Ramvardhan
Last Activity: 9 Years ago
P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C).
Here, let the the event of the random ticket being divisible by 3 is A, by 4 is B and by 5 is C.
You can find the number of times an event is occuring by finding how much time the multiples of that number to be divisible are occuring from 1-100.
P(A) = n(A)/n(s) = 33/100
P(B) = 25/100
P(C) = 20/100
P(A ∩ B) = 8/100 [ This is the probabitlity of that random number to be divisible by both 3 and 4]
P(A ∩ C) = 6/100 [ This is the probabitlity of that random number to be divisible by both 3 and 5]
P(B ∩ C) = 5/100 [ This is the probabitlity of that random number to be divisible by both 4 and 5]
P(A ∩ B ∩ C) = 1/100 [ This is the probabitlity of that random number to be divisible by 3, 4 and 5]
Then,
P(A U B U C) = 33/100 + 25/100 + 20/100 – 8/100 – 6/100 – 5/100 + 1/100
= 0.6
P(A U B U C) is the probability of the random number to be divisible by 3,4,5. Then the complement of this will be the probability of the randmo number to be not divisble by 3,4,5.
Complement of P(A U B U C) = 1 – P(A U B U C) = 1 – 0.6 = 0.4
Thus the final answer is 0.4.
