Question

Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 4; B: the numbers add to 5; C: at least one of the numbers is 3; and D: the numbers do not add to 8. Express the given event in symbolic form

Answer 1

`A:\\R=4\hspace{8}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z`

`B:\\R+G=5\hspace{8}R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{5}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{3}and\hspace{3}R\\eq G`

`C:\\R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5}R =3 \hspace{3}or\hspace{3}G=3`

D:

`R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5};\hspace{3}R+G\\eq8`

Explanation:

Let:

`R=The\hspace{3}number\hspace{3}shows\hspace{3}by\hspace{3}the\hspace{3}red\hspace{3}dice\\G=The\hspace{3}number\hspace{3}shows\hspace{3}by\hspace{3}the\hspace{3}green\hspace{3}dice`

A:

This is easy, simply R=4, and we don't know the value of G, so:

`R=4\hspace{8}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z`

B: Be careful here, we know the numbers add to 5, so we don't know the exact value of R or G, because R could be 4 and G could be 1 or R could be 2 and green could be 3. However we can be sure that they can´t have the same value, because 5 is an odd number, if we add the same number to a number the result is always even: n+n=2n. So:

`R+G=5\hspace{8}R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{5}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{3}and\hspace{3}R\\eq G`

C: R is 3 or G is 3, that's the only thing we know, hence:

`R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5}R =3 \hspace{3}or\hspace{3}G=3`

D: Similar to B, we only know that the sum of R and G isn't 8:

`R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5};\hspace{3}R+G\\eq8`