Subject: General Aptitude · Chapter: Probability · Exam: · Marks: · Difficulty:
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3 ?
(a)$\frac{3}{10}$
(b)$\frac{3}{20}$
(c)$\frac{2}{5}$
(d)$\frac{1}{2}$
Answer
Answer (as printed): A
Explanation
Here, $S=\{1,2,3,4, \ldots . ., 19,20\}$. Let $E=$ event of getting a multiple of $3=\{3,6,9,12,15,18\}$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{6}{20}=\frac{3}{10} .$$
Explanation as extracted from the printed page; notation may be imperfect.