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Exemplar Solution Class 11th

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?

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Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Class 11th

The accompanying Venn diagram shows three events, A, B, and C, and also the probabilities of the various intersections (for instance, P (A ∩ B) = .07). Determine

(a) P (A)

(b) P (B ∩ C )

(c) P (A ∪ B)

(d) P (A ∩ B )

(e) P (B ∩ C)

(f) Probability of exactly one of the three occurs.

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Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} F r o m t h e g i v e n V e n n d i a g r a m \\ (a) P (A) = 0.13 + 0.07 = 0.20 \\ (b) P (B \cap \bar{C}) = P (B) - P (B \cap C) \\ = 0.07 + 0.10 + 0.15 - 0.15 = 0.07 + 0.10 = 0.17 \\ (c) P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ = 0.13 + 0.07 + 0.07 + 0.10 + 0.15 - 0.07 \\ = 0.13 + 0.07 + 0.10 + 0.15 = 0.45 \\ (d) P (A \cap \bar{B}) = P (A) - P (A \cap B) \\ = 0.13 + 0.07 - 0.07 = 0.13 \\ (e) P (B \cap C) = 0.15 \\ (f) P (e x a c t l y o n e o f t h e t h r e e o c c u r s) = 0.13 + 0.10 + 0.28 = 0.51 \end{array}$

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Exemplar Solution Class 11th

One of the four persons, John, Rita, Aslam, or Gurpreet, will be promoted next month. Consequently, the sample space consists of four elementary outcomes:

S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}

You are told that the chances of John’s promotion are the same as that of Gurpreet. Rita’s chances of promotion are twice as likely as John’s. Aslam’s chances are four times that of John.

(a) Determine P (John promoted),

P (Rita promoted),

P (Aslam promoted), and

P (Gurpreet promoted).

(b) If A = {John promoted or Gurpreet promoted}, find P(A).

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Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t E_{1}, E_{2}, E_{3} a n d E_{4} b e t h e e v e n t s t h a t J o h n p r o m o t e d, R i t a p r o m o t e d, A s l a m p r o m o t e d \\ a n d G u r p r e e t p r o m o t e d r e s p e c t i v e l y . \\ ∴ S a m p l e s p a c e S = {E_{1}, E_{2}, E_{3}, E_{4}} \\ G i v e n t h a t p r o b a b i l i t y o f J o h n' s p r o m o t i o n i s s a m e a s t h a t o f G u r p r e e t \\ ∴ P (E_{1}) = P (E_{4}) \\ R i t a' s c h a n c e s o f p r o m o t i o n a r e t w i c e a s l i k e l y a s J o h n \\ ∴ P (E_{2}) = 2 P (E_{1}) \\ A n d A s l a m' s c h a n c e s o f p r o m o t i o n a r e 4 t i m e s t h a t o f J o h n \\ ∴ P (E_{3}) = 4 P (E_{1}) \\ Since, t h e s u m o f a l l p r o b a b i l i t i e s = 1 \\ ∴ P (E_{1}) + P (E_{2}) + P (E_{3}) + P (E_{4}) = 1 \\ \Rightarrow P (E_{1}) + 2 P (E_{1}) + 4 P (E_{1}) + P (E_{1}) = 1 \\ \Rightarrow 8 P (E_{1}) = 1 \\ \Rightarrow P (E_{1}) = \frac{1}{8} \\ (a) P (J o h n p r o m o t e d) = P (E_{1}) = \frac{1}{8} \\ P (R i t a p r o m o t e d) = P (E_{2}) = 2 P (E_{1}) = 2 \times \frac{1}{8} = \frac{2}{8} \\ P (A s l a m p r o m o t e d) = P (E_{3}) = 4 P (E_{1}) = 4 \times \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \\ P (G u r p r e e t p r o m o t e d) = P (E_{4}) = P (E_{1}) = \frac{1}{8} \\ (b) P (J o h n p r o m o t e d o r G u r p r e e t p r o m o t e d) = P (E_{1} \cup E_{4}) \\ \Rightarrow &th \end{array}$

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Exemplar Solution Class 11th

A team of medical students doing their internship has to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple, or very simple are respectively, 0.15, 0.20, 0.31, 0.26, 0.08. Find the probabilities that a particular surgery will be rated:

(a) complex or very complex;

(b) neither very complex nor very simple;

(c) routine or complex;

(d) routine or simple.

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Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t E_{1}, E_{2}, E_{3}, E_{4} a n d E_{5} b e t h e e v e n t s t h a t t h e s u r g e r i e s a r e r a t e d a s v e r y c o m p l e x, c o m p l e x, \\ r o u t i n e, s i m p l e a n d v e r y s i m p l e r e s p e c t i v e l y . \\ ∴ P (E_{1}) = 0.15, P (E_{2}) = 0.20, P (E_{3}) = 0.31, P (E_{4}) = 0.26 a n d P (E_{5}) = 0.08 \\ (a) P (c o m p l e x a n d v e r y c o m p l e x) = P (E_{1} o r E_{2}) \\ \Rightarrow P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2}) \\ = 0.15 + 0.20 - 0 \\ = 0.35 [? A l l e v e n t a r e i n d e p e n d e n t] \\ (b) P (n e i t h e r v e r y c o m p l e x n o r v e r y s i m p l e) = P (E_{1}^{'} \cap E_{5}^{'}) \\ = P (E_{1} \cup E_{5})' = 1 - P (E_{1} \cup E_{5}) \\ = 1 - [P (E_{1}) + P (E_{5})] = 1 - [0.15 + 0.08] \\ = 1 - 0.23 = 0.77 \\ (c) P (r o u t i n e o r c o m p l e x) = P (E_{3} o r E_{2}) \\ \Rightarrow P (E_{3} \cup E_{2}) = P (E_{3}) + P (E_{2}) \\ = 0.31 + 0.20 = 0.51 \\ (d) P (r o u t i n e o r s i m p l e) = P (E_{3} o r E_{4}) \\ \Rightarrow P (E_{3} \cup E_{4}) = P (E_{3}) + P (E_{4}) \\ = 0.31 + 0.26 = 0.57 \end{array}$

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Exemplar Solution Class 11th

If  $A$ and  $B$ are mutually exclusive events, $P (A) = 0.35$  and $P (B) = 0.45$ , find:

(a) $P (A^{'})$ 

(b) $P (B^{'})$ 

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

(d) $P (A \cap B)$ 

(e) $P (A \cap B^{'})$ 

(f) $P (A^{'} \cap B^{'})$

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t P (A) = 0.35 a n d P (B) = 0.45 \\ Since t h e e v e n t s A a n d B a r e m u t u a l l y e x c l u s i v e t h e n P (A \cap B) = 0 \\ (a) P (A') = 1 - P (A) = 1 - 0.35 = 0.65 \\ (b) P (B') = 1 - P (B) = 1 - 0.45 = 0.55 \\ (c) P (A \cup B) = P (A) + P (B) - P (A \cap B) = 0.35 + 0.45 - 0 = 0.80 \\ (d) P (A \cap B) = 0 [? A a n d B a r e m u t u a l l y e x c l u s i v e e v e n t s] \\ (e) P (A \cap B') = P (A) - P (A \cap B) = 0.35 - 0 = 0.35 \\ (f) P (A' \cap B') = 1 - P (A \cup B) = 1 - 0.80 = 0.20 \end{array}$

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Exemplar Solution Class 11th

In a large metropolitan area, the probabilities are 0.87, 0.36, 0.30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either any one or both kinds of sets?

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t E_{1} b e t h e e v e n t t h a t a f a m i l y o w n s c o l o u r t e l e v i s i o n s e t . \\ a n d E_{2} b e t h e e v e n t t h a t a f a m i l y o w n s b l a c k a n d w h i t e t e l e v i s i o n s e t . \\ G i v e n t h a t P (E_{1}) = 0.87, P (E_{2}) = 0.36 \\ a n d P (E_{1} \cap E_{2}) = 0.30 \\ ∴ T h e p r o b a b i l i t y t h a t a f a m i l y o w n s e i t h e r c o l o u r o r b l a c k a n d w h i t e t e l e v i s i o n s e t . \\ ∴ P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2}) \\ = 0.87 + 0.36 - 0.30 = 0.93 \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = 0.93 \end{array}$

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Exemplar Solution Class 11th

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find $P (G)$ , where $G$  is the event that a number greater than 3 occurs on a single roll of the die.

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Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t p r o b a b i l i t y o f e v e n n u m b e r s = \frac{1}{2} \times p r o b a b i l i t y o f o d d n u m b e r s \\ \Rightarrow P (O d d) : P (E v e n) = 2 : 1 \\ ∴ P (o d d n u m b e r) = \frac{2}{2 + 1} = \frac{2}{3} \\ a n d P (e v e n n u m b e r) = \frac{1}{2 + 1} = \frac{1}{3} \\ A l s o, g i v e n t h a t, G t h e e v e n t t h a t a n u m b e r g r e a t e r t h a n 3 o c c u r s i n a single t h r o w o f d i e . \\ ∴ T h e p o s s i b l e o u t c o m e a r e 4, 5 a n d 6 o u t o f w h i c h t w o a r e e v e n a n d o n e i s o d d . \\ ∴ R e q u i r e d p r o b a b i l i t y = P (G) \\ = 2 \times P (e v e n) \times P (o d d) \\ = 2 \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{9} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = \frac{4}{9} . \end{array}$

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Exemplar Solution Class 11th

An experiment consists of rolling a die until a 2 appears.

(i) How many elements of the sample space correspond to the event that the 2 appears on the $k^{t h}$ roll of the die?

(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the $k^{t h}$ roll of the die?

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Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} N u m b e r o f S a m p l e s p a c e = 6 \\ (i) G i v e n t h a t 2 a p p e a r s o n t h e k t h r o l l o f t h e d i e . \\ S o f i r s t (k - 1) t h r o l l h a v e 5 o u t c o m e s e a c h a n d k t h r o l l r e s u l t s 2 i . e . o n l y 1 o u t c o m e . \\ ∴ N u m b e r o f e l e m e n t o f s a m p l e s p a c e c o r r e s p o n d t o t h e e v e n t t h a t 2 a p p e a r s o n t h e k t h \\ r o l l o f t h e d i e = 5^{k - 1} . \\ (i i) I n t h i s c a s e, 2 a p p e a r s n o t l a t e r t h a n k t h r o l l o f t h e d i e, t h e n i t i s p o s s i b l e t h a t 2 c o m e s \\ i n f i r s t t h r o w t h e n o u t c o m e s w i l l b e 5 a n d 2 o u t c o m e s i n second t h r o w i . e . 1 o u t c o m e . \\ ∴ P o s s i b l e o u t c o m e = 5 \times 1 = 5 \\ S i m i l a r l y, i f 2 d o e s n o t a p p e a r i n second t h r o w a n d a p p e a r s i n t h i r d t h r o w \\ ∴ P o s s i b l e o u t c o m e = 5 \times 5 \times 1 \\ N o w w e h a v e t h e s e r i e s : \\ = 1 + 5 + 5 \times 5 + 5 \times 5 \times 5 + \dots + 5^{k - 1} \\ = 1 + 5 + 5^{2} + 5^{3} + \dots + 5^{k - 1} \\ = \frac{1 . (r^{k} - 1)}{r - 1} = \frac{5^{k - 1}}{5 - 1} \\ = \frac{5^{k - 1}}{4} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = \frac{5^{k - 1}}{4} . \end{array}$

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Exemplar Solution Class 11th

Suppose an integer from 1 through 1000 is chosen at random. Find the probability that the integer is a multiple of 2 or a multiple of 9.

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Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} W e h a v e m u l t i p l e s o f 2, f r o m 1 t o 1000 a r e 2, 4, 6, 8, \dots, 1000 \\ L e t n b e t h e n u m b e r o f t e r m s \\ H e r e a = 2, d = 2, a_{n} = 1000 \\ a_{n} = a + (n - 1) d \\ 1000 = 2 + (n - 1) 2 \\ \Rightarrow 1000 = 2 n \Rightarrow n = 500 \\ N o w m u l t i p l e s o f 9, f r o m 1 t o 1000 a r e 9, 18, 27, \dots, 999 \\ H e r e a = 9, d = 9, a_{m} = 999 [m b e t h e n u m b e r o f t e r m s] \\ a_{m} = a + (m - 1) d \\ \Rightarrow 999 = 9 + (m - 1) 9 \\ \Rightarrow 999 = 9 m \Rightarrow m = 111 \\ N o w m u l t i p l e s o f 2 a n d 9 a r e 18, 36, 54, \dots, 990 \\ H e r e a = 18, d = 18, a_{p} = 999 [p b e t h e n u m b e r o f t e r m s] \\ ∴ a_{p} = a + (p - 1) d \\ \Rightarrow 990 = 18 + (p - 1) 18 \\ \Rightarrow 990 = 18 p \Rightarrow p = 55 \\ ∴ N u m b e r o f m u l t i p l e s o f 2 o r 9 = 500 + 111 - 55 = 556 \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{P (E)}{n (E)} = \frac{556}{1000} = 0.556 \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = 0.556 \end{array}$

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Exemplar Solution Class 11th

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.]

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Payal Gupta | Contributor-Level 10

a year ago

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

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The accompanying Venn diagram shows three events, A, B, and C, and also the probabilities of the various intersections (for instance, P (A ∩ B) = .07). Determine

(a) P (A)

(b) P (B ∩ C )

(c) P (A ∪ B)

(d) P (A ∩ B )

(e) P (B ∩ C)

(f) Probability of exactly one of the three occurs.

In a large metropolitan area, the probabilities are 0.87, 0.36, 0.30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either any one or both kinds of sets?

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find $P (G)$ , where $G$  is the event that a number greater than 3 occurs on a single roll of the die.

Suppose an integer from 1 through 1000 is chosen at random. Find the probability that the integer is a multiple of 2 or a multiple of 9.