Conditional Probability

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

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”. Find (i) P(L) (ii) P(F ∩ L) (iii) P(F L)

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

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

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

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

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

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. P( male ) = 58% P( male and jogs ) = 20% Let A = male. Let B = jogs. Write: P( A | B ) = P( A and B ) P( A ) = Substitute 0.2 for P(A and B) and 0.58 for P(A). 0.344 Simplify. 0.2 0.58 The probability that a male respondent jogs is about 34%.

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%.

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%.

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.

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

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

(a) Calculate the probability of him arriving on time.

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

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

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

Example Walk Bus Bike Car Men 10 12 6 26 Women 7 18 4 17 Total 100 A sample of 100 adults were asked how they travelled to work. The results are shown in the table. Walk Bus Bike Car Men 10 12 6 26 Women 7 18 4 17 Total 100 Find (i) (ii) (iii) (iv) P(M) P(M/ and C) 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 C) P(C M/)

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

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

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

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

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.