The probability that the written 3 digit number starts with 3 is 1/9
The total number of 3-digit numbers starting from 100 till 999, including both these endpoints, is 900. Since the first digit can be any number from 1 to 9; tens (second digit) and hundreds’ (third digit) can be any number from 0 to 9;( 9×10×10 = 900). Therefore, three-digit numbers in between 100 and 999 = Total possible number of outcomes = 900.
For 3-digit number to start with 3, the first digit (hundreds’ place) has to be 3. The tens place (2nd digit) and units place can be any number from 0 to 9. Therefore, total number of favorable outcomes for the three digit number to start from 3 is 1×10×10 = 100 (numbers from 300 to 399)
Probability = Total number of favorable outcomes / Total possible outcomes = 100/900 = 1/9
The probability of 3 digit number starts with 3 is [tex]\frac{1}{9}[/tex] .
The total number of 3-digit numbers starting from 100 till 999 , including both these endpoints is 900.
Since the first digit can be any number from 1 to 9;
Tens (second digit) and third hundreds’ (digit) can be any number from 0 to 9; ( 9×10×10 = 900) .
Therefore, it possible outcomes between three digit number is 100 and 999 = Total possible number of outcomes = 900.
According to given condition;
For 3-digit number to start with 3,
The first digit (hundreds’ place) has to be 3.
The tens place (2nd digit) and units place can be any number from 0 to 9.
Therefore, total number of favorable outcomes for the three digit number to start from 3 is 1×10×10 = 100 (numbers from 300 to 399).
[tex]Probability = \frac{Total no. of favorable outcomes }{Total no. of possible outcomes} \\\\P = \frac{100}{900} \\\\P = \frac{1}{9}[/tex]
The probability of 3 digit number starts with 3 is [tex]\frac{1}{9}[/tex] .
For more details about probability distribution click the link given below.
https://brainly.in/question/34543433
