Question

Ticket numbered from 1 to 20 are mixed up and a ticket is drawn at random. what is the probability that the ticket drawn has a number which is multiple of 3 or 7 ?

Answer 1

total number of cards =20

P (a multiple of 3 or 7)=8/20 = 2/5

Answer 2

Given:

• Ticket numbered from 1 to 20 are mixed up and a ticket is drawn at random.

To calculate:

• The probability that the ticket drawn has a number which is multiple of 3 or 7.

Calculation:

Here, we are given that ticket numbered from 1 to 20 are mixed up and a ticket is drawn at random & we have to find out the probability that the ticket drawn has a number which is multiple of 3 or 7.

So, as we know that :

$\bigstar \: \boxed{ \sf { P(E) = \dfrac{n(E)}{n(S)} }}$

• P(E) = Probability of the occurrence of the event.

• n(E) = Number of favourable outcomes.

• n(S) = Number of possible outcomes.

⇒ Number of possible outcomes = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20} ⇒ 20

⇒ Number of favourable outcomes = Multiples of 3 or 7 from 1 to 20.

$\longrightarrow$ Multiples of 3 = 3,6,9,12,15,18,21 . . . .n

$\longrightarrow$ Multiples of 7 = 7,14,21 . . . .n

• Multiples of 3 upto 20 = {3,6,9,12,15,18} = 6
• Multiples of 7 upto 20 = {7,14} = 2

⇒ Number of favourable outcomes = {3,6,9,12,15,18,7,14} = 8

Now, calculating the probability that the ticket drawn has a number which is multiple of 3 or 7.

⇒ $\bigstar \: \boxed{ \sf { P(E) = \dfrac{n(E)}{n(S)} }}$

⇒ $\sf{ Probability = \cancel{\dfrac{8}{20}} }$

⇒ $\boxed{ \sf \red{ Probability = \dfrac{2}{5} }}$

[Note : Always write your answer in simplest form.]

Therefore, the probability that the ticket drawn has a number which is multiple of 3 or 7 is $\bf\red {\dfrac{2}{5}}$.