A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters is -Maths 9th

A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters is -Maths 9th
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1 Answer

Answer :

(c) 0.8645Required probability = P(X not defective and Y not defective) = P(\(\bar{X}\)) x P(\(\bar{Y}\))= (1 – P(X)) (1 – P(Y))= \(\bigg(1-rac{9}{100}\bigg)\)\(\bigg(1-rac{5}{100}\bigg)\)= \(rac{91}{100}\) x \(rac{95}{100}\) = \(rac{8645}{10000}\) = 0.8645.
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Related questions

The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Description : The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Last Answer : There are 7 letters in the word SOCIETY. ∴ Total number of ways of arranging all the 7 letters = n(S) = 7!. When the case of three vowels being together is taken, then the three vowels are considered as one unit, so the ... = 5! 3! ∴ Required probability = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’the two I‘s do not come together is -Maths 9th

Description : The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’the two I‘s do not come together is -Maths 9th

Last Answer : (b) \(rac{4}{5}\)Let S be the sample space. Then, n(S) = Total number of waysin which the letters of the word UNIVERSITY' can be arranged = \(rac{10!}{2!}\) (∵ There are 2I s) ... ! imes36}{rac{10!}{2!}}\) = \(rac{ ot8! imes36 imes2!}{10 imes9 imes ot8!}\) = \(rac{4}{5}\).

The letters ofthe word ‘NATIONAL’are arranged atrandom. What is the probability that the last letter will be T ? -Maths 9th

Description : The letters ofthe word ‘NATIONAL’are arranged atrandom. What is the probability that the last letter will be T ? -Maths 9th

Last Answer : (c) \(rac{1}{8}\)Let S be the sample space.Then n(S) = Total number of ways in which the letters of word NATIONAL can be arranged= \(rac{8!}{2!\,2!}\) (∵There are 2A s and 2N's in 8 letters)Let E : Event of ... !}{2!\,2!}}\) = \(rac{7!}{8!}\) = \(rac{7!}{8 imes7!}\) = \(rac{1}{8}\).

If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9 is -Maths 9th

Description : If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9 is -Maths 9th

Last Answer : (d) 226 × 52C26 | 104C26Since there are 52 distinct cards in a deck and each distinct card is 2 in number.∴2 decks will also contain only 52 distinct cards, two each.∴ Probability that the player gets all distinct cards = \(rac{^{52}C_{26} imes2^{26}}{^{104}C_{26}}\).

A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Description : A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Last Answer : (d) \(rac{31}{66}\)Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls.Thenn(S) = 12C2 = \(rac{12 imes11}{2}\) = 66∴ Let A : Event of drawing two red balls⇒ ... \(rac{n(B)}{n(S)}\) = \(rac{21}{66}\) + \(rac{10}{66}\) = \(rac{31}{66}\).

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Description : An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Last Answer : As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, n(S) = 200 Let A : Event of integer chosen from 1 to 200 being divisible by 6⇒ n(A) = 33 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Description : What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Last Answer : (a) \(rac{1}{25}\) Let us assume S as the sample space in all questions. S means the set denoting the total number of outcomes possible. Let S = {1, 2, 3, , 100} be the sample space. Then, n(S) = 100 Let A : ... ∴Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Description : From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Last Answer : (b) \(rac{7}{10}\)Total number of ways of selecting 2 persons at random out of 5 persons = 5C2 ∴ n(S) = 5C2 = \(rac{|\underline5}{|\underline3|\underline2}\) = \(rac{5 imes4}{2 imes1}\) = 10Let A : Event of selecting ... = 2 3 + 1 = 7 ∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).

A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is -Maths 9th

Description : A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is -Maths 9th

Last Answer : (b) \(rac{7}{99}\)There are (7 + 5) = 12 balls in the bag. 4 balls can be drawn at random from 12 balls in 12C4 ways. ∴ n(S) = 12C4 = \(rac{|\underline{7}}{|\underline3|\underline4}\) = \(rac{7 imes6 imes5}{3 ... ) = 35∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{35}{495}\) = \(rac{7}{99}\).

Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Description : Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Last Answer : (b) \(rac{2}{5}\)As each horse has equal chance of winning the race, Number of ways in which one of the five horses wins the race = 5C1 ∴ n(S) =5C1 = \(rac{|\underline5}{|\underline4|\underline1}\) 5To find the chance ... n(E) = 2C1 = 2 ∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that either both are black or both are kings ? -Maths 9th

Description : Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that either both are black or both are kings ? -Maths 9th

Last Answer : (b) \(rac{55}{221}\)S : Drawing 2 cards out of 52 cards ⇒ n(S) = 52C2 = \(rac{|\underline{52}}{|\underline{52}|\underline2}\) = \(rac{52 imes51}{2}\) = 1326A : Event of drawing 2 black cards out of 26 black cards⇒ n ... ) + \(rac{6}{1326}\) - \(rac{1}{1326}\) = \(rac{330}{1326}\) = \(rac{55}{221}\).

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

Description : A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

Last Answer : (d) \(rac{7}{13}\)Here n(S) = 52 Let A, B, C be the events of getting a red card, a diamond and a jack respectively. ∵ There are 26 red cards, 13 diamonds and 4 jacks, n(A) = 26, n(B) = 13, n(C) = 4 ⇒ n(A ∩ B) = ... rac{1}{52}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .

A natural number is chosen at random from amongst the first 300. What is the probability that the number chosen is a multiple of 2 or 3 or 5 ? -Maths 9th

Description : A natural number is chosen at random from amongst the first 300. What is the probability that the number chosen is a multiple of 2 or 3 or 5 ? -Maths 9th

Last Answer : (b) \(rac{11}{15}\)n(S) = 300 Let A : Event of getting a number divisible by 2 B : Event of getting a number divisible by 3 C : Event of getting a number divisible by 5 ∴ A ∩ B : Event of getting a number divisible by ... \(rac{320}{300}\) - \(rac{100}{300}\) = \(rac{220}{300}\) = \(rac{11}{15}\).

A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Description : A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Last Answer : (d) \(rac{1}{4}\)The box contains 100 balls numbered from 1 to 100. Therefore, there are 50 even and 50 odd numbered balls. The sum of the three numbers drawn will be odd, if all three are odd or one is even and 2 are odd. ∴ Required probability = P(odd) × P(odd) × P(odd) + P(even) × P(odd) × P(odd)

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Description : Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Last Answer : (c) \(rac{1}{10}\)Let S be the sample space.Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon= 6C3 = \(rac{6 imes5 imes4}{3 imes2}\) = 20.Let A : Event that the ... Required probability = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

Description : If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

Last Answer : answer:

A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that -Maths 9th

Description : A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that -Maths 9th

Last Answer : (d) None of theseThe month having 3 days less than 31 days has 28 days, i.e, it is the month of February. P(Choosing February) = \(rac{1}{12}\).

There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, -Maths 9th

Description : There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, -Maths 9th

Last Answer : Total number of ways of placing n letters in n envelopes = n! All the letters can be placed correctly in only 1 way ∴ Probability of placing all the letters in the right envelopes = \(rac{1}{n!}\) ∴ Probability that all the letters are not placed in the right envelope = 1 – \(rac{1}{n!}\) .

The odds against certain event are 5:2 and the odds in favour of another in dependent event are 6:5. The probability that at least one of -Maths 9th

Description : The odds against certain event are 5:2 and the odds in favour of another in dependent event are 6:5. The probability that at least one of -Maths 9th

Last Answer : (c) \(rac{52}{77}\)Given, odds against Event 1 = 5 : 2⇒ P(Event 1 not happening) = \(rac{5}{5+2}\) = \(rac{5}{7}\)Odds in favour of Event 2 = 6 : 5⇒ P(Event 2 happens) = \(rac{6}{6+5}\) = \( ... }\)(∵ Both event are independent)⇒ P(At least one event happens) = 1 - \(rac{25}{77}\) = \(rac{52}{77}\).

Mark correct option a) Bill of franking machine used is prepared in form MS10 b) In RMS offices, letter box key should be kept in custody of Sorting Assistant c) Greeting cards are treated as first class mails in sorting. During the greeting season they are given air lift also without air surcharge otherwise surface mode.d) Label bundle should not contain more than 50 to 60 letters & post cards.e) All of the above

Description : Mark correct option a) Bill of franking machine used is prepared in form MS10 b) In RMS offices, letter box key should be kept in custody of Sorting Assistant c) Greeting cards are treated as first class ... Label bundle should not contain more than 50 to 60 letters & post cards. e) All of the above

Last Answer : ) All of the above

MCQ Questions for Class 9 Maths Chapter 14 Statistics with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 14 Statistics with answers -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 14 Statistics Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive exams. These ... . These Statistics MCQ Questions will help you in practising more and more questions in less time.

Cbqs (case base study ) of chapter 14 Statistics of maths class 9th -Maths 9th

Description : Cbqs (case base study ) of chapter 14 Statistics of maths class 9th -Maths 9th

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A husband and a wife appear in an interview for two vacancies in the same post. The probability of husband‘s -Maths 9th

Description : A husband and a wife appear in an interview for two vacancies in the same post. The probability of husband‘s -Maths 9th

Last Answer : Let A : Event of husband being selected B : Event of wife being selectedThen P(A) = \(rac{1}{7},\) P(B) = \(rac{1}{5}\)P(\(\bar{A}\)) = 1 - \(rac{1}{7}\) = \(rac{6}{7}\), P(\(\bar{B}\)) = 1 - ... {24}{35}\)(iv) P(at least one selected) = 1 - P(none selected) = 1 - \(rac{24}{35}\) = \(rac{11}{35}\)

Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Description : Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Last Answer : (d) \(rac{1}{36}\)Total number of outcomes when three identical dice are rolled, n(S) = 6 6 6 = 216 Let A : Event of rolling same number or each dice ⇒ A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), ... 6)} ⇒ n(A) = 6 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{6}{216}\) = \(rac{1}{36}\).

When a population is divided into distinct groups based on some particular characteristic and a probability sample is taken from each group, this exemplifies: A)area sampling B)quota sampling C)stratified sampling D)cluster sampling E)simple random sampling

Description : When a population is divided into distinct groups based on some particular characteristic and a probability sample is taken from each group, this exemplifies: A)area sampling B)quota sampling C)stratified sampling D)cluster sampling E)simple random sampling

Last Answer : D)cluster sampling

There are 2 vessels. 1st vessel contains 5white and 5 blue thread roll. 2nd vessel contains 4 white and 6 black thread roll. One roll is taken at random from first vessel and put to second vessel without noticing its color. Now a roll is chosen at random from 2nd vessel. What is the probability of the second roll being a white colored roll? A) 11/13 B) 9/11 C) 13/11 D) 5/12

Description : There are 2 vessels. 1st vessel contains 5white and 5 blue thread roll. 2nd vessel contains 4 white and 6 black thread roll. One roll is taken at random from first vessel and put to second vessel without noticing its color ... second roll being a white colored roll? A) 11/13 B) 9/11 C) 13/11 D) 5/12

Last Answer : Answer: B) Case 1: first was a white roll Now it is put in second vessel, so total white rolls in second vessel = 4+1 = 5, and total rolls in second vessel = 10+1 = 11 So probability of white roll ... = 5/11+4/11 = 9/11 (added the cases because we want one of these cases to happen and not both)

Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Description : Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Last Answer : Total number of ways of arranging 4 boys and 3 girls, i.e., 7 people in a queue (row) = n(S) = 7! Let A : Event in which the 4 boys and 3 girls occupy alternate position. This is possible when the ... {4 imes3 imes2 imes1 imes3 imes2 imes1}{7 imes6 imes5 imes4 imes3 imes2 imes1}\) = \(rac{1}{35}.\)

In a batch, 40% of the students offered Maths, 30% offered science and 15% offered both. If a student is selected at random, what is the probability that they has offered science or maths? A) 0.55 B) 0.65 C) 0.45 D) 0.75

Description : In a batch, 40% of the students offered Maths, 30% offered science and 15% offered both. If a student is selected at random, what is the probability that they has offered science or maths? A) 0.55 B) 0.65 C) 0.45 D) 0.75

Last Answer : Answer: A) P(M) = 0.40 P(S) =0.30 and P(M∩S) = 0.15 P(M∪S) = P(M) + P(S) - P(M∩S) = 0.55

The probability of guessing the correct answer to a certain question is x/ 2. -Maths 9th

Description : The probability of guessing the correct answer to a certain question is x/ 2. -Maths 9th

Last Answer : Here, probability of guessing the correct answer = x / 2 And probability of not guessing the correct answer = 2 / 3 Now, x / 2 + 2 / 3 = 1 ⇒ 3x + 4 = 6 ⇒ 3x = 2 ⇒ x = 2 / 3

In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Description : In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Last Answer : Required probability = 1 – P(4) – P(5) =1- 1 / 6 - 1 / 6 = 4 / 6 = 2 / 3

If the probability of winning a race of an athlete is 1 / 6 less than the twice the probability of losing the race. -Maths 9th

Description : If the probability of winning a race of an athlete is 1 / 6 less than the twice the probability of losing the race. -Maths 9th

Last Answer : Let probability of winning the race be p Probability of losing the race = 1 - p According to the statement of question, we have p = 2 (1 - p) - 1 / 6 ⇒ 6p = 12-12p -1 ⇒ 18p = 11 ⇒ p = 11 / 18 Hence, probability of winning the race is 11 / 18

The probability of guessing the correct answer to a certain question is x/ 2. -Maths 9th

Description : The probability of guessing the correct answer to a certain question is x/ 2. -Maths 9th

Last Answer : Here, probability of guessing the correct answer = x / 2 And probability of not guessing the correct answer = 2 / 3 Now, x / 2 + 2 / 3 = 1 ⇒ 3x + 4 = 6 ⇒ 3x = 2 ⇒ x = 2 / 3

In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Description : In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Last Answer : Required probability = 1 – P(4) – P(5) =1- 1 / 6 - 1 / 6 = 4 / 6 = 2 / 3

If the probability of winning a race of an athlete is 1 / 6 less than the twice the probability of losing the race. -Maths 9th

Description : If the probability of winning a race of an athlete is 1 / 6 less than the twice the probability of losing the race. -Maths 9th

Last Answer : Let probability of winning the race be p Probability of losing the race = 1 - p According to the statement of question, we have p = 2 (1 - p) - 1 / 6 ⇒ 6p = 12-12p -1 ⇒ 18p = 11 ⇒ p = 11 / 18 Hence, probability of winning the race is 11 / 18

Can the experimental probability of an event be a negative number? -Maths 9th

Description : Can the experimental probability of an event be a negative number? -Maths 9th

Last Answer : No, because the number of trials in which the event can happen cannot be negative and the total number of trials is always positive.

Can the experimental probability of an event be greater than 1? -Maths 9th

Description : Can the experimental probability of an event be greater than 1? -Maths 9th

Last Answer : No, as the number of trials can't be greater than the total number of trials.

In a throw of a die, find the probability of getting an even number. -Maths 9th

Description : In a throw of a die, find the probability of getting an even number. -Maths 9th

Last Answer : Total even number on a die = 3 P (getting an even numbers) = 3/6 = 1/2

A die is thrown.What is the probability of getting a multiple of 3 on the upper face ? -Maths 9th

Description : A die is thrown.What is the probability of getting a multiple of 3 on the upper face ? -Maths 9th

Last Answer : Multiple of 3 on a die = 3, 6 ∴ P (a multiple of 3) = 2/6 = 1/3.

In a single throw of two dice, what is the probability of getting a sum of 9? -Maths 9th

Description : In a single throw of two dice, what is the probability of getting a sum of 9? -Maths 9th

Last Answer : Outcomes with sum of 9 = { (3, 6), (4, 5), (5, 4), (6, 3) } P ( getting a sum of 9 is ) = 4/36 = 1/9

MCQ Questions for Class 9 Maths Chapter 15 Probability with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 15 Probability with answers -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 15 Probability Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive exams. These ... . These Probability MCQ Questions will help you in practising more and more questions in less time.

What do you mean by probability? -Maths 9th

Description : What do you mean by probability? -Maths 9th

Last Answer : It is the chance of happening of an event when measured quantitatively.

Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. -Maths 9th

Description : Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. -Maths 9th

Last Answer : Let A : Getting an odd number on first die; B : Getting a multiple of 3 on second die, thenA = {1, 3, 5}, B = {3, 6} ∴ P(A) = \(rac{3}{6}=rac{1}{2}\), P(B) = \(rac{2}{6}=rac{1}{3}\) ... B are independent∴ Required probability = P (A) . P (B) = \(rac{1}{2}\) x \(rac{1}{3}\) = \(rac{1}{6}\)