Question

How many positive three-digit integers are multiples of 6?

(b) What is the probability that a randomly chosen positive three-digit integer is a multiple of 6?

(c) What is the probability that a randomly chosen positive three-digit integer is a multiple of 7?

Answer 1

Answer with Step-by-step explanation:

a.Three digit number which is multiple of 6

102,108,...996

a=102, $a_n=996$

d=108-102=6

It is an A.P because the difference between two consecutive terms is constant.

nth term of A.P is given by

$a_n=a+(n-1)d$

Substitute the values then we get

$996=102+(n-1)\times 6$

$996-102=(n-1)\times 6$

$894=6(n-1)$

$n-1=\frac{894}{6}=149$

$n=149+1=150$

Numbers of positive three-digit integers which are multiple of 6=150

b.Three digits number lying between 100 and 999 therefore,

Total three digit positive integers=$999-100+1=900$

Probability,P(E)=$\frac{favorable\;cases}{total\;number\;of\;cases}$

The probability that a random chosen positive three- digit integer is a multiple of 6=$\frac{150}{900}=\frac{1}{6}=0.167$

c.Positive integers of multiply of 7 are

105,112,119,....,994

$a=105,a_n=994$

d=112-105=7

Using nth term of A.P formula

$994=105+(n-1)\times 7$

$994-105=7(n-1)$

$7(n-1)=889$

$n-1=\frac{889}{7}=127$

$n=127+1=128$

The probability that a random chosen positive three digit integer is a multiple of 7=$\frac{128}{900}=0.142$