Question

Math Questions!

Answer 1

$Q1.\\\Omega=\{(x;\ y)|\ x,\ y\in\{1;\ 2;\ 3;\ 4;\ 5;\ ;6\}\}\\\\\overline{\overline{\Omega}}=6\cdot6=36\\\\A=\{(1;\ 6);\ (2;\ 5);\ (3;\ 4);\ (4;\ 3);\ (5;\ 2);\ (6;\ 1)\}\\\\\overline{\overline{A}}=6\\\\P(A)=\dfrac{\overline{\overline{A}}}{\overline{\overline{\Omega}}}\Rightarrow P(A)=\dfrac{6}{36}\leftarrow Answer$


$Q2.\\3+4+8=15-number\ of\ all\ chips\\\\8\ white\ chips\\\\P(B)=\dfrac{8}{15}\cdot\dfrac{8}{15}=\dfrac{64}{225}\leftarrow Answer$


$Q3.\\25-number\ of\ 2\\56-number\ of\ heat\\100-number\ of\ all\\\\\dfrac{25}{100}\cdot\dfrac{56}{100}=\dfrac{1400}{10000}\leftarrow Answer$

Answer 2

$Question \ 1)$ 

$Two \ numbers \ written \ in \ a \ certain \ order \ are \ known \ as$
$ordered \ pairs.$ 

$Ordered \ pair \ that \ add \ up \ to \ 7 \ (6,1) (5,2) (4,3) (3,4) (2,5) (1,6)$

$P(A)= \dfrac{6}{36}$ 

$Question \ 2)$ $3+4+8=15 \ Total \ number \ of \ chips$

$Where \ 8 \ represents \ = white$ 

$P(B)= \dfrac{8}{15}* \dfrac{8}{15} = \dfrac{64}{225} = Solution$ 

$Question \ 3)$

$Convert \ to \ fraction \ and \ muliply$

$The \ numerator \ in \ a \ fraction \ represents \ the \ number \ of \ pieces$ 
$selected.$$If \ the \numerator \ is \ larger \ than \ the \ denominator, the \ number \ is \ larger$$than \ one.$

$The \ denominator \ in \ a \ fraction \ represents \ the \ total \ number$
$of \ equal \ size \ pieces \ that \ make \ up \ a \ whole.$

$\dfrac{25}{100} * \dfrac{56}{100}= \dfrac{1400}{10000}$