Question

Find the probability that a number selected at random from the numbers $1,2,3,4,5,...,34,35$ is a
$\left(i\right)$ Prime number
$\left(ii\right)$ Multiple of $7$
$\left(iii\right)$ Multiple of $3$ or $5$

• A. $\left(i\right)\frac{12}{17}\left(ii\right)\frac{1}{7}\left(iii\right)\frac{1}{5}$
• B. $\left(i\right)\frac{11}{35}\left(ii\right)\frac{1}{7}\left(iii\right)\frac{16}{35}$
• C. $\left(i\right)\frac{17}{35}\left(ii\right)\frac{1}{7}\left(iii\right)\frac{19}{36}$
• D. $\left(i\right)\frac{14}{17}\left(ii\right)\frac{1}{7}\left(iii\right)\frac{11}{23}$

Answer 1

Answer: (B)

$\left(i\right)\frac{11}{35}\left(ii\right)\frac{1}{7}\left(iii\right)\frac{16}{35}$

Since, total numbers $=35$
(1) prime numbers from $1$ to $35$ are:
$2,3,5,7,11,13,17,19,23,29,31$
Total prime no.'s $=11$
$\therefore$ favorable outcomes $=11$
$\therefore$ probability $=$ $\frac{11}{35}$
(2) Multiples of $7$ from $1$ to $35$ are:
$7,14,21,28,35$
Total multiples of $7=5$
$\therefore$ favorable outcomes $=5$
$\therefore$ probability = $\frac{5}{35}=\frac{1}{7}$
(3) Multiples of $3$ or $5$ from $1$ to $35$ are:
$3,5,6,9,10,12,15,18,20,21,24,25,27,30,33,35$
Total multiples of $3$ or $5=16$
$\therefore$ favorable outcomes $=16$
$\therefore$ probability $=$ $\frac{16}{35}$