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New answer posted

a year ago

Maths Exemplar Solution

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 $B$ .

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Maths Exemplar Solution

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

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Maths Exemplar Solution

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.

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

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!} * \frac{3! 4! 2! 2!}{13!} \\ = \frac{10! 4!}{13!} = \frac{10! * 4 * 3 * 2}{13 * 12 * 11 * 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!} * \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!} * \frac{4!}{2! 2!}}{\frac{13!}{3! 4! 2! 2!}} = \frac{10!}{3! 4!} * \frac{4!}{2! 2!} * \frac{3! 4! 2! 2!}{13!} \\ = \frac{4! 10!}{13!} = \frac{4 * 3 * 2 * 10!}{13 * 12 * 11 * 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!} * \frac{4! 3! 2! 2!}{13!} \\ = \frac{11! 3!}{13!} = \frac{6}{13 * 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!} * \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!} * \frac{10!}{4! 2! 2!} * \frac{4! 3! 2! 2!}{13!} \\ = \frac{10!}{8! * 13 * 12} = \frac{10 * 9 * 8!}{8! * 13 * 12} = \frac{10 * 9}{13 * 12} = \frac{15}{26} \end{array}$

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New answer posted

a year ago

Maths Exemplar Solution

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.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Exemplar Solution

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?

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Maths Exemplar Solution

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 )

(d) P (A ∩ B )

(e) P (B ∩ C)

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

0 Follower 17 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Maths Exemplar Solution

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).

0 Follower 26 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Class 11th

Four candidates $A, B, C, D$ have applied for the assignment to coach a school cricket team. If $A$ is twice as likely to be selected as $B$ , and $B$ and $C$ are given about the same chance of being selected, while $C$ is twice as likely to be selected as $D$ , what are the probabilities that:

(a) C will be selected?

(b) A will not be selected?

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t A i s t w i c e a s l i k e l y t o b e s e l e c t e d a s B \\ i . e ., P (A) = 2 P (B) \\ a n d C i s t w i c e a s l i k e l y t o b e s e l e c t e d a s D \\ ∴ P (C) = 2 P (D) \Rightarrow P (B) = 2 P (D) \\ \Rightarrow \frac{P (A)}{2} = 2 P (D) \Rightarrow P (D) = \frac{1}{4} P (A) \\ N o w B a n d C a r e g i v e n a b o u t t h e s a m e c h a n c e \\ ∴ P (B) = P (C) \\ Since, s u m o f a l l p r o b a b i l i t i e s = 1 \\ ∴ P (A) + P (B) + P (C) + P (D) = 1 \\ \Rightarrow P (A) + \frac{P (A)}{2} + \frac{P (A)}{2} + \frac{P (A)}{4} = 1 \\ \Rightarrow 4 P (A) + 2 P (A) + 2 P (A) + P (A) = 4 \\ \Rightarrow 9 P (A) = 4 \Rightarrow P (A) = \frac{4}{9} \\ (a) P (C w i l l b e s e l e c t e d) = P (C) = P (B) = \frac{P (A)}{2} \\ = \frac{4}{9} * \frac{1}{2} = \frac{2}{9} \\ (b) P (A w i l l n o t b e s e l e c t e d) = P (A') = 1 - P (A) \\ &thi \end{array}$

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New answer posted

a year ago

Maths Exemplar Solution

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.

0 Follower 10 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Maths Exemplar Solution

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

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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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