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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) $\frac{5}{9}$

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

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Exemplar Solution Class 11th

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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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Class 11th

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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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Class 11th

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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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Class 11th

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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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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Exemplar Solution Class 11th

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.

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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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Exemplar Solution Class 11th

A sample space consists of 9 elementary outcomes  $e_{1}, e_{2}, \dots, e_{9}$ whose probabilities are:

$P (e_{1}) = P (e_{2}) = 0.08$ , $P (e_{3}) = P (e_{4}) = P (e_{5}) = 0.1$ , $P (e_{6}) = P (e_{7}) = 0.2$ , $P (e_{8}) = P (e_{9}) = 0.07$ .

Suppose $A = {e_{1}, e_{5}, e_{8}}$ , $B = {e_{2}, e_{5}, e_{8}, e_{9}}$ .

(a) Calculate $P (A)$ , $P (B)$ , and $P (A \cap B)$ .

(b) Using the addition law of probability, calculate $P (A \cup B)$ .

(c) List the composition of the event $A \cup B$ , and calculate  $P (A \cup B)$ by adding the probabilities of the elementary outcomes.

(d) Calculate  $P (B^{'})$ from $P (B)$ , also calculate  $P (B^{'})$ directly from the elementary outcomes of $B$ .

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

$\begin{array}{l} G i v e n t h a t : S = {e_{1}, e_{2}, e_{3}, e_{4}, e_{5}, e_{6}, e_{7}, e_{8}, e_{9}} \\ A = {e_{1}, e_{5}, e_{5}} a n d B = {e_{2}, e_{5}, e_{5}, e_{9}} \\ P (e_{1}) = P (e_{2}) = 0.08 \\ P (e_{3}) = P (e_{4}) = P (e_{5}) = 0.1 \\ P (e_{6}) = P (e_{7}) = 0.2 \\ P (e_{8}) = P (e_{9}) = 0.07 \\ (a) P (A) = P (e_{1}) + P (e_{5}) + P (e_{8}) \\ = 0.08 + 0.1 + 0.07 = 0.25 \\ (b) P (A \cup B) = P (A) + P (B) - P (A \cap B) \dots (i) \\ B u t P (B) = P (e_{2}) + P (e_{5}) + P (e_{8}) + P (e_{9}) \\ = 0.08 + 0.1 + 0.07 + 0.07 = 0.32 \\ a n d P (A \cap B) = {e_{5}, e_{8}} = P (e_{5}) + P (e_{8}) \\ = 0.1 + 0.07 = 0.17 \\ P u t t i n g t h e v a l u e s i n e q n . (i) w e g e t \\ P (A \cup B) = 0.25 + 0.32 - 0.17 = 0.40 \\ (c) P (A \cup B) = {e_{1}, e_{2}, e_{5}, e_{8}, e_{9}} \\ ∴ P (A \cup B) = P (e_{1}) + P (e_{2}) + P (e_{5}) + P (e_{8}) + P (e_{9}) \\ = 0.08 + 0.08 + 0.1 + 0.07 + 0.07 = 0.40 \\ (d) &t \end{array}$

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Exemplar Solution Class 11th

A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card.

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

$\begin{array}{l} T o t a l n u m b e r o f c a r d s = 52 \\ F a v o u r a b l e e v e n t s = 4 k i n g s + 13 h e a r t s + 26 r e d - 13 - 2 = 28 \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{28}{52} = \frac{7}{13} . \end{array}$

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Exemplar Solution Class 11th

**If the letters of the word ASSASSINATION are arranged at random, find the probability that:

(a) Four S’s come consecutively in the word.

(b) Two I’s and two N’s come together.

(c) All A’s are not coming together.

(d)** No two A’s are coming together.

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

$\begin{array}{l} T o t a l n u m b e r o f w o r d i s A S S A S S I N A T I O N a r e 13 . \\ W h e r e, w e h a v e 3 A' s, 4 S' s, 2 I' s, 2 N' s, 1 T' s a n d 1 O' s . \\ (a) I f 4 S' s c o m e consecutively i n t h e w o r d, t h e n a r r a n g e m e n t m a y b e a s f o l l o w s : \\ \underset{1 G r o u p}{S S S S} \underset{9 o t h e r s}{\underset{?}{A A A I I N N T O}} \\ ∴ N u m b e r o f w o r d s w h e n a l l S' s a r e t o g e t h e r = \frac{10!}{3! 2! 2!} \end{array}$

$\begin{array}{l} a n d t h e t o t a l n u m b e r o f w o r d f o r m e d f r o m t h e w o r d s A S S A S S I N A T I O N = \frac{13!}{3! 4! 2! 2!} \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{\frac{10!}{3! 2! 2!}}{\frac{13!}{3! 4! 2! 2!}} = \frac{10!}{3! 2! 2!} \times \frac{3! 4! 2! 2!}{13!} \\ = \frac{10! 4!}{13!} = \frac{10! \times 4 \times 3 \times 2}{13 \times 12 \times 11 \times 10!} = \frac{2}{143} \\ (b) I f 2 I' s a n d 2 N' s c o m e t o g e t h e r t h e n t h e r e a r e 10 a l p h a b e t s \\ N u m b e r o f w o r d s w h e n 2 I' s a n d 2 N' s c o m e t o g e t h e r = \frac{10!}{3! 4!} \times \frac{4!}{2! 2!} \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{\frac{10!}{3! 4!} \times \frac{4!}{2! 2!}}{\frac{13!}{3! 4! 2! 2!}} = \frac{10!}{3! 4!} \times \frac{4!}{2! 2!} \times \frac{3! 4! 2! 2!}{13!} \\ = \frac{4! 10!}{13!} = \frac{4 \times 3 \times 2 \times 10!}{13 \times 12 \times 11 \times 10!} = \frac{2}{143} \\ (c) I f a l l A' s a r e coming t o g e t h e r, t h e n t h e r e a r e 11 a l p h a b e t s \\ N u m b e r o f w o r d s w h e n a l l A' s c o m e t o g e t h e r = \frac{11!}{4! 2! 2!} \\ ∴ P r o b a b i l i t y w h e n a l l A' s c o m e t o g e t h e r = \frac{\frac{11!}{4! 2! 2!}}{\frac{13!}{4! 3! 2! 2!}} = \frac{11!}{4! 2! 2!} \times \frac{4! 3! 2! 2!}{13!} \\ = \frac{11! 3!}{13!} = \frac{6}{13 \times 12} = \frac{1}{26} \\ ∴ R e q u i r e d p r o b a b i l i t y w h e n a l l A' s d o n o t c o m e t o g e t h e r = 1 - \frac{1}{26} = \frac{25}{26} \\ (d) I f n o t w o A' s a r e t o g e t h e r, t h e n a r r a n g i n g t h e a l p h a b e t s e x c e p t A' s \\ - S - S - S - S - I - N - T - I - O - N - \\ N u m b e r o f w a y s o f a r r a n g i n g a l l a l p h a b e t s e x c e p t A' s = \frac{10!}{4! 2! 2!} \\ T h e r e a r e 11 v a c a n t p l a c e s b e t w e e n t h e s e a l p h a b e t s . \\ ∴ 3 A' s c a n b e p l a c e d i n 11 p l a c e s i n {}_{11}{C_{3}} w a y s = \frac{11!}{3! 8!} \\ ∴ T o t a l n u m b e r o f w o r d s w h e n n o t w o A' s c o m e t o g e t h e r = \frac{11!}{3! 8!} \times \frac{10!}{4! 2! 2!} \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{11!}{3! 8!} \times \frac{10!}{4! 2! 2!} \times \frac{4! 3! 2! 2!}{13!} \\ = \frac{10!}{8! \times 13 \times 12} = \frac{10 \times 9 \times 8!}{8! \times 13 \times 12} = \frac{10 \times 9}{13 \times 12} = \frac{15}{26} \end{array}$

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Exemplar Solution Class 11th

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:

(a) All the three balls are white.

(b) All the three balls are red.

(c) One ball is red and two balls are white.

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

$\begin{array}{l} G i v e n t h a t : N u m b e r o f r e d b a l l s = 8 \\ N u m b e r o f w h i t e b a l l s = 5 \\ (a) P (a l l t h e t h r e e b a l l s a r e w h i t e) \\ = \frac{{}_{5}{C_{3}}}{{}_{13}{C_{3}}} = \frac{\frac{5!}{3! 2!}}{\frac{13!}{3! 10!}} = \frac{5!}{3! 2!} \times \frac{3! 10!}{13!} \\ = \frac{5!}{2!} \times \frac{10!}{13 \times 12 \times 11 \times 10!} \\ = \frac{5 \times 4 \times 3 \times 2!}{2!} \times \frac{1}{13 \times 12 \times 11} = \frac{5 \times 4 \times 3}{13 \times 12 \times 11} = \frac{5}{143} \\ (b) P (a l l t h e t h r e e b a l l s a r e r e d) \\ = \frac{{}_{8}{C_{3}}}{{}_{13}{C_{3}}} = \frac{\frac{8!}{3! 5!}}{\frac{13!}{3! 10!}} = \frac{8!}{3! 5!} \times \frac{3! 10!}{13!} \\ = \frac{8 \times 7 \times 6 \times 5!}{5!} \times \frac{10!}{13 \times 12 \times 11 \times 10!} \\ = \frac{8 \times 7 \times 6}{13 \times 12 \times 11} = \frac{28}{143} \\ (c) P (o n e b a l l i s r e d a n d t w o b a l l s a r e w h i t e) \\ = \frac{{}_{8}{C_{1}} \times {}_{5}{C_{2}}}{{}_{13}{C_{3}}} = \frac{\frac{8!}{7! 1!} \times \frac{5!}{2! 3!}}{\frac{13!}{3! 10!}} = \frac{8 \times 7!}{7!} \times \frac{5 \times 4 \times 3 \times 2!}{2! \times 3!} \times \frac{3! \times 10!}{13 \times 12 \times 11 \times 10!} \\ = \frac{8 \times 5}{13 \times 11} = \frac{40}{143} \end{array}$

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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

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

A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card.

**If the letters of the word ASSASSINATION are arranged at random, find the probability that:

(a) Four S’s come consecutively in the word.

(b) Two I’s and two N’s come together.

(c) All A’s are not coming together.

(d)** No two A’s are coming together.

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:

(a) All the three balls are white.

(b) All the three balls are red.

(c) One ball is red and two balls are white.