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

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

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 = 25 The 2–digit numbers formed by 1, 2, 3, 4 and 5 that are divisible by 4 are {12, 24, 32, 44, 52}, i.e, 5 in number. ∴ Required probability = \(rac{5}{25}\) = \(rac{1}{5}.\)
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Related questions

A five digit number is formed by the digits 0, 1, 2, 3, 4 (without repetition). Find the probability that the number formed is divisible by 4 ? -Maths 9th

Description : A five digit number is formed by the digits 0, 1, 2, 3, 4 (without repetition). Find the probability that the number formed is divisible by 4 ? -Maths 9th

Last Answer : Without repetition, a five -digit number can be formed using the five digits in 5! ways (5 4 3 2 1) Out of these 5! numbers, 4! numbers will be starting with digit 0. (0 (fixed) 4 3 2 1) ∴ Total ... + 6 + 6 + 4 + 4 + 4 = 30∴ Required probability = \(rac{30}{96}\) = \(rac{5}{16}.\)

A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd is -Maths 9th

Description : A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd is -Maths 9th

Last Answer : (a) \(rac{1}{2}\) Let S be the sample spaceThen n(S) = Number of four digit numbers that can be formed with out repetition1 of 4 digits1 of 3 digits1 of 2 digits1 of 1 digitThHTO= 4! = 4 3 2 1 ways = 24 ways. Let E : ... 1 = 12∴ P(E) = \(rac{n(E)}{n(S)}\) = \(rac{12}{24}\) = \(rac{1}{2}\).

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

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

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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}\) = \(rac{16 imes15 imes14}{100 imes99 imes98}\) = \(rac{4}{1155}\).

Probability of all 4–digit numbers having all the digitssame is -Maths 9th

Description : Probability of all 4–digit numbers having all the digitssame is -Maths 9th

Last Answer : (c) \(rac{1}{1000}\)Let S be the sample space where 4 digit numbers are formed using digits 0 to 9 repetition. Then, n(S) = Total number of 4 digit number formed = 9 10 10 10 = 9000 (∵ The thousands place rest can only be ... )∴ P(E) = \(rac{n(E)}{n(S)}\) = \(rac{9}{1000}\) = \(rac{1}{1000}\).

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Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

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Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

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What are all of the different ways to choose the ones digit A in 271854A so that the number will be divisible by 9?

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Why the divisibility test for 9 is valid and a way to determine whether a three-digit counting number ABC is divisible by 9?

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What are all of the different ways to choose the ones digit A in 271854A so that the number will be divisible by 9?

Description : What are all of the different ways to choose the ones digit A in 271854A so that the number will be divisible by 9?

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What are all of the different ways to choose the tens digit A and the ones digit B in the number 631872AB so that the number will be divisible by 9?

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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 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?

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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

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Description : Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Last Answer : (c) \(rac{7}{12}\)Total number of ways in which 2 dice are rolled = 6 6 = 36 ⇒ n(S) = 36 Let A : Event of rolling an even number of 1st dice B : Event of rolling a total of 7 ⇒ A = {(2, 1), (2, 2) , (2, 6), (4 ... (rac{18}{36}\) + \(rac{6}{36}\) - \(rac{3}{36}\) = \(rac{21}{36}\) = \(rac{7}{12}\).

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Description : A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 3 ? -Maths 9th

Last Answer : (b) \(rac{11}{36}\)Let S = total ways in which two dice can be rolled ⇒ n(S) = 6 6 = 36 Let A : Event of throwing 3 with 1st dice, B : Event of throwing 3 with 2nd dice. Then, A = {(3, 1), (3, 2), (3, 3), (3, 4), ... ) - P(A ∩ B)= \(rac{6}{36}\) + \(rac{6}{36}\) - \(rac{1}{36}\) = \(rac{11}{36}\).

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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.

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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

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) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Description : Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Last Answer : (d) \(rac{1}{2}\)Let S be the sample space. Then, S = {HHH, HHT, HTH, HTT, THH, THT,TTH, TTT} ⇒ n(S) = 8 Let A : Event of getting more heads than number of tails. Then, A = {HHH, HHT, HTH, THH} ⇒ n(A) = 4∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}.\)

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) = \(rac{n(A)}{n(S)}\) = \(rac{6}{216}\) = \(rac{1}{36}\).

A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

Description : A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

Last Answer : (b) \(rac{5}{24}\)Total number of ways three die can be rolled = 6 6 6 = 216 A larger number than the previous number can be got in the three throws as (1, 2, 3), (1, 2, 4), (1, 2, 5) ( ... , 5, 6). ∴ Total number of favourable cases = 20∴ Required probability =\(rac{20}{216}\) = \(rac{5}{24}\).

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

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

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

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