Question Number 216323 by Red1ight last updated on 04/Feb/25

$$\mathrm{Let}\:\mathrm{10}\geqslant{x},{y}\geqslant\mathrm{0}\:\mathrm{and}\:{x},{y}\in\mathbb{R} \\ $$$$\mathrm{Find} \\ $$$$\left.{a}\right){P}\left({x}−\mathrm{2}>{y}\right) \\ $$$$\left.{b}\right){P}\left({x}+\mathrm{2}<{y}\right) \\ $$
Commented by AntonCWX last updated on 04/Feb/25

$${is}\:{P}\:{probability}? \\ $$
Commented by Red1ight last updated on 04/Feb/25

$${yes}. \\ $$
Answered by mehdee7396 last updated on 04/Feb/25

$${answer}\:\left({a}\right)\&\left({b}\right)\:=\frac{{A}}{{S}}=\:\mathrm{0}/\mathrm{32}\: \\ $$
Commented by mehdee7396 last updated on 04/Feb/25

Commented by mehdee7396 last updated on 04/Feb/25
