Separating Joint and Conditional Probability in a Survey
Ms. Grant’s class surveys 200 students about two activities. They find that 120 students are in the after-school program, 80 play a sport, and 50 students are in both. A student writes: ‘The probability a student plays a sport given that they are in the program is 50/80 = 62.5%, which is the same as the probability they are in the program given that they play a sport.’ Another student says the answer is 50/200 = 25% because 50 is the overlap. The class is preparing a proposal to adjust the after-school schedule and wants to use conditional probabilities accurately. Ms. Grant has a two-way table outline but some students fill it incorrectly, placing 50 in the total column. She wants to use the survey to distinguish joint, marginal, and conditional probabilities before the unit assessment.
1. Construct the correct two-way table and use it to compute P(sport | program) and P(program | sport).
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Let the rows be program: yes/no and the columns be sport: yes/no. Program yes total is 120; sport yes total is 80; both is 50. Therefore program yes and sport no is 120 − 50 = 70. Sport yes and program no is 80 − 50 = 30. Program no total is 200 − 120 = 80, so neither is 80 − 30 = 50. The table is: program yes/sport yes 50; program yes/sport no 70; program no/sport yes 30; program no/sport no 50. Then P(sport | program) = 50/120 = 5/12 ≈ 41.7%. P(program | sport) = 50/80 = 5/8 = 62.5%. The two conditional probabilities are different because they restrict to different groups. The joint probability P(program and sport) is 50/200 = 25%.
2. Explain the error in each student’s statement and identify which denominator is appropriate for each conditional question.
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The first student reverses the condition. The phrase ‘given that they are in the program’ restricts the sample space to the 120 program students, so the denominator is 120, not 80. The value 50/80 = 62.5% is actually P(program | sport), because it restricts to the 80 sport players. The second student computes 50/200 = 25%, which is the joint probability that a randomly selected student is in both activities. A conditional probability is not a fraction of the whole survey once the condition is known; it is a fraction of the conditioned group. For P(sport | program), use 50/120. For P(program | sport), use 50/80. For P(sport), use 80/200. Keeping track of whether the denominator is a row total, column total, or grand total prevents the reversal.
3. Design one follow-up question about the schedule proposal that requires a conditional probability and a caution about causation. What would a strong response include?
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Follow-up: ‘Among students in the program, 50 of 120 play a sport. Among students not in the program, 30 of 80 play a sport. Compare the two conditional probabilities and advise whether the program should be expanded to increase sports participation.’ A strong response computes 50/120 ≈ 41.7% and 30/80 = 37.5%, then notes that the difference is only about 4.2 percentage points. It uses the correct denominators for each group. It also cautions that this is an observational survey, so an association between program participation and sport participation does not establish that the program causes students to play sports. Other factors could explain the difference. The recommendation should be tentative and may call for more data rather than claiming that expanding the program will produce a large increase in sports participation.