Answer:
70%
Step-by-step explanation:
Given
Number of Siblings: || 0 || 1 || 2 || 3
Number of Students: || 4 || 18 || 10 || 8
Required
Determine the probability of a student having at least one but not more than 2 siblings
First, we have to determine the total number of 10th grade students
[tex]Total = 4 + 18 + 10 + 8[/tex]
[tex]Total = 40[/tex]
The probability of a student having at least one but not more than 2 siblings = P(1) + P(2)
Solving for P(1)
P(1) = number of students with 1 sibling / total number of students
From the given parameters, we have that:
Number of students with 1 sibling = 18
So:
[tex]P(1) = \frac{18}{40}[/tex]
Solving for P(2)
P(2) = number of students with 2 siblings / total number of students
From the given parameters, we have that:
Number of students with 2 siblings = 10
So:
[tex]P(2) = \frac{10}{40}[/tex]
[tex]P(1) + P(2) = \frac{18}{40} + \frac{10}{40}[/tex]
Take LCM
[tex]P(1) + P(2) = \frac{18 + 10}{40}[/tex]
[tex]P(1) + P(2) = \frac{28}{40}[/tex]
Divide numerator and denominator by 4
[tex]P(1) + P(2) = \frac{7}{10}[/tex]
[tex]P(1) + P(2) = 0.7[/tex]
Convert to percentage
[tex]P(1) + P(2) = 70\%[/tex]
Hence, the required probability is 70%
