48.3k views
4 votes
All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be proficient in reading, 78% were found to be proficient in mathematics, and 65% were found to be proficient in both reading and mathematics. A student is chosen at random. a. What is the probability that the student is proficient in mathematics but not in reading? b. What is the probability that the student is proficient in reading but not in mathematics? c. What is the probability that the student is proficient in neither reading nor mathematics?

2 Answers

4 votes

Final answer:

The probability of a student being proficient in mathematics but not in reading is 13%, in reading but not in mathematics is 20%, and in neither reading nor mathematics is 2%.

Step-by-step explanation:

To solve the student's query, we'll use the principle that the probability of an event is the number of favorable outcomes divided by the total number of outcomes. We can apply the addition rule for probabilities, which states that the probability of either of two mutually exclusive events occurring is the sum of their individual probabilities, minus the probability of both events happening.

a. Mathematics but not Reading

Let P(M) be the probability the student is proficient in mathematics, P(R) be the probability the student is proficient in reading, and P(M & R) be the probability the student is proficient in both. The question asks for P(M) - P(M & R), the probability of being proficient in mathematics but not in reading. That is 78% - 65% = 13%.

b. Reading but not Mathematics

Similarly, the probability of a student being proficient in reading but not mathematics is P(R) - P(M & R), which equals 85% - 65% = 20%.

c. Proficient in Neither

To find the probability of a student being proficient in neither subject, we can find the probability of a student being proficient in at least one subject and then subtracting this from 100%. The probability of being proficient in at least one subject is P(R) + P(M) - P(M & R), or 85% + 78% - 65% = 98%. Thus, the probability of being proficient in neither is 100% - 98% = 2%.

User HungryArthur
by
5.3k points
6 votes

Answer:

a. 13%

b. 20%

c. 2%

Step-by-step explanation:

The best way to solve this problem is by drawing a Venn diagram. Draw a rectangle representing all the fourth-graders. Draw two overlapping circles inside the rectangle. Let one circle represent proficiency in reading. This circle is 85% of the total area (including the overlap). And let the other circle represent proficiency in math. This circle is 78% of the total area (including the overlap). The overlap is 65% of the total area.

a. Since the overlap is 65%, and 78% are proficient in math, then the percent of all students who are proficient in math but not reading is the difference:

78% − 65% = 13%

b. Since the overlap is 65%, and 85% are proficient in reading, then the percent of all students who are proficient in reading but not math is the difference:

85% − 65% = 20%

c. The percent of students not proficient in reading or math is 100% minus the percent proficient in only reading minus the percent proficient in only math minus the percent proficient in both.

100% − 20% − 13% − 65% = 2%

See attached illustration (not to scale).

All the fourth-graders in a certain elementary school took a standardized test. A-example-1
User Tushar Mishra
by
5.3k points