The Multiplication Rule: Complements and Conditional Probability Section 4-5 The Multiplication Rule: Complements and Conditional Probability

Learning Targets Probability of “at least one”: Find the probability that among several trials, we get at least one of some specified event. Conditional probability: Find the probability of an event when we have additional information that some other event has already occurred.

Learning Objectives Probability of “at least 1” Formula for Conditional Probability “At least one” is equivalent to “one or more.” The complement of getting at least one item of a particular type is that you get no items of that type.

Practice Problem You are considering purchasing 3 new TV’s for your college apartment. The size options at the store are 20”, 32” and 56” (assume there is an endless supply of all 3). If you randomly choose what sizes you will purchase, what is the probability that you don’t buy any 32” TV? (20, 20, 20), (20, 20, 56), (20, 56, 56), (56, 56, 56), (32, 20, 20), (32, 56, 56), (32, 20, 56), (32, 32, 20), (32, 32, 56), (32, 32, 32) P(Not Buy 32”) = 4 10 = 2 5 What is the probability that you buy at least 1 32”? P(at least 1 32”) = 6 10 = 3 5

Finding the Probability of “At Least One” To find the probability of at least one of something, calculate the probability of none, then subtract that result from 1. That is, P(at least one) = 1 – P(none).

Practice Problem You are considering purchasing 3 new TV’s for your college apartment. The size options at the store are 20”, 32” and 56” (assume there is an endless supply of all 3). If you randomly choose what sizes you will purchase, what is the probability that you don’t buy any 32” TV? P(Not Buy 32”) = 4 10 = 2 5 What is the probability that you buy at least 1 32”? P(at least 1 32”) = 1 – P(Not buy 32”) = 1 – 2 5 = 3 5

Example If a couple is going to have 4 children, what is the probability that at least one of them is boy? [Method 1] The total possibilities of the 4 children could be: bbbb, bbbg, bbgb, bgbb, gbbb, bbgg, bggb, ggbb, bgbg, gbgb, gbbg, bggg, gbgg, ggbg, gggb, gggg P(At least 1 boy) = 15 16 [Method 2] No boy = All girls = “gggg” P(No boy) = 1 16 P(At least 1 boy) = 1 – 1 16 = 15 16

Conditional Probability A conditional probability of an event is a probability obtained with the additional information that some other event has already occurred. P(B | A) denotes the conditional probability of event B occurring, given that event A has already occurred, and it can be found by dividing the probability of events A and B both occurring by the probability of event A: 𝑃 𝐵 𝐴 = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴)

Intuitive Approach to Conditional Probability The conditional probability of B given A can be found by assuming that event A has occurred, and then calculating the probability that event B will occur.

Confusion of the Inverse To incorrectly believe that P(A|B) and P(B|A) are the same, or to incorrectly use one value for the other, is often called confusion of the inverse.

Example 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? A – Friday, B – Student is absent P(B|A) = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴) = 0.03 0.2 = 3 20 = 0.15

Example A math teacher gave his 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 second test given that they passed the second first test? A – pass 1st test, B – pass 2nd test P(B|A) = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴) = 0.25 0.42 = 25 42  0.5952

More Practice In New York State, 48% of all teenagers own a skateboard and 39% of all teenagers own a skateboard and roller blades. What is the probability that a teenager owns roller blades given that the teenager owns a skateboard? A – skateboard, B – roller blade P(B|A) = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴) = 0.39 0.48 = 39 48 = 13 16 = 0.8125

Example: Find the probability of a couple having a baby girl when their fourth child is born, given that the first three children were all girls. Is the result the same as the probability of getting four girls among four children? 0.5. No.

Example: An experiment with fruit flies involves one parent with normal wings and one parent with vestigial wings. When this parents have an offspring, there is a ¾ probability that the offspring has normal wings and a ¼ probability of vestigial wings. If the parents give birth to 10 offspring, what is the probability that at least 1 of the offspring has vestigial wings? If researchers need at least one offspring with vestigial wings, can they be reasonably confident of getting one? A – All Normal Wing, 𝐴 – at least one Vestigial Wing P( 𝐴 ) = 1 – P(A) = 1 – (0.75)10 = 0.944 Yes, the probability is quite high, so they can be reasonably confident of getting at least one offspring with vestigial wings.

Challenge Problem Use the table below to determine the probability that a person tests positive given that they actually lied. A – lied, B – positive P(B|A) = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴) = 42/98 57/98 = 42 57 Results from Experiments with Polygraph Instruments Did the subjects actually lie? No (did Not Lie) Yes (Lied) Positive test result 15 42 (Polygraph test indicated that the subject lied.) (false positive) (true positive) Negative test result 32 9 (Polygraph test indicated that the subject did not lie.) (true negative) (false negative)

More Practice - Your Turn Use the information in the table below to answer the following question. Assume that 1 of the 98 test subjects is randomly selected. Find the probability of selecting a subject with a positive test result, given that the subject did not lie. Results from Experiments with Polygraph Instruments Did the subjects actually lie? No (did Not Lie) Yes (Lied) Positive test result 15 42 (Polygraph test indicated that the subject lied.) (false positive) (true positive) Negative test result 32 9 (Polygraph test indicated that the subject did not lie.) (true negative) (false negative) A – not lied, B – positive P(B|A) = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴) = 15/98 47/98 = 15 47

Of those 47 people, only 15 of them had positive test results. Results from Experiments with Polygraph Instruments Did the subjects actually lie? No (did Not Lie) Yes (Lied) Positive test result 15 42 (Polygraph test indicated that the subject lied.) (false positive) (true positive) Negative test result 32 9 (Polygraph test indicated that the subject did not lie.) (true negative) (false negative) Knowing that they did not lie means that we will only look at those people. Of those 47 people, only 15 of them had positive test results. Therefore, the probability is:

More Practice 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? A – Technology, B – Spanish P(B|A) = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴) = 0.087 0.68 = 87 680  0.1279 What is the probability of getting at least one 5-answer multiple choice question right out of 4? A – 1 correct, 𝐴 – 1 not correct, P( 𝐴 ) = 4 5 P(At least 1 correct) = 1 – P(all wrong)= 1 −[𝑃 𝐴 ] 4 = 1− 4 5 4 =1 − 256 625 = 369 625

More Practice A company tests a batch of altimeters without replacement to see if they are acceptable for distribution. If one altimeter is faulty, then the whole batch is denied. Given that 3% of altimeters are faulty, what is the probability that a batch of 400 altimeters will be denied based on the first two items selected? Of the 400 altimeters, 0.03(400) = 12 are faulty. A – 1st is not faulty, B – 2nd is not faulty P(At least 1 is faulty) = 1 – P(None of 2 is faulty) = 1 – P(A) P(B|A) = 1 – 388 400 387 399 = 1 – 150156 159600  0.0592

Example: Use the data in the table below Example: Use the data in the table below. Instead of summarizing observed results, the entries reflect the actual probabilities based on births of twins. Identical twins come from a single egg that splits into two embryos, and fraternal twins are from separate fertilized eggs. The table entries reflect the principle that among sets of twins, 1/3 are identical and 2/3 are fraternal. Also, identical twins must be of the same sex and the sexes are equally likely (approximately), and sexes of fraternal twins are equally likely. a) After having a sonogram, a pregnant woman learns that she will have twins. What is the probability that she will have identical twins? P Identical Twin = 𝟏 𝟑 Sexes of Twins boy/boy boy/girl girl/boy girl/girl Identical Twins 5 Fraternal Twins

A – twin of 2 boys, B – identical twin Example continued: Use the data in the table below. Instead of summarizing observed results, the entries reflect the actual probabilities based on births of twins. Identical twins come from a single egg that splits into two embryos, and fraternal twins are from separate fertilized eggs. The table entries reflect the principle that among sets of twins, 1/3 are identical and 2/3 are fraternal. Also, identical twins must be of the same sex and the sexes are equally likely (approximately), and sexes of fraternal twins are equally likely. b) After studying the sonogram more closely, the physician tells the pregnant woman that she will give birth to twin boys. What is the probability that she will have identical twins? That is, find the probability of identical twins given that the twins consist of two boys. Sexes of Twins boy/boy boy/girl girl/boy girl/girl Identical Twins 5 Fraternal Twins A – twin of 2 boys, B – identical twin P(B|A) = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴) = 5/30 10/30 = 5 10 = 1 2

A – fraternal twin, B – one child of each sex Example 8 continued: Use the data in the table below. Instead of summarizing observed results, the entries reflect the actual probabilities based on births of twins. Identical twins come from a single egg that splits into two embryos, and fraternal twins are from separate fertilized eggs. The table entries reflect the principle that among sets of twins, 1/3 are identical and 2/3 are fraternal. Also, identical twins must be of the same sex and the sexes are equally likely (approximately), and sexes of fraternal twins are equally likely. c) If a pregnant woman is told that she will give birth to fraternal twins, what is the probability that she will have one child of each sex? Sexes of Twins boy/boy boy/girl girl/boy girl/girl Identical Twins 5 Fraternal Twins A – fraternal twin, B – one child of each sex P(B|A) = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴) = 10/30 20/30 = 10 20 = 1 2

Example: When testing blood samples for HIV infections, the procedure can be made more efficient and less expensive by combining samples of blood specimens. If samples from three people are combined and the mixture tests negative, we know that all three individual samples are negative. Find the probability of a positive result for three samples combined into one mixture, assuming the probability of an individual blood sample testing positive is 0.1 (the probability for the “at-risk” population, based on data from the New York State Health Department). A – individual testing positive, 𝑨 – individual testing negative P( A ) = 0.1 P( 𝑨 ) = 0.9 P(a positive in mixture) = P(at least 1 positive in mixture) = 1 – P(𝐚𝐥𝐥 𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞)= 𝟏 −[𝑷 𝑨 ] 𝟑 = 𝟏− 𝟎.𝟗 𝟑 = 0. 271

Recap In this section we have discussed: Concept of “at least one.” Conditional probability. Intuitive approach to conditional probability.

Homework P.175: 12, 15, 17, 19-22