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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 25 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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New answer posted

a year ago

Maths Exemplar Solution

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?

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Exemplar Solution

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.

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Exemplar Solution

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?

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Exemplar Solution

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.

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Maths Exemplar Solution

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.]

0 Follower 12 Views

Payal Gupta

Contributor-Level 10

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

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New answer posted

a year ago

Maths Exemplar Solution

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?

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

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

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