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Class 12th Probability

If $P (A \cap B) = \frac{7}{10}$ and $P (B) = \frac{17}{20}$ , then $P (A ? B)$ equals:

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (B) = \frac{17}{20}, a n d P (A \cap B) = \frac{7}{10} \\ ∴ P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{\frac{7}{10}}{\frac{17}{20}} = \frac{14}{17} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Choose the correct answer from the given four options in each of the Exercises 1 to 38:

Q1. If $P (A) = \frac{4}{5}$ , and $P (A \cap B) = \frac{7}{10}$ , then $P (B ? A)$ is equal to:

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = \frac{4}{5}, a n d P (A \cap B) = \frac{7}{10} \\ ∴ P (B / A) = \frac{P (A \cap B)}{P (A)} = \frac{\frac{7}{10}}{\frac{4}{5}} = \frac{7}{8} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Class 12th Probability

An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on $k$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t A b e t h e e v e n t h a v i n g m w h i t e a n d n b l a c k b a l l s . \\ E_{1} = {f i r s t b a l l d r a w n o f w h i t e c o l o u r} \\ E_{2} = {f i r s t b a l l d r a w n o f b l a c k c o l o u r} \\ E_{3} = {second b a l l d r a w n o f w h i t e c o l o u r} \\ P (E_{1}) = \frac{m}{m + n}, P (E_{2}) = \frac{n}{m + n}, P (E_{3} / E_{1}) = \frac{m + k}{m + n + k}, P (E_{3} / E_{2}) = \frac{m}{m + n + k} \\ N o w, P (E_{3}) = P (E_{1}) . P (E_{3} / E_{1}) + P (E_{2}) . P (E_{3} / E_{2}) \\ = \frac{m}{m + n} * \frac{m + k}{m + n + k} + \frac{n}{m + n} * \frac{m}{m + n + k} \\ = \frac{m}{m + n + k} [\frac{m + k}{m + n} + \frac{n}{m + n}] \\ = \frac{m}{m + n + k} [\frac{m + n + k}{m + n}] = \frac{m}{m + n} \\ H e n c e, t h e p r o b a b i l i t y o f d r a w n i n g a w h i t e b a l l d o e s n o t d e p e n d u p o n k . \end{array}$

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New answer posted

a year ago

Class 12th Probability

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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alok kumar singh

Contributor-Level 10

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

Class 12th Probability

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.

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alok kumar singh

Contributor-Level 10

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

Class 12th Probability

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 12th Probability

The random variable $X$ can take only the values 0, 1, 2. Given that $P (X = 0) = P (X = 1) = p$ and that $E (X^{2}) = E [X]$ , find the value of $p$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

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New answer posted

a year ago

Class 12th Probability

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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alok kumar singh

Contributor-Level 10

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

Sol:

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New answer posted

a year ago

Class 12th Probability

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

0 Follower 10 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 12th Probability

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?

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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