The probability of selecting a red ball at random from the jar that contains only red,

It is given that,

$P($ getting a red ball $)=\frac{1}{4}$ and $P($ getting a blue ball $)=\frac{1}{3}$

Let P(getting an orange ball) be x.

Since, there are only 3 types of balls in the jar, the sum of probabilities of all the three balls must be 1.

$\therefore \frac{1}{4}+\frac{1}{3}+x=1 \backslash$

$\Rightarrow x=1-\frac{1}{4}-\frac{1}{3}$

$\Rightarrow x=\frac{12-3-4}{12}$

$\Rightarrow x=\frac{5}{12}$

$\therefore \mathrm{P}($ getting an orange ball $)=\frac{5}{12}$

Let the total number of balls in the jar be n.

$\therefore \mathrm{P}($ getting an orange ball $)=\frac{10}{n}$

$\Rightarrow \frac{10}{n}=\frac{5}{12}$

$\Rightarrow n=24$

Thus, the total number of balls in the jar is 24.