let A denote the event that the number is divisible by 4 and B denote the event that the number is divisible by 4.
To find : Probability that the number is both divisible by 4 or 6 = P(A or B)
The formula used : Probability =
= \(\frac{favourable\,number\,of\,outcomes}{Total\,no.of\,outcomes}\)
P(A or B) = P(A) + P(B) - P(A and B)
Numbers from 1 to 100 divisible by 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100.
There are 25 numbers from 1 to 100 divisible by 4
Favourable number of outcomes = 25
Total number of outcomes = 100 as there are 100 numbers from 1 to 100
P(A) = \(\frac{25}{100}\)
Numbers from 1 to 100 divisible by 6 are 6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96
There are 16 numbers from 1 to 100 divisible by 6
Favourable number of outcomes = 16
Total number of outcomes = 100 as there are 100 numbers from 1 to 100
P(B) = \(\frac{16}{100}\)
Numbers from 1 to 100 divisible by both 4 and 6 are 12,24,36,48,60,72,84,96
There are 8 numbers from 1 to 100 divisible by both 4 and 6
Favourable number of outcomes = 8
P(A and B) = \(\frac{8}{100}\)
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = \(\frac{25}{100}+\frac{16}{100}-\frac{8}{100}\)
P(A or B) = \(\frac{25+16-8}{100}\) = \(\frac{33}{100}\)
P(A or B) = \(\frac{33}{100}\)
The probability that the number is both divisible by 4 or 6 = P(A or B) = \(\frac{33}{100}\)
