Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A box contains 3 orange balls, 3 green balls, and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is

17 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} P robability o f d r a w i n g 2 g r e e n a n d 1 b l u e b a l l s \\ = P (G) . P (G) . P (B) + P (G) . P (B) . P (G) + P (B) . P (G) . P (G) \\ = \frac{3}{8} . \frac{2}{7} . \frac{2}{6} + \frac{3}{8} . \frac{2}{7} . \frac{2}{6} + \frac{2}{8} . \frac{3}{7} . \frac{2}{6} = \frac{12}{336} + \frac{12}{336} + \frac{12}{336} = \frac{36}{336} = \frac{3}{28} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t o f g e t t i n g e v e n n u m b e r o n t h e d i e . E_{2} b e t h e e v e n t o f s e l e c t i n g a s p a d e c a r d . \\ ∴ P (E_{1}) = \frac{3}{6} = \frac{1}{2} a n d P (E_{2}) = \frac{13}{52} = \frac{1}{4} \\ S o, P (E_{1} \cap E_{2}) = P (E_{1}) . P (E_{2}) = \frac{1}{2} . \frac{1}{4} = \frac{1}{8} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl, given that the family has at least one girl is

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t G d e n o t e s t h e g i r l a n d B d e n o t e s t h e b o y o f t h e g i v e n f a m i l y . \\ S o, n (S) = {(B G G), (G B G), (G G B), (G B B), (B G B), (B B G), (B B B), (G G G)} \\ L e t E_{1} b e t h e e v e n t t h a t t h e f a m i l y h a s a t l e a s t o n e g i r l \\ ∴ E_{1} = {(B G G), (G B G), (G G B), (G B B), (B G B), (B B G), (G G G)} \\ . E_{2} b e t h e e v e n t t h a t t h e e l d e s t c h i l d i s a g i r l . \\ ∴ E_{2} = {(G B G), (G G B), (G B B), (G G G)} \\ (E_{1} \cap E_{2}) = {(G B B), (G G B), (G B G), (G G G)} \\ ∴ P (E_{2} / E_{1}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{4 / 8}{7 / 8} = \frac{4}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Three persons, A, B, and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3, and 0.2 respectively. The probability of exactly two hits is:

(a) 0.024

(b) 0.188

(c) 0.336

(d) 0.452

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = 0.4, P (B) = 0.3, P (C) = 0.2 \\ A l s o, P (\bar{A}) = 1 - 0.4 = 0.6, P (\bar{B}) = 1 - 0.3 = 0.7, P (\bar{C}) = 1 - 0.2 = 0.8 \\ ∴ Probabilities o f t w o h i t s \\ = P (A) . P (B) . P (\bar{C}) + P (A) . P (\bar{B}) . P (C) + P (\bar{A}) . P (B) . P (C) \\ = 0.4 \times 0.3 \times 0.8 + 0.4 \times 0.7 \times 0.2 + 0.6 \times 0.3 \times 0.2 \\ = 0.096 + 0.056 + 0.036 = 0.188 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Refer to Question 74 above. The probability that exactly two of the three balls were red, with the first ball being red, is

17 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A c c o r d i n g t o Q u e s t i o n 74, \\ L e t E_{1} b e e v e n t t h a t f i r s t b a l l i s r e d . \\ E_{2} b e e v e n t t h a t e x a c t l y t w o o f t h e t h r e e b a l l s a r e r e d . \\ ∴ P (E_{1}) = P (R) . P (R) . P (B) + P (R) . P (R) . P (R) + P (R) . P (B) . P (R) + P (R) . P (B) . P (B) \\ = \frac{5}{8} . \frac{4}{7} . \frac{3}{6} + \frac{5}{8} . \frac{4}{7} . \frac{3}{6} + \frac{5}{8} . \frac{3}{7} . \frac{4}{6} + \frac{5}{8} . \frac{3}{7} . \frac{2}{6} \\ = \frac{60}{336} + \frac{60}{336} + \frac{60}{336} + \frac{30}{336} = \frac{210}{336} \\ P (E_{1} \cap E_{2}) = P (R) . P (B) . P (R) + P (R) . P (R) . P (B) \\ = \frac{5}{8} . \frac{3}{7} . \frac{4}{6} + \frac{5}{8} . \frac{4}{7} . \frac{3}{6} = \frac{60}{336} + \frac{60}{336} = \frac{120}{336} \\ ∴ P (E_{2} / E_{1}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{120 / 336}{210 / 336} = \frac{4}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, the probability of getting exactly one red ball is

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : B a g c o n t a i n s 5 r e d a n d 3 b l u e b a l l s . \\ o f g e t t i n g e x a c t l y o n e r e d b a l l i f 3 b a l l s a r e r a n d o m l y d r a w n w i t h o u t r e p l a c e m e n t . \\ P (R) . P (B) . P (B) + P (B) . P (R) . P (B) + P (B) . P (B) . P (R) \\ = \frac{5}{8} . \frac{3}{7} . \frac{2}{6} + \frac{3}{8} . \frac{5}{7} . \frac{2}{6} + \frac{3}{8} . \frac{2}{7} . \frac{5}{6} = \frac{30}{336} + \frac{30}{336} + \frac{30}{336} = \frac{90}{336} = \frac{15}{56} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Two events $E$ and $F$ are independent. If $P (E) = 0.3$ , $P (E \cup F) = 0.5$ , then $P (E ∣ F) - P (F ∣ E)$ equals:

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : E a n d F a r e t w o i n d e p e n d e n t e v e n t s s u c h t h a t \\ P (E) = 0.3, a n d P (E \cup F) = 0.5 \\ ∴ P (E \cup F) = P (E) + P (F) - P (E \cap F) \\ 0.5 = 0.3 + P (F) - P (E) . P (F) \\ \Rightarrow 0.5 - 0.3 = P (F) [1 - P (E)] \Rightarrow 0.2 = P (F) (1 - 0.3) \\ \Rightarrow 0.2 = P (F) . (0.7) \\ ∴ P (F) = \frac{0.2}{0.7} = \frac{2}{7} \\ N o w P (E / F) - P (F / E) = \frac{P (E \cap F)}{P (F)} . \frac{P (E \cap F)}{P (E)} \\ = \frac{P (E) . P (F)}{P (F)} . \frac{P (E) . P (F)}{P (E)} = P (E) - P (F) = \frac{3}{10} - \frac{2}{7} = \frac{1}{70} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

If the events $A$ and $B$ are independent, then $P (A \cap B)$ is equal to:

$(a) P(A) +P (B)$

$(b) P(A) -P (B)$

$(c) P(A) \cdot P(B)$

$(d) P(A) ? P(B)$

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A a n d B a r e t w o i n d e p e n d e n t e v e n t s \\ ∴ P (A \cap B) = P (A) . P (B) \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Let $A$ and $B$ be two events such that $P (A) = \frac{3}{8}$ , $P (B) = \frac{5}{8}$ , and $P (A \cup B) = \frac{3}{4}$ . Then $P (A ∣ B) \cdot P (A^{'} ∣ B)$ is equal to:

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = \frac{3}{8}, P (B) = \frac{5}{8} a n d P (A \cup B) = \frac{3}{4} \\ ∴ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \frac{3}{4} = \frac{3}{8} + \frac{5}{8} - P (A \cap B) \\ \Rightarrow P (A \cap B) = \frac{3}{8} + \frac{5}{8} - \frac{3}{4} = \frac{1}{4} \\ N o w P (A / B) . P (A' / B) = \frac{P (A \cap B)}{P (B)} . \frac{P (A' \cap B)}{P (B)} \\ = \frac{P (A \cap B)}{P (B)} . \frac{P (B) - P (A \cap B)}{P (B)} \\ = \frac{\frac{1}{4}}{5} . \frac{(\frac{5}{8} - \frac{1}{4})}{5} = \frac{2}{5} . \frac{3}{5} = \frac{6}{25} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$