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Conditional Probability

Conditional Probability. P (A B) means. “the probability that event A occurs given that B has occurred”. Find ( i ) P (L) (ii) P (F ∩ L) (iii) P (F L). Ex 1. The following table gives data on the type of car, grouped by gas consumption, owned by 100 people. Low. Medium. High.

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Conditional Probability

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  1. Conditional Probability

  2. P(A B) means “the probability that event A occurs given that B has occurred”.

  3. Find (i) P(L) (ii) P(F ∩L)(iii) P(FL) Ex 1. The following table gives data on the type of car, grouped by gas consumption, owned by 100 people. Low Medium High Total Male 12 33 7 Female 23 21 4 100 One person is selected at random. L is the event “the person owns a low rated car” F is the event “a female is chosen”.

  4. 12 23 Find (i) P(L) (ii) P(F ∩L)(iii) P(FL) Low Low Medium High Total Solution: Male 12 33 7 Female 23 21 4 35 100 100 (leave the answer as fraction (i) P(L) =

  5. Find (i) P(L) (ii) P(F and L)(iii) P(FL) Low Medium High Total Solution: Male 12 33 7 Female 23 23 21 4 100 100 (i) P(L) = The probability of selecting a female with a low rated car. (ii) P(F ∩L) =

  6. Find (i) P(L) (ii) P(F and L)(iii) P(FL) (iii) P(F L) = Low Medium High Total Solution: Male 12 12 33 7 Female 23 23 21 4 35 100 (i) P(L) = (ii) P(F ∩ L) = The probability of selecting a female given the car is low rated.

  7. Find (i) P(L) (ii) P(F and L)(iii) P(FL) (iii) P(F L) = Low Medium High Total Solution: Male 12 33 7 Female 23 21 4 100 (i) P(L) = (ii) P(F ∩ L) =

  8. If you don’t have a table, you can use this formula

  9. P( male ) = 58% P( male and jogs ) = 20% Let A = male. Let B = jogs. P( A and B ) P( A ) Write: P( A | B ) = 0.2 0.58 = Substitute 0.2 for P(A and B) and 0.58 for P(A). 0.344 Simplify. Conditional Probability Researchers asked people who exercise regularly whether they jog or walk. Fifty-eight percent of the respondents were male. Twenty percent of all respondents were males who said they jog. Find the probability that a male respondent jogs. The probability that a male respondent jogs is about 34%.

  10. Using Tree Diagrams Jim created the tree diagram   after examining years of weather observations in his hometown. The diagram shows the probability of whether a day will begin clear or cloudy, and then the probability of rain on days that begin clear and cloudy. a. Find the probability that a day will start out clear, and then will rain. The path containing clear and rain represents days that start out clear and then will rain. P(clear and rain) = P(rain | clear) • P(clear) = 0.04 • 0.28 = 0.011 The probability that a day will start out clear and then rain is about 1%.

  11. Conditional Probability (continued) b. Find the probability that it will not rain on any given day. The paths containing clear and no rain and cloudy and no rain both represent a day when it will not rain. Find the probability for both paths and add them. P(clear and no rain) + P(cloudy and no rain) = P(clear) • P(no rain | clear) + P(cloudy) • P(no rain | cloudy) = 0.28(.96) + .72(.69) = 0.7656 The probability that it will not rain on any given day is about 77%.

  12. Ex 3 In November, the probability of a man getting to work on time if there is fog on 285 is . If the visibility is good, the probability is . The probability of fog at the time he travels is . (a) Calculate the probability of him arriving on time. (b) Calculate the probability that there was fog given that he arrives on time.

  13. P(T F) P(T F/) P(F) On time T F Fog Not on time T/ On time T No Fog F/ Not on time T/

  14. T F T/ P(T F) T P(T F/) F/ T/ P(F) Because we only reach the 2nd set of branches after the 1st set has occurred, the 2nd set must represent conditional probabilities.

  15. T F T/ T F/ T/ (a) Calculate the probability of him arriving on time.

  16. T F T/ T F/ T/ (a) Calculate the probability of him arriving on time. ( foggy and he arrives on time )

  17. T F T/ T F/ T/ (a) Calculate the probability of him arriving on time. ( not foggy and he arrives on time )

  18. We need Getting to work Fog on 285 T F From part (a), (b) Calculate the probability that there was fog given that he arrives on time.

  19. One person is picked at random. M is the event “the person is a man” C is the event “the person travels by car” P(M) P(M/ and C) P(C M/) Find (i) (ii) (iii) (iv) P(M C) Example A sample of 100 adults were asked how they travelled to work. The results are shown in the table.

  20. (i) P(M) = (ii) P(M C) = P(M C) Solution: Walk Bus Bike Car Men 10 12 6 26 54 Women 7 18 4 17 Total 100 100 P(M) P(M/ and C) P(C M/) Find (i) (ii) (iii) (iv) (i)

  21. (iii) P(M/ and C) = P(M C) = P(M C) Solution: Walk Bus Bike Car Men 10 12 6 26 54 Women 7 18 4 17 Total 43 100 P(M) P(M/ and C) P(C M/) Find (i) (ii) (iii) (iv) (i) P(M) = (ii)

  22. P(C M/) = (iv) P(M C) = P(M C) Solution: Walk Bus Bike Car Men 10 12 6 26 54 Women 7 18 4 17 Total 43 100 100 P(M) P(M/ and C) P(C M/) Find (i) (ii) (iii) (iv) (i) P(M) = (ii) (iii) P(M/ and C) =

  23. (iv) P(M C) = P(M C) Solution: Walk Bus Bike Car Men 10 12 6 26 54 Women 7 18 4 17 46 Total 43 100 P(M) P(M/ and C) P(C M/) Find (i) (ii) (iii) (iv) (i) P(M) = (ii) P(C M/) = (iii) P(M/ and C) =

  24. Independent Events If the probability of the occurrence of event A is the same regardless of whether or not an outcome B occurs, then the outcomes A and B are said to be independent of one another. Symbolically, if then A and B are independent events.

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