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Probability (2) Conditional Probability

Probability (2) Conditional Probability. Second ball (from 5). First ball (from 6). For these 6 balls, if 2 are chosen randomly …. What is the probability they are a green and a red?. P(R) = 1/5. P(G) = 2/6. P(G n R) = 2/6 x 1/5 = 1/15. then ->. P(G) = 2/5. P(R) = 1/6.

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Probability (2) Conditional Probability

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  1. Probability (2)Conditional Probability

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

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

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

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

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

  7. 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? P(S|F) = P(F S) P(F) 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) = 0.25 / 0.42 = 0.60 = 60%

  8. 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) = 0.34 = 0.72 = 72% P(Black) 0.47

  9. 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

  10. 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) = 0.087 = 0.13 = 13% P(Technology) 0.68

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