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

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
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1 Answer

Answer :

Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)
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Related questions

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)

In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Description : In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Last Answer : 2 boys, 1 gi(Prl = (15c2×10c1) / 25c3 = 1050/2300 1 boy, 2 girls = (15c1×10c2) / 25c3 = 675/2300 Difference = (1050 - 675)/2300 = 375/2300 = 15/92 Answer: c)

In a batch, there are 22 boys and 18 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is: a) 3754/8854 b) 4158/9880 c) 8514/9880 d) 2078/4920

Description : In a batch, there are 22 boys and 18 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is: a) 3754/8854 b) 4158/9880 c) 8514/9880 d) 2078/4920

Last Answer : Answer: B) Let , S - sample space E - event of selecting 1 girl and 2 boys. Then, n(S) = Number ways of selecting 3 students out of 40 = 40C3 = 9880 n(E) = 18C1 *22C2 = 18*231  = 4158 P(E) = n(E)/n(s) = 4158/9880 

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

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

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 = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).

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 = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

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)

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}\) = \(rac{1}{35}.\)

6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is -Maths 9th

Description : 6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is -Maths 9th

Last Answer : (b) \(rac{1}{462}\)Let S be the sample space. Then, n(S) = Number of ways in which 6 boys and 6 girls can sit in a row = 12! Let E : Event of 6 girls and 6 boys sitting alternately. Then, the ... )= \(rac{2 imes6 imes5 imes4 imes3 imes2 imes1}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{462}\).

6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

Description : 6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

Last Answer : (d) \(rac{1}{132}\)Let S be the sample space for arranging 6 boys and 6 girls in a row. Then, n(S) = 12! If all 6 boys are to sit together, then consider the 6 boys as one entity. Now the ... ) = \(rac{7 imes6 imes5 imes4 imes3 imes2}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{132}\).

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

A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

Description : A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

Last Answer : (c) \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)Total number of boys and girls = 2n + 2n = 4n Since, there are two equal batches, each batch has 2n members ∴ Let S (Sample space) : Selecting one batch out of 2 ⇒ S : ... )2∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

In a hostel, 40% of the students play cricket, 20% play chess and 10% both. If a student is selected at random, then the probability that he plays cricket or chess is: a) 1/2 b) 3/5 c) 1/4 d) 4/7

Description : In a hostel, 40% of the students play cricket, 20% play chess and 10% both. If a student is selected at random, then the probability that he plays cricket or chess is: a) 1/2 b) 3/5 c) 1/4 d) 4/7

Last Answer : Answer: A) Given that, 40% play cricket; that is, P(C) = 40/100=4/10 20% play chess; that is, P(c) = 20/100 =2/10 And, 10% play both cricket and chess; that is, P(C And c) = 10/100 = 1/10 Now, we have ... P(C) + P(c) - P(C And c) = 4/20+2/10-1/10 =5/10=1/2 Hence, the required probability 1/2

The ratio of girls and boys in a class is 1: 3. Set up an equation between the students of a class and boys and then draw its graph. Also find the number of boys in a class of 40 students from the graph. -Maths 9th

Description : The ratio of girls and boys in a class is 1: 3. Set up an equation between the students of a class and boys and then draw its graph. Also find the number of boys in a class of 40 students from the graph. -Maths 9th

Last Answer : Total number of boys and girl = 40, Ratio = 1 : 3 Number of girls be A and Number of boys be B. Ratio of number of girls and boys is 1 : 3, so Therefore 3A=B To find number of boys we ... the number 30 represents the number of girls. 40 as total on the line A = 10, which is the common equation.

A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Description : A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Last Answer : Let P (A) = Probability of selecting a fair complexioned person. ThenP(A) = \(rac{20}{80}\) = \(rac{1}{4}\)Let P(B) = Probability of selecting a rich person. Then P(B) = \(rac{10}{80}\) = \(rac{1}{8}\)Let P (C) = ... ) = \(rac{1}{4}\)x \(rac{1}{8}\)x \(rac{5}{16}\) = \(rac{5}{512}\) = 0.009.

If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Description : If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Last Answer : (c) \(rac{4}{1155}\)Let n(S) = Number of ways of selecting 3 numbers from 100 numbers = 100C3 Let E : Event of selecting three numbers divisible by both 2 and 3 from numbers 1 to 100 = Event of selecting three ... C_3}{^{100}C_3}\) = \(rac{16 imes15 imes14}{100 imes99 imes98}\) = \(rac{4}{1155}\).

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 Class IX English book contains 200 pages. A page is selected at random. What is the probability that the number on the page is divisible by 10?

Description : A Class IX English book contains 200 pages. A page is selected at random. What is the probability that the number on the page is divisible by 10?

Last Answer : A Class IX English book contains 200 pages. A page is selected at random. What is the probability that the number on the page is divisible by 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

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80 bulbs are selected at random from a lot -Maths 9th

Description : 80 bulbs are selected at random from a lot -Maths 9th

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In a group there are 3 women and 3 men. 4 people are selected at random from this group -Maths 9th

Description : In a group there are 3 women and 3 men. 4 people are selected at random from this group -Maths 9th

Last Answer : A : Selected 3 women and 1 man B : Selected 1 women and 3 men S : Selected 4 people from 6 people (3 + 3) Then n(A) = 3C3 3C1, n(B) = 3C1 3C3, n(S) = 6C4∴ Required probability = P(A) + P(B) = \(rac{ ... 3C_1}{^6C_4}\) + \(rac{^3C_1 imes^3C_3}{^6C_4}\)= \(rac{2 imes1 imes3}{15}\) = \(rac{2}{5}.\)

A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Description : A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Last Answer : (a) \(rac{5}{6}\)P(At least one junior professor is selected) = P(Selecting 1 Junior) P(Selecting 2 Seniors) + P(Selecting 2 Junior) P(Selecting 1 Senior) + P(Selecting all 3 Juniors)∴ Required probability = \(rac{^4C_1 imes^ ... }{30}\) = \(rac{15+9+1}{30}\) = \(rac{25}{30}\) = \(rac{5}{6}\).

A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Description : A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Last Answer : answer:

There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

Description : There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

Last Answer : Have the 55 boys stand in a line. This can be done in 5!5! ways. For the moment, add a boy at each end, who will be removed when we're done. Now send the 33 girls, one at a time, to ... that each girl can either wedge herself between two boys or else stand next to a boy at one end or the other.

There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

Description : There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

Last Answer : Since the chairs are numbered, so they are distinguishable. Therefore 3 boys can be arranged on 3 alternate chairs in 3! ways. 3 girls can be arrenged in 3! ways Also, the girls can be seated before the boys. Total number of required ways = 3! × 3! + 3! × 3! = 2 × (3!)2

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

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

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 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

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 = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

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 = \(rac{n(A)}{n(S)}\) = \(rac{35}{495}\) = \(rac{7}{99}\).

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 ... \(rac{320}{300}\) - \(rac{100}{300}\) = \(rac{220}{300}\) = \(rac{11}{15}\).

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!}}\) = \(rac{ ot8! imes36 imes2!}{10 imes9 imes ot8!}\) = \(rac{4}{5}\).

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⇒ ... \(rac{n(B)}{n(S)}\) = \(rac{21}{66}\) + \(rac{10}{66}\) = \(rac{31}{66}\).

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 = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

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) = \(rac{1}{12}\).

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

Description : 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

Last 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.

What number is chosen at random from 1 to 10. Find the probability of not selecting a multiple of 2 or a multiple of3?

Description : What number is chosen at random from 1 to 10. Find the probability of not selecting a multiple of 2 or a multiple of3?

Last Answer : Need answer

there are 24 students in a class 8 of the student are boys a) write the amount of boys in the class as a fraction in its simplest form b) write the amount of girls in the class as a fraction in its simplest form?

Description : there are 24 students in a class 8 of the student are boys a) write the amount of boys in the class as a fraction in its simplest form b) write the amount of girls in the class as a fraction in its simplest form?

Last Answer : two thirds

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 \(rac{5}{7}\) = \(rac{26}{49}.\)

Form, a cosmetic shop containing perfumes and does, a pair is selected at random. The probability that the selected pair will consist of one perfume a

Description : Form, a cosmetic shop containing perfumes and does, a pair is selected at random. The probability that the selected pair will consist of one perfume a

Last Answer : Form, a cosmetic shop containing perfumes and does, a pair is selected at random. The probability ... of perfumes and does the shop can contain ?

A mathematics book contains 250 pages. A page is selected at random. What is the probability that the number on the page selected is a perfect square?

Description : A mathematics book contains 250 pages. A page is selected at random. What is the probability that the number on the page selected is a perfect square?

Last Answer : A mathematics book contains 250 pages. A page is selected at random. What is the probability that the number on the page ... `(3)/(50)` D. `(7)/(125)`

A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random, find the probability that the number on the ticket

Description : A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random, find the probability that the number on the ticket

Last Answer : A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random ... number on the ticket is not a perfect square.

What is the probability that a leap year selected at random will have 33 Sunday?

Description : What is the probability that a leap year selected at random will have 33 Sunday?

Last Answer : Need answer

Nine playing cards are numbered 2 to 10. A card is selected at random. What is the probability that the card will be an odd number? a. 1/9 b. 2/9 c. 4/9 d. 3/7

Description : Nine playing cards are numbered 2 to 10. A card is selected at random. What is the probability that the card will be an odd number? a. 1/9 b. 2/9 c. 4/9 d. 3/7

Last Answer : c. 4/9