10.
[tex]P(\text{ice skating}\mid\text{6th grade})=\dfrac{P(\text{ice skating AND 6th grade})}{P(\text{6th grade})}[/tex]
155 of the total 545 students belong to the 6th grade. 46 of the total 545 students belong to the 6th grade and like ice skating. In terms of probability,
[tex]P(\text{6th grade})=\dfrac{155}{545}[/tex]
[tex]P(\text{ice skating AND 6th grade})=\dfrac{46}{545}[/tex]
So
[tex]P(\text{ice skating}\mid\text{6th grade})=\dfrac{\frac{46}{545}}{\frac{155}{545}}=\dfrac{46}{155}[/tex]
11. Similar setup. We know 171 of 545 students like skiing, and there are 74 8th graders that like skiing. So
[tex]P(\text{8th grade}\mid\text{skiing})=\dfrac{\frac{74}{545}}{\frac{171}{545}}=\dfrac{74}{171}[/tex]
