Question

One number is choosen at random from numbers 1 to 100. Find the probability that it is

divisible by 4 or 6​

Answer 1

$\huge\bold\pink{AnSwEr.....≧∇≦}$

Total number of numbers n(S)=100

Numbers divisible be 4 between 1 to 100 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}

Number divisible by 6 between 1 to 100 are {6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96}

6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96}Hence numbers divisible by 4 or 6 between 1 to 100 are {4,6,8,12,16,18,20,24,28,30,32,36,40,42,44,48,52,54,56,60,64,66,68,72,76,78,80,84,88,90,92,96,100}, n(E)=33

4,6,8,12,16,18,20,24,28,30,32,36,40,42,44,48,52,54,56,60,64,66,68,72,76,78,80,84,88,90,92,96,100}, n(E)=33Hence the probability that it is divisible by 3 or 7 is n(S)/ n(E)

= 100/ 33....♥

Answer 2

Total number of numbers n(S)=100

Numbers divisible be 4 between 1 to 100 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}

Number divisible by 6 between 1 to 100 are {6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96}

Hence numbers divisible by 4 or 6 between 1 to 100 are {4,6,8,12,16,18,20,24,28,30,32,36,40,42,44,48,52,54,56,60,64,66,68,72,76,78,80,84,88,90,92,96,100}, n(E)=33

Hence the probability that it is divisible by 3 or 7 is

n(S)

n(E)

=

100

33