Probability to choose house number with each number being equally likely and no repeated digits in a number, as P(A∩B) = [tex]\frac{1}{24}[/tex] (simplest form ).
" Probability is defined as the ratio of number of favourable outcomes to the total number of outcomes."
Formula used
Probability = [tex]\frac{Number of favourable outcomes}{total number of outcomes}[/tex]
According to the question,
House number is made up of nonzero digits and are of two digits
Numbers from 1 to 9.
As per the condition given,
First digit can be filled in 9 ways
Second digit can be filled in 8 ways
Total number of ways = 9 × 8
= 72 ways ___(1)
Event A is defined as choosing 6 as the first digit.
Possible number of ways to fill first digit = 1
Event B is defined as choosing a number less than 4 as the second digit, that is 1, 2, 3
Possible number of ways to fill second digit = 3
Possible number of ways to form two digit number 'A∩B '= 1 × 3
= 3 ____(2)
Substitute the values in the formula to get the required probability,
Probability (A∩B) = [tex]\frac{3}{72}[/tex]
= [tex]\frac{1}{24}[/tex]
Hence, the required probability to choose a house number P(A∩B) = [tex]\frac{1}{24}[/tex] (simplest form ).
Learn more about probability here
https://brainly.com/question/11034287
#SPJ3
