1. IT 102: SOLVING CONDITIONAL PROBABILITY QUESTIONS Created by Jonathan Hsu

2. GEORGE MASON UNIVERSITY WHAT IS A CONDITIONAL PROBABILITY? Conditional probability is a probability calculation that is contingent on an initial condition which limits the sample space. Conditional probability is written as P(2 nd Event | 1 st Event), read as the probability of the second event occurring given the first event. Alternatively the conditional probability may be read as: Assuming the first event, what is the probability of the second event occurring? For example, the probability of a randomly selected day being Saturday is 1/7; however, assuming the randomly selected day is a weekend, the new conditional probability is 1/2. This scenario is written as P(Saturday | Weekend) The conditional statement reduces the sample space from 7 to 2 because there are two days that are elements of the set of days that are weekends {Saturday, Sunday} Once a condition is set, the new sample space will have a total probability of 1.0. So, P(Not Saturday | Weekend) + P(Saturday | Weekend) = 1 This conditional probability only exists within the assumption of the condition, it does not represent the overall probability of a randomly selected day being a weekend and Saturday. P(A | B) ≠ P(A ∩ B)

3. GEORGE MASON UNIVERSITY UNDERSTANDING A TREE DIAGRAM The tree diagram outlines the previous scenario… Weekend Weekday Saturday Not Saturday Saturday Not Saturday Initial Probability P(A) Conditional Probability P(B | A) Intersectional Probability P(A ∩ B) P(Weekend) = 2/7 P(Saturday | Weekend) = 1/2 P(Not Saturday | Weekend) 1/2 P(Weekday) = 5/7 P(Saturday | Weekday) = 0/5 P(Not Saturday | Weekday) = 5/5 Note: The sum of each conditional probability sample space is 1.0

4. GEORGE MASON UNIVERSITY UNDERSTANDING A TREE DIAGRAM, CONTD. The tree diagram outlines the previous scenario… Weekend Weekday Saturday Not Saturday Saturday Not Saturday Initial Probability P(A) Conditional Probability P(B | A) Intersectional Probability P(A ∩ B) P(Weekend ∩ Saturday) = 1/7 P(Weekend ∩ Not Saturday) =1/7 P(Weekday ∩ Saturday) = 0/7 P(Weekday ∩ Not Saturday) = 5/7 Note: The sum of all intersectional probabilities is 1.0

5. GEORGE MASON UNIVERSITY SOLVING A NEW PROBLEM What is the probability that the sum of two dice will be greater than 8 given that the first die is a 6? P(D 1 + D 2 > 8 | D 1 is 6)

6. GEORGE MASON UNIVERSITY ESTABLISHING THE INITIAL SAMPLE SPACE 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 What is the probability that the sum of two dice will be greater than 8 given that the first die is a 6? P(D 1 + D 2 > 8 | D 1 is 6) Second Roll First Roll Sum of Both Rolls Sample Space = 36

7. GEORGE MASON UNIVERSITY REDUCING SAMPLE SPACE BASED ON CONDITION 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 What is the probability that the sum of two dice will be greater than 8 given that the first die is a 6? P(D 1 + D 2 > 8 | D 1 is 6) Second Roll First Roll Sum of Both Rolls Conditional Sample Space = 6

8. GEORGE MASON UNIVERSITY IDENTIFYING FAVORABLE AND TOTAL OUTCOMES 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 What is the probability that the sum of two dice will be greater than 8 given that the first die is a 6? P(D 1 + D 2 > 8 | D 1 is 6) Second Roll First Roll Sum of Both Rolls Conditional Sample Space = 6 Favorable Total 4646 =

9. GEORGE MASON UNIVERSITY What if we flip the probability? P(D 1 is 6 | D 1 + D 2 > 8) This is Bayes’ Theorem…

10. GEORGE MASON UNIVERSITY BAYES’ THEOREM Bayes’ Theorem is no different than any other probability question, it is simply However, because the total (or sample space) is across multiple branches, they must be added together. If it helps, our normal probability equation can be reimagined as Favorable Total Favorable Favorable + Unfavorable 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 Second Roll First Roll 4 4 + 6 2525 =

11. GEORGE MASON UNIVERSITY SOLVING FROM A TREE DIAGRAM Mute City Elementary has 25 1 st grade students, 27 2 nd grade students, and 31 3 rd grade students. 15 of the 1 st graders are boys, 14 of the 2 nd graders are boys, and 20 of the 3 rd graders are boys. What is the probability a student is in second grade given they are a girl? 1 st (25) 2 nd (27) 3 rd (31) Boy (15) Girl (10) Boy (14) Girl (13) Boy (20) Girl (11) 1 st ∩ Boy = (25/83)(15/25) ≈ 0.1807 1 st ∩ Girl = (25/83)(10/25)≈ 0.1204 2 nd ∩ Boy = (27/83)(14/27) ≈ 0.1686 2 nd ∩ Girl = (27/83)(13/27)≈ 0.1566 3 rd ∩ Boy = (31/83)(20/31)≈ 0.2409 3 rd ∩ Girl = (31/83)(11/31)≈ 0.1325 83 Total 0.9997 ++

12. GEORGE MASON UNIVERSITY SOLVING FROM A TREE DIAGRAM P(2 nd Grade | G) The initial event (girl) sets the sample space. The second event (2 nd grade) sets the favorable outcomes for the probability calculation. 1 st (25) 2 nd (27) 3 rd (31) Boy (15) Girl (10) Boy (14) Girl (13) Boy (20) Girl (11) 1 st ∩ Boy = (25/83)(15/25) ≈ 0.1807 1 st ∩ Girl = (25/83)(10/25)≈ 0.1204 2 nd ∩ Boy = (27/83)(14/27) ≈ 0.1686 2 nd ∩ Girl = (27/83)(13/27)≈ 0.1566 3 rd ∩ Boy = (31/83)(20/31)≈ 0.2409 3 rd ∩ Girl = (31/83)(11/31)≈ 0.1325.1566.1566 +.1204 +.1325 Favorable Favorable + Unfavorable.1566.4095.1566.4095.3824 ====