Question

Tickets numbered from 1 to 20 are mixed up together and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3?

Answer 1

Answer:

multiples of 3 - 3 , 9, 12, 15 , 18 .

multiples of 5 - 5 , 10 , 15 , 20.

multiples of both 15.

total - 20

probability of multiple of 3 n(a) - 5/20

probability of multiple of 5 n(b) - 4/20

probability of both 3 and 5 n(a and b) - 1/20

probability of mulitple 3 or 5 n(a) + n(b) - n(a and b)

= 5/20 + 4/20 - 1/20

=8/20

= 2/5

Answer 2

Answer:

3/10

Step-by-step explanation:

Tickets numbered from 1 to 20 are mixed up.

Since one ticket is drawn from a lot of mixed number, total possible outcomes are,

n(S) = C(20,1) = 20.

let E be the event of getting ticket which has number that is multiple of 3,

E = {3, 6, 9, 12, 15, 18}

n(E) = 6.

∴ P(E) = n(E)/n(S)

         = 6/20

        = 3/10

Hope it helps!