Question

Se considera multimea A=(x∈N* / 2(17-2x)-5 ≥-31-x) Atunci probabilitatea ca alegand la intamplare un element din multime,aceasta sa fie numar prim este egala cu..

Answer 1

$A=\{ x \in N^* \ / \ 2(17-2x)-5 \geq -31-x \} \\\\\\ 2(17-2x)-5 \geq -31-x \\\\ 34-4x-5 \geq -31 -x \\\\ 29 -4x \geq -31-x \\\\ -4x+x \geq -31-29 \\\\ -3x \geq -60 \ \ \ |:(-3) \\\\ x \leq 20 \\ x \in N^* \ \ \ \Longrightarrow x \in \{1;2;3;...;20 \} -C_{p} \\\\\\ Probabilitatea=\frac{Nr. \ cazuri \ favorabile}{Nr. \ cazuri \ posibile} \\\\\\ \hbox{Numerele prime din multimea data sunt:} \\\\ S= \{2;3;5;7;11;13;17;19 \} \longrightarrow 9 \ nr. \ in \ total \\\\\\ \boxed{P=\frac{9}{20} \% }$

Answer 2

$\it A=\{x\in\mathbb{N}^*| 2(17-2x)-5\geq-31-x\}$

$\it 2(17-2x)-5\geq-31-x \Leftrightarrow 34-4x-5\geq-31-x \Leftrightarrow$

$\it \Leftrightarrow 34-5+31 \geq 4x-x \Leftrightarrow 60 \geq 3x|_{:3} \Leftrightarrow 20 \geq x \Leftrightarrow x \leq 20$

$\it A=\{1, 2, 3, ... , 20\}$

$\it Cazuri\ \ favorabile: \ 2, \ 3,\ 5,\ 7, \ 11,\ 13,\ 17,\ 19$

$\it Avem\ 8\ cazuri\ favorabile\ si\ 20\ cazuri\ posibile$

$\it p = \dfrac{8}{20} = \dfrac{2}{5}$

$\it p = \dfrac{8}{20} =\dfrac{40}{100} =40\%$