What do you mean by Exhaustive Events? -Maths 9th

What do you mean by Exhaustive Events? -Maths 9th
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The total number of possible outcomes of a random experiment is called exhaustive events. Ex. (i) In tossing a coin, exhaustive events are 2 (Head or tail) (ii) In throwing a die, the exhaustive number of cases is 6, since any of the faces marked with 1, 2, 3, 4, 5 or 6 may come uppermost. (iii) In drawing 4 cards from a well shuffled pack of 52 cards, the exhaustive number of cases is 52 C4, since 4 cards can be drawn out of 52 cards in 52 C4 ways.
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A and B are two mutually exclusive and exhaustive events with P(B) = 3P(A). What is the value of P (bar(B)) ? -Maths 9th

Description : A and B are two mutually exclusive and exhaustive events with P(B) = 3P(A). What is the value of P (bar(B)) ? -Maths 9th

Last Answer : (c) 43 : 34Given, odds against A = 8 : 3⇒ P(not A) = \(rac{8}{8+3}\) = \(rac{8}{11}\) ⇒ P(A happens) = \(rac{3}{11}\)Odds against B = 5 : 2⇒ P(not B) = \(rac{5}{5+2}\) = \(rac{5}{7}\) ⇒ P(B happens ... {43}{77}\)odds against C = \(rac{P(not\,c)}{P(c)}\) = \(rac{rac{43}{77}}{rac{34}{77}}\) = 43 : 34.

What do you mean by Mutually Exclusive Events? -Maths 9th

Description : What do you mean by Mutually Exclusive Events? -Maths 9th

Last Answer : The events are said to be mutually exclusive if they cannot occur simultaneously in a single draw. Events such as tossing a head or a tail with a coin, drawing a queen or a jack from a pack of cards, ... the card can be the king of spades. Similarly for B and C, the card can be king of hearts.

A coin is tossed thrice and all eight outcomes are assumed equally likely. Find whether the events E -Maths 9th

Description : A coin is tossed thrice and all eight outcomes are assumed equally likely. Find whether the events E -Maths 9th

Last Answer : When a coin is tossed three times, the sample space is given by S = [HHH, HHT, HTH, THT, THH, HTT, TTH, TTT] E = {HHH, HTT, THT, TTH}, F = {TTT, HTH, THH, HHT}E ∩ F = ϕP(E) = \(rac{4}{8}\) = \(rac{1}{2}\ ... rac{1}{2}\) x \(rac{1}{2}\) x \(rac{1}{4}\) ≠ P(E ∩ F) ∴ E and F are not independent events.

Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Description : Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Last Answer : ThenP (A) = P (B) = P (C) = \(rac{2}{4}\) = \(rac{1}{2}\) andP (A ∩ B) = P ({HH}) = \(rac{1}{4}\), P (A ∩ C) = P ({HT}) = \(rac{1}{4}\)P ( ... C)Thus condition (i) is satisfied, i.e., the events are pairwise independent. But condition (ii) is not satisfied and so the three events are not independent

The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

Description : The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

Last Answer : p1 = P(A) = \(rac{3}{7}\), p2 = P(B) = \(rac{5}{7}\) ∴ q1 = P(\(\bar{A}\)) = 1 - P(A) = 1 - \(rac{3}{7}\) = \(rac{4}{7}\). q2 = P(\(\bar{B}\)) = 1 - P(B) = 1 - \(rac{5}{7}\) = \(rac{2}{7}\) ... passes) = p1 q2 + q1 p2 = \(rac{3}{7}\) x \(rac{2}{7}\) + \(rac{4}{7}\) x \(rac{5}{7}\) = \(rac{26}{49}.\)

Define : Algebra of Events. -Maths 9th

Description : Define : Algebra of Events. -Maths 9th

Last Answer : Let A, B and C be any two events associated with a random experiment whose sample space is S. Then, (i) A ∪ B. (Union of A and B) is the event that occurs if A occurs or B occurs or both A and B occur ... a mutually exclusive and exhaustive set of events. ∴ A∩B = ϕ, B∩C = ϕ, A ∩C = ϕ and A∪B∪C = S.

There are three events E1, E2 and E3, one of which is must and only one can happen. -Maths 9th

Description : There are three events E1, E2 and E3, one of which is must and only one can happen. -Maths 9th

Last Answer : Since one and only one of the three events E1, E2 and E3 can happen, i.e, they are mutually exclusive. Therefore, P(E1) + P(E2) + P(E3) = 1 ...(i) Odds against E1 are 7 : 4 ⇒ Odds in favour of E1 are 4 : 7⇒ ... \(rac{1-rac{23}{88}}{rac{23}{88}}\) = \(rac{rac{65}{88}}{rac{23}{88}}\) = 65 : 23.

If the three independent events E1, E2 and E3, the probability that only E1 occurs is α, E2 occurs -Maths 9th

Description : If the three independent events E1, E2 and E3, the probability that only E1 occurs is α, E2 occurs -Maths 9th

Last Answer : (c) 6Let \(x\), y, z be the probabilities of happening of events E1, E2 and E3 respectively. Then, \(\alpha\) = P (occurrence of E1 only) = x (1 - y) (1 - z) \(\beta\) = P (occurrence of E2 only) = (1 - \(x\)) y (1 - ... (x\) = 6z (From (i) and (ii))⇒ \(rac{x}{z}\) = 6 ⇒ \(rac{P(E_1)}{P(E_3)}\) = 6.

There are three events A, B and C, one of which must and can only happen. If the odds one 8 : 3 against A, 5 : 2 against -Maths 9th

Description : There are three events A, B and C, one of which must and can only happen. If the odds one 8 : 3 against A, 5 : 2 against -Maths 9th

Last Answer : (c) \(rac{3}{8}\)Total number of ways of checking in the 4 hotels by 3 men = 4 4 4 = 43. Number of ways in which each man checks into a different hotel = 4 3 2 (As for the 1st person, ... P(Each person checks into a different hotel)= \(rac{4 imes3 imes2}{4 imes4 imes4}\) = \(rac{3}{8}\).

The nitrogen atom in the following cyclic compounds can be removed as trimethylamine by successive Hoffmann eliminations (involving exhaustive methyla

Description : The nitrogen atom in the following cyclic compounds can be removed as trimethylamine by successive Hoffmann eliminations (involving exhaustive methyla

Last Answer : The nitrogen atom in the following cyclic compounds can be removed as trimethylamine by successive Hoffmann eliminations ... is : A. B. C. D.

The more nutrient exhaustive family is

Description : The more nutrient exhaustive family is

Last Answer : Ans.Poaceae (Graminae)

The crops leave the field exhaustive after growing?

Description : The crops leave the field exhaustive after growing?

Last Answer : Ans.Exhaustive Crops

The Preamble is useful in constitutional interpretation because it - (1) uses value loaded words (2) contains the real objective and philosophy of the constitution makers (3) is a source of power and limitation (4) gives and exhaustive list of basic features of the Constitution

Description : The Preamble is useful in constitutional interpretation because it - (1) uses value loaded words (2) contains the real objective and philosophy of the constitution makers (3) is a source of power and limitation (4) gives and exhaustive list of basic features of the Constitution

Last Answer : (2) contains the real objective and philosophy of the constitution makers Explanation: The preamble is useful in constitutional interpretation because it contains the real objective and philosophy of the constitution makers.

Which one of the following is the best statement about the Indian Contract Act? (a) It is an exhaustive code containing the entire law of contract. (b) It is an Act to amend certain parts of the law relating to contracts (c) It is an Act to define certain parts of the law relating to contracts and contains only the general principles of contract. (d) It is not an exhaustive code containing the entire law of contracts being an Act to define and amend certain parts of law relating to contract.

Description : Which one of the following is the best statement about the Indian Contract Act? (a) It is an exhaustive code containing the entire law of contract. (b) It is an Act to amend certain parts of ... the entire law of contracts being an Act to define and amend certain parts of law relating to contract.

Last Answer : (c) It is an Act to define certain parts of the law relating to contracts and contains only the general principles of contract.

Exhaustive testing is a) always possible b) practically possible c) impractical but possible d) impractical and impossible

Description : Exhaustive testing is a) always possible b) practically possible c) impractical but possible d) impractical and impossible

Last Answer : Ans :c

find the mean of first five multiples of 10 -Maths 9th

Description : find the mean of first five multiples of 10 -Maths 9th

Last Answer : This is the correct one...

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Description : The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Last Answer : Here's ur answer..

2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Description : Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Last Answer : Since mean of 20 observations is 17 Sum of the 20 observations = 17 x 20 = 340 New sum of 20 observations = 340 – 40 + 12 = 312 New mean=312 / 20 =15.6

Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Description : Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Last Answer : Mean of 36 observations = 12 Total of 36 observations = 36 x 12 = 432 Correct sum of 36 observations = 432 – 74 + 47 = 405 Correct mean of 36 observations = 405/ 36 =11.25

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Description : If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 11 5x + 20 = 55 5x = 35 ⇒ x = 7

Determine the mean of first 10 natural numbers. -Maths 9th

Description : Determine the mean of first 10 natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 / 10 = 55 / 10 = 5.5

Find the mean of x, x + 2, x + 4, x + 6, x + 8. -Maths 9th

Description : Find the mean of x, x + 2, x + 4, x + 6, x + 8. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 5x + 20 / 5 = x + 4

The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Description : The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Last Answer : Let the 8 observations are x1, x2, x3, x4, x5, x6, x7, x8 ∴ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = 40 × 8 = 320 New mean = 320 + 5 × 8 = 360 / 8 = 45

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

Find the mean of the following distribution : -Maths 9th

Description : Find the mean of the following distribution : -Maths 9th

Last Answer : Now, mean =(x̅) = Σfx / Σf = 1900 / 100 = 19

The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Description : The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Last Answer : x̅ = 1 / n (Σxi) ⇒ 55 = x1 + x2 + .......+ ⇒ x7 / 7 ⇒ x1 + x2 + ...... + x7 = 55 × 7 = 385 x1 + x2 + ...... + x6 = 52 + 54 + 55 + 53 + 56 + 54 = 324 ∴ x7 = 385 - 324 = 61 kg ∴ weight of the seventh student is 61 kg.

Obtain the mean of the following distribution and also find the mode. -Maths 9th

Description : Obtain the mean of the following distribution and also find the mode. -Maths 9th

Last Answer : fixi/fi 270/55 Mean=4.9

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Description : The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Last Answer : Here's ur answer..

find the mean of first five multiples of 10 -Maths 9th

Description : find the mean of first five multiples of 10 -Maths 9th

Last Answer : This is the correct one...

2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Description : Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Last Answer : Since mean of 20 observations is 17 Sum of the 20 observations = 17 x 20 = 340 New sum of 20 observations = 340 – 40 + 12 = 312 New mean=312 / 20 =15.6

Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Description : Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Last Answer : Mean of 36 observations = 12 Total of 36 observations = 36 x 12 = 432 Correct sum of 36 observations = 432 – 74 + 47 = 405 Correct mean of 36 observations = 405/ 36 =11.25

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Description : If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 11 5x + 20 = 55 5x = 35 ⇒ x = 7

Determine the mean of first 10 natural numbers. -Maths 9th

Description : Determine the mean of first 10 natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 / 10 = 55 / 10 = 5.5

Find the mean of x, x + 2, x + 4, x + 6, x + 8. -Maths 9th

Description : Find the mean of x, x + 2, x + 4, x + 6, x + 8. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 5x + 20 / 5 = x + 4

The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Description : The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Last Answer : Let the 8 observations are x1, x2, x3, x4, x5, x6, x7, x8 ∴ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = 40 × 8 = 320 New mean = 320 + 5 × 8 = 360 / 8 = 45

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

Find the mean of the following distribution : -Maths 9th

Description : Find the mean of the following distribution : -Maths 9th

Last Answer : Now, mean =(x̅) = Σfx / Σf = 1900 / 100 = 19

The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Description : The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Last Answer : x̅ = 1 / n (Σxi) ⇒ 55 = x1 + x2 + .......+ ⇒ x7 / 7 ⇒ x1 + x2 + ...... + x7 = 55 × 7 = 385 x1 + x2 + ...... + x6 = 52 + 54 + 55 + 53 + 56 + 54 = 324 ∴ x7 = 385 - 324 = 61 kg ∴ weight of the seventh student is 61 kg.

Obtain the mean of the following distribution and also find the mode. -Maths 9th

Description : Obtain the mean of the following distribution and also find the mode. -Maths 9th

Last Answer : fixi/fi 270/55 Mean=4.9

The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Description : The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Last Answer : NEED ANSWER

If the mean of the observations x, x + 3, x + 5, x + 7and x + 10 is 9, then mean of the last three observations is -Maths 9th

Description : If the mean of the observations x, x + 3, x + 5, x + 7and x + 10 is 9, then mean of the last three observations is -Maths 9th

Last Answer : NEED ANSWER

If each observation of the data is increased by 5, then their mean -Maths 9th

Description : If each observation of the data is increased by 5, then their mean -Maths 9th

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Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Description : Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

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The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be -Maths 9th

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Last Answer : NEED ANSWER

There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Last Answer : NEED ANSWER

The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

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What is the mean of 12,34,15,3,16? -Maths 9th

Description : What is the mean of 12,34,15,3,16? -Maths 9th

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The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Description : The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

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