7th grader needs help

Hello,Nostradamus!

Quote:

A die is rolled. What's the probability that the number is odd?

With one die, there are six outcomes: . $1, 2, 3, 4, 5, 6$

Three of them $\{1,3,5\}$ are odd.

Therefore: . $P(\text{odd}) \:=\:\frac{3}{6}\:=\:\frac{1}{2}$

Quote:

Draw a chart for when two dice are rolled.

. . $\begin{array}{cccccc}(1,1)&(1,2)&(1,3)&(1,4)&(1,5) &(1,6)\\(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\(3,1) &(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\(4,1)&(4,2)&(4,3)& (4,4)&(4,5)&(4,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&( 5,6) \\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{array}$

Quote:

Find these probabilities:
P(both odd)
P(both prime)
P(sum of 7)

There are 36 possible outcomes.

Both odd
There are nine cases: . $\begin{array}{ccc}(1,1)&(1,3)&(1,5)\\(3,1)&(3,3)&( 3,5)\\(5,1)&(5,3)&(5,5)\end{array}$

$P(\text{both odd}) \:=\:\frac{9}{36} \:=\:\frac{1}{4}$

Both prime
There are nine cases: . $\begin{array}{ccc}(2,2)&(2,3)&(2,5)\\(3,2)&(3,3)&( 3,5)\\(5,2)&(5,3)&(5,5)\end{array}$

$P(\text{both prime}) \:=\:\frac{9}{36} \:=\:\frac{1}{4}$

Sum of 7

There are six cases: . $\begin{array}{cccccc}(1,6)&(2,5)&(3,4)&(4,3)&(5,2) &(6,1)\end{array}$

$P(\text{sum of 7}) \:=\:\frac{6}{36}\:=\:\frac{1}{6}$