Question

If a positive integer $n$ is picked at random from the positive integers less than or equal to $10$, what is the probability that $5n+3\le 14$ ?

• A. $0$
• B. $\frac{1}{10}$
• C. $\frac{1}{5}$
• D. $\frac{3}{10}$
• E. $\frac{2}{5}$

Answer 1

The correct option is C $\frac{1}{5}$

Given, $5n$ $+$ $3$ $\le$ $14$

$\Rightarrow 5n$ $\le$ $14$ $-$ $3$

$\Rightarrow (5$ $\times$ $n)$ $\le$ $11$

$\Rightarrow n$ $\le$ $\frac{11}{5}$

$\Rightarrow n$ $\le$ $2.2$

As ${}^{\prime }{n}^{\prime }$ is a positive integer, $n$ $>$ $0$

$0$ $<$ $n$ $\le$ $2.2$

${}^{\prime }{n}^{\prime }$ can only take integer values.

Hence, $n$ $=$ $1$ $or$ $n$ $=$ $2$

If ${}^{\prime }{n}^{\prime }$ is picked at random from the $10$ positive integers, only two values can satisfy the given condition.

Number of favorable outcomes $=$ $2$

Total number of possible outcomes $=$ $10$

From the above, probability that the given condition is satisfied $=$ $\frac{Numberoffavorableoutcomes}{Totalnumberofpossibleoutcomes}$

$=$ $\frac{2}{10}$

$=$ $\frac{1}{5}$

Therefore, the probability for ${}^{\prime }{n}^{\prime }$ that $5n$ $+$ $3$ $\le$ $14$ is ${}^{\prime }$$\frac{1}{5}$${}^{\prime }$.