1) 150 students were surveyed.
2) From the table sketched above we can deduce that 4 pets were surveyed the leasest.
[tex]\text{Probability}=\frac{Number\text{ of required outcomes}}{Total\text{ number of outcomes}}[/tex]
3) The total number of students with 2pets= 39
while the total number of pets of all the grades= 150
[tex]P(\text{student with 2pets)=}\frac{39}{150}=\text{ }\frac{\text{13}}{50}=0.26[/tex]
4) P(student with 8th grade)= ?
The total number of student in 8th grade= 50
while the total number of students= 150
[tex]\begin{gathered} P(\text{student in the 8th grade)= }\frac{50}{150}=\frac{1}{3}\text{ } \\ =\text{ 0.33} \end{gathered}[/tex]
5) The total number of students in 7th grade=50
Total number of students= 150
[tex]\begin{gathered} P(\text{student in 7th grade or has 1 pet)= P(student in 7th grade)} \\ \text{plus the P( student with 1 pet)} \end{gathered}[/tex]
The total number of student that has one pet= 62
[tex]\begin{gathered} P(7th\text{ grade)= }\frac{50}{150}=\frac{1}{3}=\text{ 0.33} \\ P(1\text{pet)}=\text{ }\frac{\text{62}}{150}=\text{ 0.41} \\ P(7th\text{ grade and 1pet) = }\frac{22}{150} \\ P(7th\text{ grade or 1 pet)= 0.33+0.41 - 0.15= }0.59 \end{gathered}[/tex]
6) Total number of students in the 6th grade with 3pets is a conditional statement.
[tex]\begin{gathered} P(6th\text{ grade with 3pets)=}\frac{P(6th\text{ grade}\cap3pets)}{\text{Total number of 3pets}} \\ =\text{ }\frac{5}{10}=\frac{1}{2}=\text{ 0.5} \end{gathered}[/tex]
7) Total number of students that has 1 pet and in 7th grade is a conditional statement.
[tex]\begin{gathered} P(1\text{ pet and in 7th grade)}=\frac{P(1pet\cap7thgrade)}{\text{Total number of students in the 7th grade}} \\ =\text{ }\frac{22}{50}=0.44 \end{gathered}[/tex]
