Question

number y is chosen from a set of positive integers less than 10. What is the probability
nat ((10 + x)/x) > X?​

Answer 1

Step-by-step explanation:

10+x/x>x

10+1/1>1

10+2/2>2

10+3/3>3

10+4/4<4........when X=4 wrong❌

3/10

Answer 2

GivenL a number y is chosen from 1 to 10.

To find: the probability that$\[\frac{{\left( {10 + x} \right)}}{x} > x\]$?

Solution:

Understand that total possible outcomes$=10$

Find the favourable outcomes.

Put $x=1$.

$\[ \Rightarrow \frac{{\left( {10 + 1} \right)}}{1} > 1 = 9 > 1\]$

Put$x=2$

$\[ \Rightarrow \frac{{\left( {10 + 2} \right)}}{2} > 2 = 6 > 2\]$

Put $x=3$.

$\[ \Rightarrow \frac{{\left( {10 + 3} \right)}}{3} > 3 = 4.33 > 3\]$

Put $x=4$

$\[ \Rightarrow \frac{{\left( {10 + 4} \right)}}{4} > 4 = 3.5 < 4\]$

Observe that, $\[\frac{{\left( {10 + x} \right)}}{x} > x\]$ hold true only when $x=1,2,3$.

Therefore, number of favourable outcome$=3$

Find the probability that $\[\frac{{\left( {10 + x} \right)}}{x} > x\]$?.

Probability$\[ = \frac{{favorable\,outcome}}{{Total\,Outcome}}\]$

                 $\[ = \frac{3}{{10}}\]$

Hence, the probability that $\[\frac{{\left( {10 + x} \right)}}{x} > x\]$ is $\[\frac{3}{{10}}\]$.