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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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.

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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?

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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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9 months ago

NIT Jalandhar: Placements, Admission 2024, Cut off, Courses, Fees, Ranking

What are the pros and cons of Jalandhar NIT?

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Nishtha Singh

Contributor-Level 10

The pros of Jalandhar NIT is its name itself. To compete with all other NITs, the college takes great efforts. Being a part of the prestigious NIT group, Jalandhar NIT is visited by top companies for placements. The cons is again the name itself, to maitain its great reputation, the college has set some rules. There are several rules that students have to follow such as mandatory attendence and others. Some students find these rules as quite strict.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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 14 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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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).

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Maths

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

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

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9 months ago

Engineering in India JEE

Which is the best engineering college that I can get with 6664 JEE Rank?

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9 months ago

KAU - Kerala Agricultural University Business Management Studies

How can I get admission get admission to MBA at Kerala Agricultural University?

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Rachit Kumar

Contributor-Level 10

The KAU MBA admission process can be divided into six broad steps. Interested students can follow the below-mentioned steps to get admission to the MBA course:

1. Eligibility check: Any candidate who has passed a bachelor's degree with a minimum of 50% aggregate is eligible to apply.

2. Application process: Those who are eligible must fill out the application form available on the official website with all the requisite documents and scores in the accepted entrance exam.

3. Merit list check: After the application process, students must stay updated with the merit list. Shortlisted students move on to the next stage.

4. GD & PI:

...more

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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?

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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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9 months ago

National Institute of Electronics and Information Technology JEE

Can I get any branch at NIELIT, Gorakhpur with JEE Rank 50000?

0 Follower 7 Views

Tanisha Jain

Contributor-Level 7

The NIELIT Gorakhpur cutoff 2025 for CSE (AI and ML) was 62927 for general, 69777 for BTech in ECE (Minor in Wearable Electronics). In both course, the cutoff rank is above 50000 as per the NIELIT Gorakhpur JEE Main cutoff 2025 for the Round 3 seat allotment results.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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.

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

KAU - Kerala Agricultural University Business Management Studies

What specialisations are offered for MBA programme at KAU?

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Piyush Jain

Contributor-Level 10

Kerala Agricultural University only offers one specialisation in the MBA course, i.e., Agri Business Management. The MBA (Agribusiness Management) programme of KAU is crafted to refashion the DNA of agribusiness in India by meeting the human capital needs of the fast-growing food and agribusiness sector. In this programme, Industrial and farm visits are undertaken to expose the students to real-life situations to unlearn and relearn management lessons.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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?

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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.

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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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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.]

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Payal Gupta

Contributor-Level 10

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

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9 months ago

Business Management Studies KMAT

Is KMAT compulsory for MBA at KAU?

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Vishakha

Contributor-Level 10

No, KMAT scores are not compulsory to get admission into the MBA course at KAU, because the university also considers relevant scores from other entrance exams, such as CMAT/ CAT. Therefore, candidates with relevant scores in KMAT can get admission to MBA courses. However, it is not compulsory. Candidates with relevant scores in any of the entrance exams- CMAT/ CAT/ KMAT can get admission to the MBA course.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 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?

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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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9 months ago

National Institute of Electronics and Information Technology, Aurangabad JEE Main

Which branch at NIELIT Aurangabad has the highest cutoff 2025?

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Nitesh Lama

Contributor-Level 7

According to NIELIT Aurangabad JEE Main cutoff 2025, the General AI category cutoff stood between 46118 and 57153. Only CSE and EE cutoffs were published by the institute. Between these two, the highest cutoff possess by BTech in CSE at a Round 3 closing rank of 46118. Likewise, for the OBC AI category, CSE remained the most competitive course at a closing rank of 18150.

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