Question

what is the probability that a randomly chosen positive three-digit integer is a multiple of 6? 150/901 incorrect: your answer is incorrect.

Answer 1

Using Probability formula,

we find that the given answer is incorrect .

The probability that a randomly chosen positive three-digit integer is a multiple of 6 = 150/900

We have ask about the three digit number .

The three digit number starts from 100 to 999

then , there are total 900 three digit numbers from 100 to 999.

Now , first 3 digit number which is multiple of 6

= 102

and the last 3 digit number which is multiple of 6 = 996.

thus , the three digit multiples of 6 are

S = { 102,108,114,120,….996 }

which is an Arithmetic progression sequence i.e A.P with

• first term of sequence ( a)= 102,
• common difference d= 6
• the last term of sequence (l) = 996.

so, we calculate total number of terms which are multiple of 6 in three digit numbers

n = (l-a)/d + 1

putting all available values of variables we get ,

n = (996–102)/6 + 1 = 149+1= 150

i.e., there is total 150 terms out of all three digit numbers which are multiple of 6 .

probability that a randomly chosen positive three-digit integer is a multiple of 6 = numbers which are multiple of 6 / total number of three digit numbers

= 150/900

Hence, the given answer i.e., 150/901 is incorrect.

To learn more about probability, refer:

https://brainly.com/question/13604758

#SPJ4