Question

suppose that one student is randomly selected during lunch time. what is the experimental probability if students the brought lunch from house is 55 and students that order school lunch is 45

Answer 1

Answer:

p(A) = 0.55

p(b) = 0.45

Step-by-step explanation:

Experimental probability = $\frac{Number of favorable outcomes}{Total number of outcomes}$

total students = 55 + 45   ⇒ 100

let A be the probability that students brought lunch from house = $\frac{55}{100}$ ⇒ 0.55

let B be the probability that students order school lunch = $\frac{45}{100}$ ⇒ 0.45