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New answer posted

a year ago

Class 11th 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.

0 Follower 2 Views

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

a year ago

Class 11th 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.

0 Follower 2 Views

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

a year ago

Class 11th 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

a year ago

Class 11th 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) $\frac{1}{2}$

0 Follower 4 Views

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

a year ago

Class 11th 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) $\frac{27}{51}$

0 Follower 6 Views

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

a year ago

Class 11th 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

0 Follower 2 Views

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

a year ago

Class 11th 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

0 Follower 4 Views

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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New answer posted

a year ago

Class 11th Exemplar Solution

Determine the probability p, for each of the following events.

(a) An odd number appears in a single toss of a fair die.

(b) At least one head appears in two tosses of a fair coin.

(c) A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.

(d) The sum of 6 appears in a single toss of a pair of fair dice.

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} (a) P o s s i b l e o u t c o m e o f a single t h r o w o f d i e \\ S = {1, 2, 3, 4, 5, 6} o u t o f w h i c h 1, 3, 5 a r e o d d \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{3}{6} = \frac{1}{2} \\ (b) W h e n a f a i r c o i n i s t o s s e d t w i c e, t h e n t h e s a m p l e s p a c e \\ S = {H H, H T, T H, T T} \\ ∴ P r o b a b i l i t y o f g e t t i n g a t l e a s t o n e h e a d (H H, H T, T H) = \frac{3}{4} \\ (c) F a v o u r a b l e e v e n t s a r e 4 k i n g s + 2 o f h e a r t s + 3 o f s p a d e s = 4 + 1 + 1 = 6 = \frac{6}{52} = \frac{3}{26} \\ (d) W h e n a p a i r o f d i c e i s r o l l e d, t h e n t o t a l n u m b e r o f s a m p l e s p a c e = 36 o u t o f w h i c h \\ (1, 5), (5, 1), (2, 4), (4, 2) a n d (3, 3) a r e t h e f a v o u r a b l e e v e n t s \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{5}{36} . \end{array}$

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