Probability Ques 34
Question-
There are 140 tickets (numbered 1 to 140) in a bowl. Find the probability of choosing a ticket which bears multiple of either 3 or 7.
(1) $\frac{4}{7}$
(2) $\frac{5}{7}$
(3) $\frac{3}{49}$
(4) $\frac{6}{7}$
(5) $\frac{3}{7}$
IBPS RRBs Officer CWE (Prelim Exam)
Show Answer
Correct Answer: (5)
Solution: (5)
From 1 to 140 :
Number of multiples of $3 \Rightarrow 46$
Number of multiples of $7 \Rightarrow 20$
Number of multiples of $21 \Rightarrow 6$
$\therefore \quad$ Number of multiples of 3 or
$7=46+20-6=60$
$\therefore$ Required probability $=\frac{60}{140}=\frac{3}{7}$
BETA
