Two numbers 'a' and 'b' are selected successively without replacement In that order from the integers 1 to 10. The probability that a/b is an integer, is. (a) 17/45 (b) 1/5 (c) 17/90 (d) 8/45?

Question

Answer

The total number of possibilities is 10∗910∗9 =4545since we are choosing two numbers without replacement and order matters..Hence the cases are(2,1),(3,1),(4,1),(5,1),(6,1),(7,1),(8,1),(9,1)and(10,1)(2,1),(3,1),(4,1),(5,1),(6,1),(7,1),(8,1),(9,1)and(10,1)(4,2),(6,2),(8,2),(10,2)(4,2),(6,2),(8,2),(10,2)(6,3),(9,3)(6,3),(9,3)(8,4)(8,4)(10,5)(10,5)For all b greater than 5 the smallest multiple is greater than 10.Hence they dont contribute to the total.Hence a total of 9+4+2+1+1=17 valid possibilities.Hence the probability is 17/90

Answer

The first number 'a' can be selected in 10 ways and the second number 'b' can be selected in 9 ways so you can select the pair (a,b) in 90 ways. The possible out comes where a/b is an integer are:

If a = 10 => b ={5,2,1}

If a = 9 => b ={3,1}

If a = 8 => b ={4,2,1}

If a = 7 => b ={1}

If a = 6 => b ={3,1}

If a = 5 => b ={1}

If a = 4 => b ={2,1}

If a = 3 => b ={1}

If a = 2 => b ={1}

If a = 1 => b ={1}

Probability = 17/90

Answer

Solution:

The total number of ways of selecting two numbers without replacement from 1 to 10 is 10 × 9 = 90.

Case 1: When the first number is divisible by the second number.

There are 5 pairs of numbers between 1 and 10 that are divisible without a remainder: (1,1), (2,1), (3,1), (4,1), and (5,1). For each of these pairs, the first number can be chosen in one way and the second number in one way.

Therefore, the total number of ways of selecting two numbers such that the first number is divisible by the second number is 5.

Case 2: When the second number is divisible by the first number.

There are 4 pairs of numbers between 1 and 10 such that the second number is divisible without a remainder by the first number: (2,4), (3,6), (4,8), and (5,10). For each of these pairs, the first number can be chosen in two ways (since there are two numbers that divide into the second number) and the second number in one way.

Therefore, the total number of ways of selecting two numbers such that the second number is divisible by the first number is 4 × 2 = 8.

Case 3: When neither number is divisible by the other.

There are 5 pairs of numbers between 1 and 10 such that neither number is divisible by the other: (2,3), (2,5), (2,7), (3,5), and (3,7). For each of these pairs, the first number can be chosen in two ways (since there are two possible values that are less than the second number and do not divide into it) and the second number can be chosen in one way.

Therefore, the total number of ways of selecting two numbers such that neither number is divisible by the other is 5 × 2 = 10.

Therefore, the total number of ways of selecting two numbers such that a/b is an integer is 5 + 8 = 13.

The probability that a/b is an integer is 13/90.

Hence, the answer is (c) 17/90.

Answer

B

Similar Class 10 Doubts

Two numbers A and B are selected successively without replacement in that order from the integers 1 to 10. The probability that A/B is an integer is???

A number is selected from the numbers 1 to 25. The probability that it is prime is. (

A number is selected from numbers 1to 25. The probability that it is prime is. (

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears 1. a two -digit number is a perfect square number2. a number divisible by 5?

A number x is selected from the numbers 1,2,3 and then a second number y is randomly selected from the numbers 1,4,9 what is the probability that product xy of the two numbers will be less than 9?

Top Courses for Class 10View all

Science Class 10

Social Studies (SST) Class 10

Mathematics (Maths) Class 10

English Grammar Advanced

English Grammar Basic

Top Courses for Class 10

Top Courses for Class 10

Suggested Free Tests

Related Content

About Class 10

Class 10 Study Material

Class 10 All Subjects

NCERT Textbook Solutions

Other Classes

Important Links

Company