k p r o p e r l y \\ W e c a n s a y t h a t l e s s t h a n 2 b u l b s a r e d e f e c t i v e \\ P (X < 2) = P (x = 0) + P (x = 1) \\ = {}_{10}{C_{0}} {(\frac{1}{50})}^{0} {(\frac{49}{50})}^{10 - 0} + {}_{10}{C_{1}} {(\frac{1}{50})}^{1} {(\frac{49}{50})}^{9} = {(\frac{49}{50})}^{10} + \frac{1}{5} {(\frac{49}{50})}^{9} \\ = {(\frac{49}{50})}^{9} (\frac{49}{50} + \frac{1}{5}) = {(\frac{49}{50})}^{9} (\frac{59}{50}) = \frac{59 {(49)}^{9}}{{(50)}^{10}} \end{array}$