Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ? -Maths 9th

1 Answer

Answer :

The chosen consists of players (6 + 5). ∴ Number of ways of selecting 11 players out of 15 players = n(S) = 15C11 Let A : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers Then, n(A) = 8C6 x 7C5∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{^8C_6 imes^7C_5}{^{15}C_{11}}\) = \(rac{rac{8 imes7}{2} imesrac{7 imes6}{2}}{rac{15 imes14 imes13 imes12}{4 imes3 imes3 imes1}}\)= \(rac{8 imes7}{2}\) x \(rac{7 imes6}{2}\) x \(rac{4 imes3 imes2 imes1}{15 imes14 imes13 imes12}\) = \(rac{28}{65}.\)

Related questions

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

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

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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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

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

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

Description : In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th

Last Answer : As each team of 11 players has one goalkeeper and 10 team members, and out of 14 players there are 2 goalkeepers and 12 team members. = 12×112×2 = 132.

Description : Two batsman Rahul and Anil while playing a cricket match scored 120 runs. For this, write a linear equation in two variables and draw the graph. -Maths 9th

Last Answer : Solution :-

Description : In cricket match, a batsman hits a boundary 12 times -Maths 9th

Last Answer : Let E be the event 'the batsman hits a boundary'. Then, E bar is the event 'the batsman does hit a boundary'. ∴ P(E) = 12/48 = 1/4 ⇒ P(E bar) = 1 - P(E) = 1 - 1/4 = 3/4

Description : In how many ways can a pack of 52 cards be divided equally among four players in order? -Maths 9th

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Last Answer : First of all number of ways. Committees can be founded =6C3 5C2 Now, we remove committees with both A and B =4 5C2 again we need to remove committees with B and not C =5C2 4. Now we shall add the committees with A, ... −4 5C2 −5C2 4+4 4C2 =124. Hence, the answer is =124.

Description : The probability of A, B, C solving a problem are 1/3, 2/7 and 3/8 respectively. -Maths 9th

Last Answer : Let E1, E2, E3 be the eventsthat the problem issolved by A, B, C respectively and let p1, p2, p3 be corresponding probabilities. Then,p1 = P(E1) = \(rac{1}{3}\), p2 = P(E2) = \(rac{2}{7}\), p3 = P(E3) = \(rac{3}{8}\) ... }{8}\) = \(rac{25}{168}\) + \(rac{5}{42}\) + \(rac{5}{28}\) = \(rac{25}{56}.\)

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.

Description : Cbqs (case base study ) of chapter 15 Probability of maths class 9th -Maths 9th

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Description : There are 6 positive numbers and 8 negative numbers. Four numbers are chosen at random and multiplied. The probability that the product is a positive number is: a) 502/1001 b) 505/1001 c) 501/1001 d) 503/1001

Last Answer : b) 505/1001

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Last Answer : Let A, B, C, D denote the events of not getting a white ball in first, second, third and fourth draw respectively. Since the balls are drawn with replacement, therefore, A, B, C, D are independent events such that P (A) = P (B) ... x \(rac{11}{16}\) x \(rac{11}{16}\) = \(\big(rac{11}{16}\big)^4.\)

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

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

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Description : A cartoon contains 15 torch lights out of which 3 are defective. Two torch light are chosen at random from this cartoon. The probability that at least one of these is defective is. A) 13/35 B) 14/35 C) 11/35 D) 17/35

Last Answer : Answer: A) P (none is defective) = 12c2/15c2 = (12*11/2*1)/(15*14/2*1) = 66/105=22/35 P (at least one is defective) = (1 – 22/35) =13/35

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Description : If a number x is chosen at random from the numbers -2, -1, 0 1, 2. What is the probability that x 2 < 2? a. 2/5 b. 3/5 c. 1/4 d. 4/7

Last Answer : b. 3/5

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

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

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

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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)

Description : A person invites 15 guests for dinner and wishes to arrange them at two round tables that can accommodate 8 persons and 7 persons respectively. -Maths 9th

Last Answer : answer:

Description : Two players A and B play a game by alternately drawing a card from a well-shuffled pack of playing cards, replacing the card each time after draw. -Maths 9th

Last Answer : (a) \(rac{13}{25}\)Let E : Event of drawing a queen in a single draw the pack of 52 cards. As there are 4 queens in a pack of 52 cards,P(E) = \(rac{4}{52}\) = \(rac{1}{13}\)P(\(\bar{E}\)) = P(not ... {25}\). [Sum of a G.P with infinite terms = \(rac{a}{1-r}\) where a = 1st term, r = common ratio.]

Description : A bag contains a white and b black balls. Two players A and B alternately draw a ball from the bag -Maths 9th

Last Answer : (c) 2 : 1Let W denote the event of drawing a white ball at any draw and B that of drawing a black ball. Then, P (W) = \(rac{a}{a+b},\) P(B) = \(rac{b}{a+b}\)∴ P (A wins the game) = P (W or BBW or BBBBW or ... ... the given condition,\(rac{a+b}{a+2b}\) = 3. \(rac{b}{a+2b}\) ⇒ a = 2b ⇒ a : b = 2 : 1.

Description : Batsman Are On 94 Notout,need To Win 7 Runs Off 2 Balls,both Hit A Century? How It Is Possible? -AI Class 9th

Last Answer : Match is 49.1 overs because of rain. Last 2 balls means 48.5 and 49.1 then. A hit six and got his century then over finish means 48.6 then new over start 49.1 balls and b is on strikes then his also hit 6 runs and got his century. Match also win two players will make his 100 runs.

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

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

Description : Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Last Answer : The two digit numbers can be formed by putting any of 5 digits at the one 's place and also one of the 5 digits at ten's place. So, Total number of 2-digit numbers that can be formed using these 5-digits = 5 5 = ... 52}, i.e, 5 in number. ∴ Required probability = \(rac{5}{25}\) = \(rac{1}{5}.\)

Description : The probability that A hits a target is 1/3 and the probability that B hits it is 2/5 . -Maths 9th

Last Answer : (b) \(rac{3}{5}\)Let A : Event that A hits the target B : Event that B hits the target Given,P(A) = \(rac{1}{3}\) ⇒ P(A not hitting the target) = P(\(\bar{A}\)) = 1 - \(rac{1}{3}\) = \(rac{2}{3}\)P(B) = \(rac{ ... = \(rac{3}{15}\) + \(rac{4}{15}\) + \(rac{2}{15}\) = \(rac{9}{15}\) = \(rac{3}{5}\).

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

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : In conclusion, the range of data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, and 20 is 26.

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : (d) In a given data, maximum value = 32 and minimum value = 6 We know, range of the data = maximum value – minimum value = 32 – 6 = 26 Hence, the range of the given data is 26.

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

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

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

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)