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Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

If the probabilities for A to fail in an examination are 0.2 and for B are 0.3, then the probability that either A or B fails is:

(a) > 0.5

(b) 0.5

(c) ≤ 0.5

(d) 0

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : P (A f a i l s) = 0.2 \\ P (B f a i l s) = 0.3 \\ ∴ P (e i t h e r A o r B f a i l s) \leq P (A f a i l s) + P (B f a i l s) \\ \leq 0.2 + 0.3 \\ \leq 0.5 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

A single letter is selected at random from the word “PROBABILITY.” The probability that it is a vowel is:

(a) $\frac{1}{3}$

(b) $\frac{1}{4}$

(c) $\frac{2}{11}$

(d) None of these.

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f a l p h a b e t s i n p r o b a b i l i t y = 11 \\ N u m b e r o f v o w e l s = 4 \\ ∴ Required p r o b a b i l i t y = \frac{4}{11} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

Six boys and six girls sit in a row at random. The probability that all the girls sit together is:

(a) $\frac{1}{432}$

(b) $\frac{12}{431}$

(c) $\frac{1}{132}$

(d) None of these.

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} I f a l l t h e g i r l s s i t t o g e t h e r, t h e n w e c o n s i d e r i t a s 1 g r o u p \\ ∴ T o t a l n u m b e r o f a r r a n g e m e n t o f 6 + 1 = 7 p e r s o n s i n a r o w = 7! \\ a n d t h e g i r l s a l s o t h e i r p l a c e s w i t h 6! w a y s . \\ ∴ Required p r o b a b i l i t y = \frac{6! 7!}{12!} = \frac{6 * 5 * 4 * 3 * 2 * 7!}{12 * 11 * 10 * 9 * 8 * 7!} = \frac{1}{132} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

$\begin{array}{l} . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

If P(A ∪ B) = P(A ∩ B) for any two events A and B, then:

(a) P(A) = P(B)

(b) P(A) > P(B)

(c) P(A) < P(B)

(d) None of these.

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : P (A \cup B) = P (A \cap B) \\ \Rightarrow P (A) + P (B) - P (A \cap B) = P (A \cap B) \\ \Rightarrow [P (A) - P (A \cap B)] + [P (B) - P (A \cap B)] = 0 \\ B u t P (A) - P (A \cap B) \geq 0 [? P (A \cap B) \leq P (A) o r P (B)] \dots (i) \\ a n d P (B) - P (A \cap B) \geq 0 \dots (i i) \\ F r o m e q n . (i) a n d (i i) w e g e t \\ P (A) = P (B) \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

If A and B are mutually exclusive events, then: (A) P

(a) ≤ P(B)

(b) P(A) ≥ P(B)

(c) P(A) < P(B)

(d) None of these.

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F o r m u t u a l l y e x c l u s i v e e v e n t s, \\ P (A \cap B) = 0 \\ ∴ P (A \cup B) = P (A) + P (B) [? P (A \cap B) = 0] \\ \Rightarrow P (A) + P (B) \leq 1 \\ \Rightarrow P (A) + 1 - P (\bar{B}) \leq 1 [? P (B) = 1 - P (\bar{B})] \\ \Rightarrow P (A) - P (\bar{B}) \leq 0 \\ \Rightarrow P (A) \leq P (\bar{B}) \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

Without repetition of the numbers, four-digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is:

(a) $\frac{1}{5}$

(b) $\frac{4}{5}$

(c) $\frac{1}{30}$

(d) $\frac{5}{9}$

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar $\begin{array}{l} F o u r d i g i t n u m b e r t h e d i g i t s 0, 2, 3, 5 w i t h o u t r e p e t i t i o n a n d d i v i s i b l e b y 5 w i t h t h e \\ g i v e n c o n d i t i o n i s \\ I f u n i t p l a c e b e f i l l e d w i t h 0 \\ T h e n t h e n u m b e r o f w a y s = 3 * 2 * 1 * 1 = 6 \\ I f u n i t p l a c e b e f i l l e d w i t h 5 \\ T h e n t h e n u m b e r o f w a y s = 2 * 2 * 1 * 1 = 4 \\ ∴ T o t a l n u m b e r o f w a y s = 6 + 4 = 10 \\ T o t a l n u m b e r o f w a y s o f a r r a n g i n g t h e d i g i t s 0, 2, 3, 5 t o f o r m 4 d i g i t n u m b e r s w i t h o u t \\ r e p e t i t i o n i s 3 * 3 * 2 * 1 = 18 \\ ∴ Required p r o b a b i l i t y = \frac{10}{18} = \frac{5}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is:

(a) $\frac{1}{3}$

(b) $\frac{1}{6}$

(c) $\frac{2}{7}$

(d) $\frac{1}{2}$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e t w o p a r t i c u l a r p e r s o n s t o b e s e a t e d n e x t e a c h o t h e r t h e n, t h e y f o r m o n e g r o u p . \\ N o w t h e p e r m u t a t i o n o f 6 p e r s o n s = 6! * 2! \\ a n d T o t a l n u m b e r o f p e r m u t a t i o n o f 7 p e r s o n s = 7! \\ ∴ Required p r o b a b i l i t y = \frac{6! * 2!}{7!} \\ = \frac{6! * 2}{7 * 6!} = \frac{2}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards are of different colors

(a) $\frac{29}{52}$

(b) $\frac{1}{2}$

(c) $\frac{26}{51}$

(d) $\frac{27}{51}$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t o u t o f 52 p l a y i n g c a r d s 26 a r e o f r e d a n d 26 a r e o f b l a c k c o l o u r . \\ ∴ P (b o t h c a r d s o f d i f f e r e n t c o l o u r) = \frac{26}{52} * \frac{26}{51} + \frac{26}{52} * \frac{26}{51} \\ = 2 * \frac{26}{52} * \frac{26}{51} = \frac{26}{51} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive:

(a) $\frac{186}{190}$

(b) $\frac{187}{190}$

(c) $\frac{188}{190}$

(d) 18/²⁰C₃

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} S e t o f t h r e e c o n s e c u t i v e n u m b e r s f r o m 1 t o 20 a r e 1, 2, 3; 2, 3, 4; 3, 4, 5; \dots; 18, 19, 20 . \\ S o, t h e p r o b a b i l i t y t h a t t h e n u m b e r s a r e consecutive \\ = \frac{18}{{}_{20}{C_{3}}} = \frac{18}{\frac{20!}{3! 7!}} = \frac{18.3! 7!}{20!} \\ = \frac{18 * 3 * 2 * 17!}{20 * 19 * 18 * 17!} = \frac{3 * 2}{20 * 19} = \frac{3}{190} \\ ∴ P (t h r e e n u m b e r s a r e n o t consecutive) = 1 - \frac{3}{190} = \frac{187}{190} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is:

(a) $\frac{1}{7}$

(b) $\frac{2}{7}$

(c) $\frac{3}{7}$

(d) None of these

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e r e a r e 365 d a y s i n a l e a p y e a r a n d t h e r e a r e 7 d a y s i n a w e e k . \\ ∴ 365 \div 7 = 52 w e e k s + 1 d a y \\ S o, t h i s d a y m a y b e T u e s d a y o r W e d n e s d a y . \\ S o, t h e r e q u i r e d p r o b a b i l i t y = \frac{1}{7} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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