Question

according to data from the unesco institute for statistics, globally, 30% of adolescents are proficient in reading and mathematics. two adolescent students are selected at random, one from a school in the united states, and one from a school in europe. knowing that the proficiency of these two students in reading and mathematics is independent of one another: (enter your answers below without rounding them) a) what is the probability that both students are proficient in reading and mathematics? b) what is the probability that at least one of the students is proficient in reading and mathematics? c) what is the probability that neither student is proficient in reading and mathematics?

Explanation:

Step1: Calculate the probability for part a

Let \(P(A)\) be the probability that a student is proficient. Given \(P(A)=0.3\). Since the two events (proficiency of the US - student and proficiency of the European - student) are independent, we use the formula for the probability of the intersection of two independent events \(P(A\cap B)=P(A)\times P(B)\).
\(P(\text{both proficient}) = 0.3\times0.3\)

Step2: Calculate the probability for part b

First, find the probability that neither is proficient. The probability that a student is not proficient is \(P(\overline{A})=1 - 0.3=0.7\). The probability that neither is proficient is \(P(\overline{A}\cap\overline{B})=P(\overline{A})\times P(\overline{B})=0.7\times0.7 = 0.49\). Then, use the formula \(P(\text{at least one proficient})=1 - P(\text{neither proficient})\)
\(P(\text{at least one proficient})=1-0.7\times0.7\)

Step3: Calculate the probability for part c

Use the formula for the probability of the intersection of two independent events for non - proficiency. The probability that a student is not proficient is \(P(\overline{A}) = 1 - 0.3=0.7\).
\(P(\text{neither proficient})=(1 - 0.3)\times(1 - 0.3)\)

Answer:

a) \(0.09\)
b) \(0.51\)
c) \(0.49\)