Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen: Overview, Questions, Preparation

Updated on Sep 8, 2025 14:16 IST

Table of content

Probability Short Answer Type Questions
Probability Long Answer Type Questions
Probability Objective Type Questions
Probability True or False Type Questions
Probability Fill in the blank Type Questions
Probability Matching Type Questions

Probability Short Answer Type Questions

1. If the letters of the word ALGORITHM are arranged at random in a row, what is the probability that the letters GOR must remain together as a unit?

Sol.

\begin{array}{l} W o r d A L G O R I T H M h a s 9 l e t t e r s . \\ I f G O R r e m a i n t o g e t h e r, t h e n i t w i l l r e m a i n t o g e t h e r . \\ ∴ N u m b e r o f l e t t e r s = A L G O R I T H M = 6 + 1 = 7 \\ N u m b e r o f w o r d s = 7! \\ a n d t h e t o t a l n u m b e r o f w o r d s f r o m A L G O R I T H M = 9! \\ 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{7!}{9!} = \frac{7!}{9.8.7!} = \frac{1}{72} \\ 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 = \frac{1}{72} . \end{array}

2. Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks?

[Hint: First find the probability that the couple has adjacent desks, and then subtract it from 1.]

Sol.

\begin{array}{l} W o r d A L G O R I T H M h a s 9 l e t t e r s . \\ N u m b e r o f d e s k o c c u p i e d b y o n e c o u p l e = 1 \\ O n l y (4 + 1) = 5 p e r s o n s t o b e a s s i g n e d . \\ ∴ N u m b e r o f w a y s o f a s s i g n i n g t h e s e 5 p e r s o n s = 5! \times 2! \\ T o t a l n u m b e r o f w a y s o f a s s i g n i n g t h e s e 6 p e r s o n s = 6! \\ ∴ Probability t h a t a c o u p l e h a s a d j a c e n t d e s k = \frac{5! \times 2!}{6!} = \frac{1}{3} \\ S o, t h e p r o b a b i l i t y t h a t t h e m a r r i e d c o u p l e w i l l h a v e n o - a d j a c e n t d e s k s = 1 - \frac{1}{3} = \frac{2}{3} \\ 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 = \frac{2}{3} . \end{array}

Probability Long Answer Type Questions

1. One urn contains two black balls (labeled $B_{1}$ and $B_{1}$ ) and one white ball. A second urn contains one black ball and two white balls (labeled $W_{1}$ and $W_{2}$ ). Suppose the following experiment is performed: one of the two urns is chosen at random. Next, a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.

(a) Write the sample space showing all possible outcomes.

(b) What is the probability that two black balls are chosen?

(c) What is the probability that two balls of opposite color are chosen?

Sol.

\begin{array}{l} G i v e n t h a t o n e o f t h e u r n s i s c h o o s e n, t h e n a b a l l i s r a n d o m l y c h o o s e n f r o m t h e u r n, t h e n \\ a second b a l l i s c h o o s e n a t r a n d o m f r o m t h e s a m e u r n w i t h o u t r e p l a c i n g t h e f i r s t b a l l \\ (a) S a m p l e s p a c e S = {B_{1} B_{2}, B_{1} W, B_{2} B_{1}, B_{2} W, W B_{1}, W B_{2}, B W_{1}, B W_{2}, W_{1} B, W_{1} W_{2}, W_{2} B, W_{2} W_{1}} \\ T o t a l n u m b e r o f S a m p l e s p a c e, S = 12 \\ (b) I f t w o b l a c k b a l l s a r e c h o o s e n \\ t h e n t h e f a v o u r a b l e e v e n t s a r e B_{1} B_{2}, B_{2} B_{1} i . e . 2 \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{2}{12} = \frac{1}{6} \\ (c) I f t w o b a l l s o f o p p o s i t e c o l o u r s a r e c h o o s e n t h e n, t h e \\ r e q u i r e d p r o b a b i l i t y = \frac{8}{12} = \frac{2}{3} \end{array}

2. 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.

Sol.

\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}

Probability Objective Type Questions

1. 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

Sol.

\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}

2. 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

Sol.

\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}

Probability True or False Type Questions

1. The probability that a person visiting a zoo will see the giraffe is 0.72, the probability that he will see the bears is 0.84, and the probability that he will see both is 0.52.

Sol.

\begin{array}{l} G i v e n t h a t : \\ P (t o s e e g i r a f f e e) = 0.72 \\ P (t o s e e b e a r) = 0.84 \\ P (t o s e e b o t h g i r a f f e e a n d b e a r) = 0.52 \\ ∴ P (t o s e e g i r a f f e e o r b e a r) = P (t o s e e g i r a f f e e) + P (t o s e e b e a r) - P (t o s e e b o t h) \\ = 0.72 + 0.84 - 0.52 \\ = 1.04 w h i c h i s n o t p o s s i b l e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}

2. The probability that a student will pass his examination is 0.73, the probability of the student getting a compartment is 0.13, and the probability that the student will either pass or get a compartment is 0.96.

Sol.

\begin{array}{l} L e t E b e t h e e v e n t t h a t t h e s t u d e n t w i l l p a s s \\ a n d F b e t h e e v e n t t h a t h e w i l l g e t c o m p a r t m e n t \\ ∴ P (E) = 0.73, P (F) = 0.13 a n d P (E \cup F) = 0.96 \\ ∴ P (E \cup F) = P (E) + P (F) - P (E \cap F) \\ = 0.73 + 0.13 - 0 [∵ P (E \cap F) = 0] \\ = 0.86 \\ B u t P (E \cup F) = 0.96 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}

Probability Fill in the blank Type Questions

1. The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is ___.

Sol.

\begin{array}{l} P (loosing t h e g a m e) = 1 - (0.77 + 0.08) \\ = 1 - 0.85 = 0.15 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 0.15 \end{array}

2. If e₁, e₂, e₃, e₄ are the four elementary outcomes in a sample space and P(e₁) = 0.1, P(e₂) = 0.5, P(e₃) = 0.1, then the probability of e₄ is __.

Sol.

\begin{array}{l} W e k n o w t h a t t h e s u m o f a l l p r o b a b i l i t i e s = 1 \\ ∴ P (e_{1}) + P (e_{2}) + P (e_{3}) + P (e_{4}) = 1 \\ \Rightarrow 0.1 + 0.5 + 0.1 + P (e_{4}) = 1 \\ \Rightarrow 0.7 + P (e_{4}) = 1 \\ \Rightarrow P (e_{4}) = 1 - 0.7 = 0.3 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 0.3 \end{array}

Probability Matching Type Questions

1. Match the proposed probability under Column C₁ with the appropriate written description under column C₂ :

C₁ C₂

Probability Written Description

(a) 0.95 (i) An incorrect assignment

(b) 0.02 (ii) No chance of happening

(c) – 0.3 (iii) As much chance of happening as not.

(d) 0.5 (iv) Very likely to happen

(e) 0 (v) Very little chance of happening

Sol.

\begin{array}{l} (a) 0.95 = V e r y l i k e l y t o h a p p e n, s o i t i s c l o s e t o 1 . \\ (b) 0.02 = V e r y l i t t l e c h a n c e o f h a p p e n i n g a s t h e p r o b a b i l i t y i s v e r y l o w . \\ (c) - 0.3 = A n i n c o r r e c t a s s i g n m e n t b e c a u s e p r o b a b i l i t y i s n e v e r n e g a t i v e . \\ (d) 0.5 = A s m u c h c h a n c e o f h a p p e n i n g a s n o t b e c a u s e s u m o f c h a n c e s o f h a p p e n i n g \\ a n d n o t h a p p e n i n g i s o n e . \\ (e) 0 = N o c h a n c e o f h a p p e n i n g . \\ H e n c e, (a) \leftrightarrow (i v), (b) \leftrightarrow (v), (c) \leftrightarrow (i), (d) \leftrightarrow (i i i), (e) \leftrightarrow (i i) \end{array}

2. Match the following

(a) If E₁ and E₂ are the two mutually (i) E₁ ∩ E₂ = E₁ exclusive events

(b) If E₁ and E₂ are the mutually (ii) (E₁ – E₂) ∪ (E₁ ∩ E₂) = E₁

exclusive and exhaustive events

(c) If E₁ and E₂ have common (iii) E₁ ∩ E₂ = φ, E₁ ∪ E₂ = S

outcomes, then

(d) If E₁ and E₂ are two events (iv) E₁ ∩ E₂ = φ

such that E₁ ⊂ E₂

Sol.

\begin{array}{l} (a) I f E_{1} a n d E_{2} a r e m u t u a l l y e x c l u s i v e e v e n t s, t h e n E_{1} \cap E_{2} = ϕ . \\ (b) I f E_{1} a n d E_{2} a r e m u t u a l l y e x c l u s i v e a n d e x h a u s t i v e e v e n t s, t h e n \\ E_{1} \cap E_{2} = ϕ a n d E_{1} \cup E_{2} = S . \\ (c) I f E_{1} a n d E_{2} h a v e c o m m o n o u t c o m e s, t h e n (E_{1} - E_{2}) \cup (E_{1} \cap E_{2}) = E_{1} \\ (d) I f E_{1} a n d E_{2} a r e t w o e v e n t s s u c h t h a t E_{1} \subset E_{2} \Rightarrow E_{1} \cap E_{2} = E_{1} \\ H e n c e, (a) \leftrightarrow (i v), (b) \leftrightarrow (i i i), (c) \leftrightarrow (i i), (d) \leftrightarrow (i) \end{array}

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exam

Student Forum

Anything you would want to ask experts?

Write here...

Other Similar chapters for you

Probability

News & Updates

Latest NewsPopular News

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen: Overview, Questions, Preparation

Probability Short Answer Type Questions

Probability Long Answer Type Questions

Probability Objective Type Questions

Probability True or False Type Questions

Probability Fill in the blank Type Questions

Probability Matching Type Questions

Student Forum

Other Similar chapters for you

News & Updates

UGC NET 2025 December Notification Shortly LIVE Updates - NTA Application Form Link, Eligibility & Age Limit

GATE Registration 2026 Last Date Today With Regular Fees Live: Steps, Eligibility, Fees, and Exam Updates

Kerala SSLC, Plus Two Time Table 2026 Soon! Live updates on Kerala 10th, 12th Time Table 2026

JKBOSE Class 10 Date Sheet 2025 for October-November Session OUT @jkbose.nic.in Live Updates; Check Annual Exam Dates

JEE Main Marks vs Percentile vs Rank 2025: Calculate Percentile and Rank Using Marks

List of NIT Colleges in India 2025 - NIRF Ranking, Courses, Seats, Cutoff, Fees and Placement

CBSE 10th Maths Question Paper 2025, 2024, 2023, 2022, 2020, 2019 PDF Download

NEET Chapter Wise Weightage 2025 PDF by NTA: Physics, Chemistry & Biology

List of Top IITs in India 2025: NIRF Ranking, Courses, Seats, Fees & Placement