Statistics: Two-Way Tables and Conditional Probability
Using counts to find conditional probabilities
Statistics: Two-Way Tables and Conditional Probability
Using counts to find conditional probabilities
Statistics - Grade 9-12
- 1
A class survey of 60 students is shown. Juniors: 18 play sports, 12 do not play sports, 30 total. Seniors: 15 play sports, 15 do not play sports, 30 total. Totals: 33 play sports, 27 do not play sports, 60 total. Find P(plays sports | junior).
Because the condition is junior, use the junior total as the denominator.
P(plays sports | junior) is 18/30, which simplifies to 3/5 or 0.60. This means 60% of the juniors in the survey play sports. - 2
Use the same class survey. Find P(senior | does not play sports).
P(senior | does not play sports) is 15/27, which simplifies to 5/9, or about 0.556. This means about 55.6% of the students who do not play sports are seniors. - 3
A reading survey of 120 students is shown. Reads daily: 28 like mystery books, 12 do not like mystery books, 40 total. Does not read daily: 32 like mystery books, 48 do not like mystery books, 80 total. Totals: 60 like mystery books, 60 do not like mystery books, 120 total. Find P(reads daily and likes mystery books) and P(likes mystery books | reads daily).
For an and probability, use the whole sample as the denominator. For a conditional probability, use the condition as the denominator.
P(reads daily and likes mystery books) is 28/120, which simplifies to 7/30. P(likes mystery books | reads daily) is 28/40, which simplifies to 7/10 or 0.70. - 4
Use the reading survey from Problem 3. Decide whether liking mystery books and reading daily appear to be independent events.
The events do not appear to be independent. P(likes mystery books) is 60/120 = 0.50, but P(likes mystery books | reads daily) is 28/40 = 0.70. Since these probabilities are not equal, the events are not independent. - 5
A transportation table for 200 students is partly missing. 9th grade: 54 use the bus, 46 do not use the bus, 100 total. 10th grade: 36 use the bus, missing number do not use the bus, 100 total. Totals: 90 use the bus, 110 do not use the bus, 200 total. Find the missing number and find P(uses the bus | 10th grade).
First complete the 10th grade row, then use the 10th grade total as the denominator.
The missing number is 64 because 100 - 36 = 64. P(uses the bus | 10th grade) is 36/100 = 0.36, so 36% of the 10th grade students use the bus. - 6
A club survey of 75 members is shown. Owns a tablet: 20 have a phone case, 10 do not have a phone case, 30 total. Does not own a tablet: 15 have a phone case, 30 do not have a phone case, 45 total. Totals: 35 have a phone case, 40 do not have a phone case, 75 total. If a member has a phone case, find the probability that the member owns a tablet.
The probability is 20/35, which simplifies to 4/7, or about 0.571. This means about 57.1% of members with a phone case own a tablet. - 7
Use the club survey from Problem 6. Decide whether owning a tablet and having a phone case appear to be independent events.
Compare the overall probability of having a phone case with the probability of having a phone case among tablet owners.
The events do not appear to be independent. P(has a phone case) is 35/75 = 7/15, but P(has a phone case | owns a tablet) is 20/30 = 2/3. Since the probabilities are not equal, the events are not independent. - 8
An after-school survey is shown. Has a quiet study space: 64 completed homework, 16 did not complete homework, 80 total. No quiet study space: 36 completed homework, 24 did not complete homework, 60 total. Totals: 100 completed homework, 40 did not complete homework, 140 total. Compare P(completed homework | has a quiet study space) and P(completed homework | no quiet study space).
P(completed homework | has a quiet study space) is 64/80 = 0.80. P(completed homework | no quiet study space) is 36/60 = 0.60. The completion rate is 20 percentage points higher for students with a quiet study space. - 9
Use the after-school survey from Problem 8. If a student completed homework, find the probability that the student had a quiet study space.
Because the condition is completed homework, use the total number who completed homework as the denominator.
The probability is 64/100 = 0.64. This means 64% of the students who completed homework had a quiet study space. - 10
In a sample of 500 drivers, 300 wear seat belts and 200 do not wear seat belts. Of the drivers who wear seat belts, 270 stop completely at a stop sign. Of the drivers who do not wear seat belts, 150 stop completely. Find P(stops completely | does not wear a seat belt) and P(does not wear a seat belt | stops completely).
P(stops completely | does not wear a seat belt) is 150/200 = 0.75. There are 270 + 150 = 420 drivers who stop completely, so P(does not wear a seat belt | stops completely) is 150/420 = 5/14, or about 0.357. - 11
A group has 160 students. There are 70 students in band and 90 students not in band. Of the band students, 80% own an instrument. Of the non-band students, 30% own an instrument. Complete the counts and find P(in band | owns an instrument).
Use the given percents to find the counts in each row before finding the conditional probability.
There are 56 band students who own an instrument because 0.80 x 70 = 56, and 14 band students who do not. There are 27 non-band students who own an instrument because 0.30 x 90 = 27, and 63 non-band students who do not. The total number who own an instrument is 83, so P(in band | owns an instrument) is 56/83, or about 0.675. - 12
A medical screening table for 1,000 people is shown. Has the condition: 45 positive test results, 5 negative test results, 50 total. Does not have the condition: 90 positive test results, 860 negative test results, 950 total. Totals: 135 positive test results, 865 negative test results, 1,000 total. Find P(has the condition | positive test result). Then explain why this is different from P(positive test result | has the condition).
P(has the condition | positive test result) is 45/135 = 1/3. This is different from P(positive test result | has the condition), which is 45/50 = 0.90, because the two probabilities use different conditions and therefore different denominators.