Question

tickets are numbered 1 to 100 tickets are reshuffled and one ticket is drawn at random . find the probability of drawing a ticket whose marked number is multiple of 3 or 5

Answer 1

$3multiple \: no \: are \: 33 \\ and \: 5 \: multiple \: no \: are \: 20 \\ 33 + 20 = 53 \\ 53 \div 100 = 0.53$

Answer 2

Answer:

$\frac{47}{100}$      

Step-by-step explanation:

Ticket are numbered 1 to 100

Total number of tickets = 100

Tickets  whose marked number is multiple of 3 = {3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99} =33

Tickets whose marked number is multiple of 5 = {5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100}

Tickets whose marked number is multiple of 5 excluding the multiple of 3={5,10,20,25,35,40,50,55,65,70,80,85,95,100}=14

So, The probability of drawing a ticket whose marked number is multiple of 3 or 5= $\frac{33}{100}+\frac{14}{100}$

      = $\frac{47}{100}$      

Hence The probability of drawing a ticket whose marked number is multiple of 3 or 5 is $\frac{47}{100}$