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Sagot :
The same probability*
Write firstly what results it's able to get and their sum:
.....1....2.....3.....4....5.....6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
As you can see there are four options to get "5" and also four options to get "9".
So the B is proper answer.
Sum 5 and sum 9 have the same probability.
Write firstly what results it's able to get and their sum:
.....1....2.....3.....4....5.....6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
As you can see there are four options to get "5" and also four options to get "9".
So the B is proper answer.
Sum 5 and sum 9 have the same probability.
[tex]\Omega=\{(x;y):x;y\in\{1;\ 2;\ 3;\ 4;\ 5;\ 6\}\}\\\\\overline{\overline{\Omega}}=6^2=36\\\\A=\{(1;2);\ (2;1);\ (1;3);\ (3;1);\ (2;\ 2)\};\ \overline{\overline{A}}=5;\ P(A)=\frac{5}{36}\\\\B=\{(1;4);\ (4;1);\ (2;3);\ (3;2);\ (3;6);\ (6;3);\ (4;5);\ (5;4)\};\\\overline{\overline{B}}=8;\ P(B)=\frac{8}{36}\\\\C=\{(1;4);\ (4;1);\ (2;3);\ (3;2);\ (2;6);\ (6;2);\ (3;5);\ (5;3);\ (4;4)\}\\\overline{\overline{C}}=9;\ P(C)=\frac{9}{36}[/tex]
[tex]D=\{(4;6);\ (4;6);\ (5;5);\ (6;6)\};\ \overline{\overline{D}}=4;\ P(D)=\frac{4}{36}[/tex]
[tex]A.\\3\to(1;2);\ (2;1)\to\frac{2}{36}\\4\to(1;3);\ (3;1);\ (2;2)\to\frac{3}{36}\\\\B.\\5\to(1;4);\ (4;1);\ (2;3);\ (3;2)\to\frac{4}{36}\\9\to(3;6);\ (6;3);\ (4;5);\ (5;4)\to\frac{4}{36}[/tex]
[tex]C.\\5\to(1;4);\ (4;1);\ (2;3);\ (3;2)\to\frac{4}{36}\\8\to(2;6);\ (6;2);\ (3;5);\ (5;3);\ (4;4)\to\frac{5}{36}\\\\D.\\10\to(4;6);\ (6;4);\ (5;5)\to\frac{3}{36}\\12\to(6;6)\to\frac{1}{36}[/tex]
[tex]D=\{(4;6);\ (4;6);\ (5;5);\ (6;6)\};\ \overline{\overline{D}}=4;\ P(D)=\frac{4}{36}[/tex]
[tex]A.\\3\to(1;2);\ (2;1)\to\frac{2}{36}\\4\to(1;3);\ (3;1);\ (2;2)\to\frac{3}{36}\\\\B.\\5\to(1;4);\ (4;1);\ (2;3);\ (3;2)\to\frac{4}{36}\\9\to(3;6);\ (6;3);\ (4;5);\ (5;4)\to\frac{4}{36}[/tex]
[tex]C.\\5\to(1;4);\ (4;1);\ (2;3);\ (3;2)\to\frac{4}{36}\\8\to(2;6);\ (6;2);\ (3;5);\ (5;3);\ (4;4)\to\frac{5}{36}\\\\D.\\10\to(4;6);\ (6;4);\ (5;5)\to\frac{3}{36}\\12\to(6;6)\to\frac{1}{36}[/tex]
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