Probability 2 Conditional Probability For these 6 balls

Probability (2) Conditional Probability

For these 6 balls, if 2 are chosen randomly …. What is the probability they are a green and a red? First ball (from 6) P(G) = 2/6 then -> Second ball (from 5) P(R) = 1/5 P(G n R) = 2/6 x 1/5 = 1/15 P(R) = 1/6 then -> P(G) = 2/5 P(R n G) = 1/6 x 2/5 = 1/15 Either way there is a 1/15 chance of Green and Red The 2 events are not independent

For these 6 balls, if 2 are chosen randomly …. If the first one is Green, what is the probability of the second being Red? Second ball (from 5) P(R) = 1/5 This probability is CONDITIONAL on the first ball being green. CONDITIONAL PROBABILITIES are written: - P(R|G) = 1/5 Meaning “the probability of event R given event G has already happened”

For these 6 balls, if 2 are chosen randomly …. If the first one is Blue, what is the probability of the second being Red? P(R|B) = 1/5 If the first one is Red, what is the probability of the second being Red? P(R|R) = 0

For these 6 balls, if 2 are chosen randomly …. What is the probability the 2 nd is Red if the 1 st is Green? First ball (from 6) P(G) = 2/6 Second ball (from 5) P(R|G) = 1/5 P(G n R) = P(G) x P(R|G) = 2/6 x 1/5 = 1/15 Since. . . P(G Then. . . R) = P(G) x P(R|G) = P(R G) P(G) G)

Conditional Probability Rule P(B|A) = P(A B) P(A) “The probability that event B occurs, given event A has occurred is : the probability they both occur divided by the probability event A occurs”

Problem: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test? Define: F the event a pupil passed the first test S the event a pupil passes the second test P(F n S) = 25% = 0. 25 P(F) = 42% = 0. 42 P(S|F) = P(F S) P(F) P(S|F) = 0. 25 / 0. 42 = 0. 60 = 60%

P(B|A) = P(A B) P(A) Example 1: A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and a white marble is 0. 34, and the probability of selecting a black marble on the first draw is 0. 47. What is the probability of selecting a white marble on the second draw, given that the first marble drawn was black? Solution: P(White|Black) = P(Black and White) = P(Black) 0. 34 0. 47 = 0. 72 = 72%

P(B|A) = P(A B) P(A) Example 2: The probability that it is Friday and that a student is absent is 0. 03. Since there are 5 school days in a week, the probability that it is Friday is 0. 2. What is the probability that a student is absent given that today is Friday? Solution: P(Absent|Friday) = P(Friday and Absent) = 0. 03 = 0. 15 = 15% P(Friday) 0. 2

P(B|A) = P(A B) P(A) Example 3: At Kennedy Middle School, the probability that a student takes Technology and Spanish is 0. 087. The probability that a student takes Technology is 0. 68. What is the probability that a student takes Spanish given that the student is taking Technology? Solution: P(Spanish|Technology) = P(Technology and Spanish) P(Technology) = 0. 087 = 0. 13 0. 68 = 13%