Notes 6.1: Conditional Probability



Learning Target Conditional Probability? I can determine conditional probabilities P(B|A) ?Conditional Probability: Determining the probability of one event after another has occurred.The probability of event B occurring knowing that event A has already occurred. This is read "the probability of B given A" and is written:?P(B|A)The formula for P(B|A) = P(A and B)P(A )(A and B means it has to be in BOTH )The probability of event A occurring knowing that event B has already occurred.This is read "the probability of A given B" and is written:?P(A|B)The formula for P(A|B) = P(A and B)P( B) (A and B means it has to be in BOTH)Since we are given that an event has occurred, we have a reduced sample space. Instead of the entire sample space S, we now have a smaller sample space (of what has already occurred). The old rule about being the number in the event divided by the number in the sample space still applies, but the sample space has been modified (to the given set).Example 1: A = rolling a sum of 10 on a pair of dice B = rolling doubles on a pair of diceDetermine P(A|B). Start by looking at the outcome of doubles only (new sample space) and then find P(10).Sample space for B: {1,12,23,34,45,56,6}Of these, only 1 has a sum of 10. So P(A|B) = P(A and B)P( B) = 16 Determine P(B|A). Start by looking only at the outcomes with sums of 10 and then find P(Doubles).Sample space for A: S ={ }Of these, how many are doubles? So P(B|A) = P(A and B)P( B) = --------Example 2: A survey given to 100 people asked the question, "Do you smoke?" Results are shown in the table..YesNoTotalMale194160Female122840Total3169100Use the results to determine the probability of the following events. a. P(Male who smokes) = b. P(Male) = c. P(Smoke) = d. P(Smoke | Male) = e. P(Male | Smoke) =Example 3: Determine the probability of each situation.A person draws one card from a deck of 52 cards and then draws a second card without putting the first card back. Calculate the probability of both cards being clubs.A person draws one card from a deck of 52 cards and then returns it to the deck and the deck is reshuffled. Another card is then drawn. Calculate the probability of both cards being clubs.You are dealt 2 cards (without replacement). Find the probability that the second card is a black jack given that the first card is an Ace.P(J|A) = Example 4: Determine the probability of each situation.If a family has two children, find P(Both children are boys | First child is a boy).If a family has two children, find P (Both children are boys | First child is a girl).If a family has two children, find P(At least one child is a boy | First child is a boy).Conditional probabilities are often indicated by words/phrases like “given that”, “if”, or by words implying a subgroup.Here are some examples of statements that refer to a conditional probability, along with their translation into mathematical language. 10 percent of those taking drugs test negative → Translation: P(Test negative | Take drugs) = 0.1The probability of a smoker dying during the year is 0.05 → Translation: P(Dying | Smoker) = 0.05A blood test indicates the presence of a disease 95% of the time the disease is actually present → Translation: P(Test positive | Has disease) = 0.95We know that P(A and B) = P(A) · P(B) when events are independent. The following rule will always work whether events are independent or dependent! General Multiplication Rule: P(A and B) = P(A) · P(B|A)Example 5: If two cards are drawn without replacement, find P(Q and then J).Setup using the General Multiplication Rule: P(Q) · P(J | Q) → Translation: Find the probability of getting a Q, then find the probability of getting a J, knowing that you have a Q.? I can determine if events are independent using P(B|A)= P(B) ?To review…The outcome of one event does NOT affect the probability of another for independent events.If A and B are independent events, then P(A and B) = P(A) · P(B)If A and B are independent events, then P(A|B) = P(A)If A and B are independent events, then P(B|A) = P(B)If two events are independent, the occurrence of one doesn't change the probability of the occurrence of the other. This means that the probability of B occurring, whether A has happened or not, is simply the probability of B occurring.Example 6: A = Rolling a die and getting an even number B = Rolling a 2 on a die P(A) = 12 P(B) = 16Find P(A|B) = 1 ≠ 12 . Since P(A|B) ≠ P(A), the events are dependent (not independent).Find P(B|A) = 13 ≠ 16 . Since P(B|A) ≠ P(B), the events are dependent (not independent).Example 7: Suppose you are given P(A) = Angel barks = 0.1, P(B) = Brutus barks = 0.2, and 548640012065000P(A|B) = Angel barks given Brutus barks = 0.3.Use the formula P(A|B) = P(A and B)P(B) to find P(A and B) = Angel and Brutus both bark.Are the two events independent or dependent?Example 8: A bag contains 12 red M&Ms, 12 blue M&Ms, and 12 green M&Ms. What is the probability of drawing two M&Ms of the same color in a row? Is this independent?482186568643500There are a total of 36 M&Ms in the bag.? You remove a blue M&M and eat it.? There are now 11 blue M&Ms remaining in the bag.? There are 35 total M&Ms now remaining.? You will now need to remove another blue M&M.? The conditional probability will be:P (Blue) = _____P(Blue?|?Blue) =?_____Since P(Blue | Blue) ≠ P(Blue), then these are _______________________ .Example 9: Suppose you put the first M&M back in the bag before you took another one? Now find the probability of drawing two M&Ms of the same color in a row? Is this independent?P(Blue | Blue)= _____P(Blue) = _____Since P(Blue | Blue) ≠ P(Blue), then these are _______________________ . ? I can use a tree diagram to compute conditional probabilities ?Tree diagrams branch with outcomes. You can attach probabilities to these as well. Because there is more than one event occurring, tree diagrams may help with probabilities. Multiply as you move across the branches (one event and then another) Add as you combine events (one event or another)Example 10: Tryouts are coming for your favorite sport. There is a 75% chance you will make the team. If you make the team there is a 25% chance you will have to quit your job. If you don’t make the team there is a 10% chance you will still quit your job. Draw a tree diagram with each outcome and probability.Determine the probability that both events will occur by multiplying each branch in your tree diagram. Then calculate the following probability of the following events occurring.P(Make team and quit) =P(Make team and don’t quit) =P(Don’t make team and quit)=P(Don’t make team and don’t quit) =486727526225500Example 11: There is a 70% chance that you will buy a coffee on Friday. If you buy coffee, you will also buy a bagel 50% of the time. If you don’t buy coffee on Friday, you will still buy a bagel 30% of the time. Make a tree diagram to find the probabilities of each outcome.53721009906000? I can use a tree diagram to compute Bayes’ Rule (reverse conditional probability) ?091440Bayes’ Rule (reverse conditional probability): Looking at the probability that the second event occurred, now find the probability of the first event.Bayes’ Rule (reverse conditional probability): Looking at the probability that the second event occurred, now find the probability of the first event.Example 12: Use the tree diagram from Example 10 to determine the following probabilities. Find P(Made the team | Quit your job) =Find P (Didn’t make the team | Quit your job) =Example 13: Use the tree diagram from Example 11 to determine the following probabilities.Find P (Bought coffee | Bought bagel) =Find P (Didn’t buy coffee| Bought bagel) =Find P (Bought coffee | Didn’t buy a bagel) =Find P (Didn’t buy coffee | Didn’t buy a bagel) = ................
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